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Maths 4 Marks

Probability · Gujarat Board (GSEB) STD 12 Science (GSEB - English Medium). 275 questions.

Questions · Page 3 of 6

4 Marks

Question 1014 Marks
From a lot of 30 bulbs that includes 6 defective bulbs, a sample of 4 bulbs is drawn at random with replacement. Find the probability distribution of the number of defective bulbs.
Answer
Let X denote the number of defective bulls in a sample of 4 bulbs drawn successively with replacement.
Then, X follows a binomial distribution with the following parameters: n = 4,
$\text{p}=\frac{6}{30}=\frac{1}{5}$ and $\text{q}=\frac{4}{5}$
Then, the distribution is given by
$\text{P(X = r})=\text{ }^4\text{C}_{\text{r}}\big(\frac{1}{5}\big)^{\text{r}}\big(\frac{4}{5}\big)^{4-\text{r}}.\text{r}=0,1,2,3,4$
$\text{P(X = } 0)=\big(\frac{4}{5}\big)^4$
$=\frac{256}{625}$
$\text{P(X}=1)=4\big(\frac{1}{5}\big)^1\big(\frac{4}{5}\big)^3$
$=\frac{256}{625}$
$\text{P(X}=2)=6\big(\frac{1}{5}\big)^2\big(\frac{4}{5}\big)^2$
$=\frac{96}{625}$
$\text{P(X}=3)=4\big(\frac{1}{5}\big)^3\big(\frac{4}{5}\big)^1$
$=\frac{16}{625}$
$\text{P(X}=4)=\big(\frac{1}{5}\big)^4$
$=\frac{1}{625}$
$\text{X}$
 $1$
 $2$
 $3$
 $4$
 $5$
$1\text{P(X)}$
 $\frac{256}{625}$
 $\frac{256}{625}$
 $\frac{96}{625}$
 $\frac{16}{625}$
 $\frac{1}{625}$
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Question 1024 Marks
A fair coin is tossed four times. Let X denote the number of heads occuring. Find the probability distribution, mean and variance of X.
Answer
We know that, in a toss of coin.

$\text{P(T)}=\frac{1}{2},\text{P(H)}=\frac{1}{2}$

Let X denote the number of accuring head in four throws of a coins.

So, X can take values from X = 0, 1, 2, 3, 4

$\text{P(X=0)}=\text{P(T)}\text{P(T)}\text{P(T)}\text{P(T)}$

$=\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}$

$=\frac{1}{16}$

$\text{P(X=1)}=\text{P(H)}\text{P(T)}\text{P(T)}\text{P(T)}\times{^{4}}\text{C}_1$

$=\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}\times4$

$=\frac{4}{16}$

$\text{P(X=2)}=\text{P(H)}\text{P(H)}\text{P(T)}\text{P(T)}\times{^{4}}\text{C}_2$

$=\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}\times6$

$=\frac{6}{16}$

$\text{P(X=3)}=\text{P(H)}\text{P(H)}\text{P(H)}\text{P(T)}\times{^{4}}\text{C}_3$

$=\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}\times4$

$=\frac{4}{16}$

$\text{P(X=4)}=\text{P(H)}\text{P(H)}\text{P(H)}\text{P(H)}$

$=\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}$

$=\frac{1}{16}$

So,

$\text{x}_\text{i}$

 $\text{p}_\text{i}$

 $\text{x}_\text{i}\text{p}_\text{i}$

 $\text{x}_\text{i}^2\text{p}_\text{i}$
$0$

 $\frac{1}{16}$

 $0$

 $0$
$1$

 $\frac{4}{16}$

 $\frac{4}{16}$

 $\frac{4}{16}$
$2$

 $\frac{6}{16}$

 $\frac{12}{16}$

 $\frac{24}{16}$
$3$

 $\frac{4}{16}$

 $\frac{12}{16}$

 $\frac{36}{16}$
$4$

 $\frac{1}{16}$

 $\frac{4}{16}$

 $\frac{16}{16}$

 

 

 $\sum\text{xp}=2$

 $\sum\text{x}^2\text{p}=5$
Mean $=\sum\text{xp}=2$

Variance $=\sum\text{x}^2\text{p}-(\text{mean})^2$

$=5-(2)^2=1$

Probability distribution is

$\text{x}:$ $0$ $1$ $2$ $3$ $4$
$\text{p(x)}:$ $\frac{1}{16}$ $\frac{4}{16}$ $\frac{6}{16}$ $\frac{4}{16}$ $\frac{1}{16}$
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Question 1034 Marks
An insurance company insured 2000 scooters and 3000 motorcycles. The probability of an accident involving a scooter is 0.01 and that of a motorcy is 0.02. An insured vehicle met with an accident. Find the probability that the accidented vehicle was a motorcycle.
Answer
Let E1, E2 and A be events ar:
E1 = Vehilcle is scooter
E2 = Vehicle is motorcycle
A = An insured met with scooter
$\text{P}(\text{E}_1)=\frac{2000}{5000}=\frac{2}{5}$
$\text{P}(\text{E}_2)=\frac{3000}{5000}=\frac{3}{5}$
P(A|E1) = P(Accident of scooter)
= 0.01
$\text{P}\Big(\frac{\text{A}}{\text{E}_2}\Big)=$ P(Accident of motorcycle)
= 0.02
To find, P(Accident vehicle was motorcycle) $=\text{P}\Big(\frac{\text{E}_2}{\text{A}}\Big)$
By Baye's theorem,
$\text{P}\Big(\frac{\text{E}_2}{\text{A}}\Big)=\frac{\text{P}(\text{E}_2)\text{P}\Big(\frac{\text{A}}{\text{E}_2}\Big)}{\text{P}(\text{E}_1)\text{P}\Big(\frac{\text{A}}{\text{E}_1}\Big)+\text{P}(\text{E}_2)\text{P}\Big(\frac{\text{A}}{\text{E}_2}\Big)}$
$=\frac{\frac{3}{5}\times\frac{2}{100}}{\frac{2}{5}\times\frac{1}{100}+\frac{3}{5}\times\frac{2}{100}}$
$=\frac{\frac{6}{500}}{\frac{2}{500}+\frac{6}{500}}$
$=\frac{6}{8}$
$=\frac{3}{4}$
Required probability $=\frac{3}{4}$
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Question 1044 Marks
Bag I contains 3 black and 2 white balls, Bag II contains 2 black and 4 white balls. A bag and a ball is selected at random. Determine the probability of selecting a black ball.
Answer
Bag I = {3B, 2W}, Bag II = {2B, $W}
Let E1= Event that bag I is selected
E2= Event that bag II is selected
And E = Event that a black ball is selected
$\Rightarrow\text{P}(\text{E}_1)=\frac{1}{2},\text{P}(\text{E}_2)=\frac{1}{2},$ $\text{P}\Big(\frac{\text{E}}{\text{E}_1}\Big)=\frac{3}{5},\text{P}\Big(\frac{\text{E}}{\text{E}_2}\Big)=\frac{2}{6}=\frac{1}{3}$
$\therefore\text{P}(\text{E})=\text{P}(\text{E}_1)\cdot\text{P}\Big(\frac{\text{E}}{\text{E}_1}\Big)+\text{P}(\text{E}_2)\cdot\text{P}\Big(\frac{\text{E}}{\text{E}_2}\Big)$
$=\frac{1}{2}\cdot\frac{3}{5}+\frac{1}{2}\cdot\frac{2}{6}=\frac{3}{10}+\frac{2}{12}$
$=\frac{18+10}{60}=\frac{28}{60}=\frac{7}{15}$
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Question 1054 Marks
A bag contains 25 tickets, numbered from 1 to 25. A ticket is drawn and then another ticket is drawn without replacement. Find the probability that both tickets will show even numbers.
Answer
Tickets are numbered from 1 to 25
⇒ Total number of tickets = 25
Number of tickets with even numbers on it
= 12 {2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24}
A = first ticket with even number
B = second ticket with even number
P (Both tickets will show even number, without replacement)
$=\text{P}(\text{A})\ \text{P}\Big(\frac{\text{B}}{\text{A}}\Big)$
$=\frac{12}{25}\times\frac{11}{24}$
$=\frac{11}{50}$
Required probability $=\frac{11}{50}$
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Question 1064 Marks
A bag contains 3 red and 5 black balls and a second bag contains 6 red and 4 black balls. A ball is drawn from each bag. Find the probability that one is red and the other is black.
Answer
There are two bags.
Bag (1) contain 3 red and 5 black balls
Bag (2) contain 6 red and 4 black balls
P (One red ball from bag 1) $=\frac{3}{8}$
$\text{P}(\text{R}_1)=\frac{3}{8}$
P (One black ball from bag 1) $=\frac{5}{8}$
$\text{P}(\text{B}_1)=\frac{5}{8}$
P (One red ball from bag 1) $=\frac{6}{10}$
$\text{P}(\text{R}_2)=\frac{3}{5}$
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Question 1074 Marks
A bag contains 10 balls, each marked with one of the digits from 0 to 9. If four balls are drawn successively with replacement from the bag, what is the probability that none is marked with the digit 0?
Answer
Let p denote the probability of getting a ball market with 0. So
$\text{p}=\frac{1}{10}$ [Since balls are market with 0, 1, 2, 3, 4, 5, 6, 7, 8, 8, 9]
$\text{q}=1-\frac{1}{10}$ [Since p + q = 1]
$\text{q}=\frac{9}{10}$
Let X denote the variable presenting the number of balls marked with 0 out of four balls drawn. probability of drawing r balls out of n balls that are marked 0 is given by
$\text{P}(\text{X = r})=\text{ }^\text{n}\text{c}_\text{r}\text{p}^{\text{r}}\text{q}^{\text{n}-\text{r}}$
$=\text{ }^{4}\text{c}_\text{r}\big(\frac{1}{10}\big)^\text{r}\big(\frac{9}{10}\big)^{4-\text{r}}\dots(1)$
Probability of getting none balls marked with 0
$=\text{P}(\text{X}=0)$
$=\text{ }^4\text{c}_0\big(\frac{1}{10}\big)^0\big(\frac{9}{10}\big)^{4-0}$
$=1.1.\big(\frac{9}{10}\big)^4$
$=\big(\frac{9}{10}\big)^4$
Probability of getting nine balls marked with $0=\big(\frac{9}{10}\big)^4$
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Question 1084 Marks
In a large bulk of items, 5 percent of the items are defective. What is the probability that a sample of 10 items will include not more than one defective item?
Answer
Let X denote the number of defective items in a sample of 10 items.
X follows a binomial distribution with n = 10;
P = Probability of detective items = 5% = 0.05; q = 1- p = 0.95
P(X = r)=Cr10(0.05)r(0.95) 10 - r
P(x=r) = 10Cr(0.05)r (0.95)10-r
Probability (sample of 10 items will include not more than one defective item)=P(X ≤ 1)
= P(X = 0) + P(X = 1)
= 10C0(0.05)0 (0.95)10-0 + 10C1(0.05)1 (0.95)10-1
= (0.95)9 (0.95+0.5)
= 1.45(0.95)9
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Question 1094 Marks
A letter is known to have come either from LONDON or CLIFTON. On the envelope just two consecutive letters ON are visible. What is the probability that the letter has come from,
LONDON.
Answer
Consider events E1, E2 and A events As:
E1 = Letters come from LONDON
E2 = Letters come from CLIFTON
E3 = Two consecutive letters visible on the envelope are on
$\text{P}(\text{E}_1)=\frac{1}{2}$
$\text{P}(\text{E}_2)=\frac{1}{2}$
[Since letters came either from LONDON or CLIFTON]
P(A | E1) = P(Two consecutive letters ON from LONDON)
$=\frac{2}{5}$
[Since LONDON has 2 - ON and 5 letters we consider one 'ON' as one letter]
$\text{P}\Big(\frac{\text{A}}{\text{E}_2}\Big)=\text{P}$ (Two consecutive letters On from CLIFTON)
$=\frac{1}{6}$
[Since CLIFTON has one 'ON' nad 6 letters considering ON as one letter]
To find, P (ON visible are from LONDON) $\text{P}\Big(\frac{\text{E}_1}{\text{A}}\Big).$
By baye's theorem,
$\text{P}\Big(\frac{\text{E}_1}{\text{A}}\Big)=\frac{\text{P}(\text{E}_1)\text{ P}\Big(\frac{\text{A}}{\text{E}_1}\Big)}{\text{P}(\text{E}_1)\text{ P}\Big(\frac{\text{A}}{\text{E}_1}\Big)+\text{P}(\text{E}_2)\text{ P}\Big(\frac{\text{A}}{\text{E}_2}\Big)}$
$=\frac{\frac{1}{2}\times\frac{2}{5}}{\frac{1}{2}\times\frac{2}{5}+\frac{1}{2}\times\frac{1}{6}}$
$=\frac{\frac{2}{10}}{\frac{2}{10}+\frac{1}{12}}$
$=\frac{2}{10}\times\frac{60}{17}$
$=\frac{12}{17}$
$\text{P}\Big(\frac{\text{E}_1}{\text{A}}\Big)=\frac{12}{17}$
Required probability $=\frac{12}{17}$
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Question 1104 Marks
The bag A contains 8 white and 7 black balls while the bag B contains 5 white and 4 black balls. One ball is randomly picked up from the bag A and mixed up with the balls in bag B. Then a ball is randomly drawn out from it. Find the probability that ball drawn is white.
Answer
Bag A contains 8 white and 7 black balls
Bag B contains 5 white and 4 black balls
Transfer can be done in two ways:
I - A white ball is transferred from bag A to bag B and then onw white ball is drawn from bag B.
II - A black ball is transferred fron bag A to bag, then one white ball is drawn from bag B.
Let E1, E2 and A be events as:
E1 = One white ball from bag A
E2 = one black ball from bag A
A = One white ball from bag B
$\text{P}(\text{E}_1)=\frac{8}{15}$
$\text{P}(\text{E}_2)=\frac{7}{15}$
$\text{P}(\text{A}|\text{E}_1)=\frac{6}{10}$
[Since E1 has increased white balls in bag B]
$\text{P}\Big(\frac{\text{A}}{\text{E}_2}\Big)=\frac{5}{10}$
[Since E2 has increased black ball in bag B]
By law of total probability,
$\text{P(A)}=\text{P}(\text{E}_1)\text{P}(\text{A}|\text{E}_1)+\text{P}(\text{E}_2)\text{ P}\Big(\frac{\text{A}}{\text{E}_2}\Big)$
$=\frac{8}{15}\times\frac{6}{10}+\frac{7}{15}\times\frac{5}{10}$
$=\frac{48}{150}+\frac{35}{150}$
$=\frac{83}{150}$
Required probability $=\frac{83}{150}$
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Question 1114 Marks
By examining the chest X ray, the probability that TB is detected when a person is actually suffering is 0.99. The probability of an healthy person diagnosed to have TB is 0.001. In a certain city, 1 in 1000 people suffers from TB. A person is selected at random and is diagnosed to have TB. What is the probability that he actually has TB?
Answer
Let E1 = Event that person has TB
E2 = Event that the person does not have TB
E = Event that the person is diagnosed to have TB
$\therefore\text{P}(\text{E}_1)=\frac{1}{1000}=0.001,$
$\text{P}(\text{E}_2)=\frac{999}{1000}=0.999$
And $\text{P}\Big(\frac{\text{E}}{\text{E}_1}\Big)=0.099$ and $\text{P}\Big(\frac{\text{E}}{\text{E}_2}\Big)=0.001$
$\therefore\text{P}\Big(\frac{\text{E}_1}{\text{E}}\Big)=\frac{\text{P}(\text{E}_1)\cdot\text{P}\Big(\frac{\text{E}}{\text{E}_1}\Big)}{\text{P}(\text{E}_1)\cdot\text{P}\Big(\frac{\text{E}}{\text{E}_1}\Big)+\text{P}(\text{E}_2)\cdot\text{P}\Big(\frac{\text{E}}{\text{E}_2}\Big)}$
$=\frac{0.001\times0.99}{0.001\times0.99+0.999\times0.001}$
$=\frac{0.000990}{0.000990+0.000999}$
$=\frac{990}{1989}=\frac{110}{221}$
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Question 1124 Marks
Assume that the chances of a patient having a heart attack is 40%. It is also assumed that a meditation and yoga course reduce the risk of heart attack by 30% and prescription of certain drug reduces its chances by 25%. At a time a patient can choose any one of the two options with equal probabilities. It is given that after going through one of the two options the patient selected at random suffers a heart attack. Find the probability that the patient followed a course of meditation and yoga?
Answer
Let E1 and E2 be the events 
E1 : Treatment of yoga and meditation
E2 : Treatment of prescription of certain drugs
$\therefore\ \text{P}(\text{E}_1)=\frac{1}{2},\ \text{P}(\text{E}_2)=\frac{1}{2}$
Let A denotes that a person has heart risk attack
$\therefore\ \text{P}(\text{A})=40\%=0.40$
Yoga and mediation reduces the heart risk by 30%
i. e. inspite of getting yoga and meditation treatment heart risk is 70% of the 0.40
$\therefore\ \text{P}(\text{A|}\text{E}_1)=0.40\times0.70=0.28$
Drug prescription reduces the heart risk by 25%
Even after adopting the drug prescription heart risk is 75% of the 0.40
$\therefore\ \text{P}(\text{A|E}_2)=0.40\times0.75=0.30$
$\therefore\ \text{P}(\text{E}_1|\text{A})=\frac{\text{P}(\text{E}_1)\text{P}(\text{A|E}_1)}{\text{P}(\text{E}_1)\text{P}(\text{A|E}_1)+\text{P}(\text{E}_2)\text{P}(\text{A|E}_2)}$
$=\frac{\frac{1}{2}\times0.28}{\frac{1}{2}\times0.28+0.30\frac{1}{2}}=\frac{0.28}{0.28+0.30}=\frac{28}{58}=\frac{14}{29}$
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Question 1134 Marks
Two cards are drawn simultaneously from a pack of 52 cards. Compute the mean and standard deviation of the number of kings.
Answer
Two cards are drawn simultaneously from a pack of 52 cards.

Let X denotes the number of kings drawn.

So, X = 0, 1, 2.

P(X = 0)

$=\frac{\text{}^{48}\text{C}_2}{\text{}^{52}\text{C}_2}$

$=\frac{1128}{1326}$

$=\frac{188}{221}$

P(X = 1)

$=\frac{\text{}^{4}\text{C}_1\times\text{}^{48}\text{C}_1}{\text{}^{52}\text{C}_2}$

$=\frac{192}{1326}$

$=\frac{32}{221}$

P(X = 2)

$=\frac{\text{}^{4}\text{C}_2}{\text{}^{52}\text{C}_2}$

$=\frac{6}{1326}$

$=\frac{1}{221}$

So,

$\text{x}_\text{i}$

 $\text{p}_\text{i}$

 $\text{x}_\text{i}\text{p}_\text{i}$

 $\text{x}_\text{i}^2\text{p}_\text{i}$
$0$

 $\frac{188}{221}$

 $0$

 $0$
$1$

 $\frac{32}{221}$

 $\frac{32}{221}$

 $\frac{32}{221}$
$2$

 $\frac{1}{221}$

 $\frac{2}{221}$

 $\frac{4}{221}$

 

 

 $\sum\text{xp}=\frac{34}{221}$

 $\sum\text{x}^2\text{p}$
Mean $=\sum\text{xp}$

Mean $=\frac{34}{221}$

Variance $=\sum\text{x}^2\text{p}-(\text{Mean})^2$

$=\frac{36}{221}-\Big(\frac{34}{221}\Big)^2$

$=\frac{7956-1156}{48841}$

$=\frac{6800}{48841}$

Variance $=\frac{400}{2873}.$

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Question 1144 Marks
Given the probability that A can solve a problem is $\frac{2}{3}$ and the probability that B can solve the same problem is $\frac{3}{5}$. Find the probability that none of the two will be able to solve the problem.
Answer
Given,
Probability that A can solve a problem $=\frac{2}{3}$
$\Rightarrow\ \text{P(A)}=\frac{2}{3}$
$=\text{P}(\overline{\text{A}})=1-\frac{2}{3}$
$\text{P}(\overline{\text{A}})=\frac{1}{3}$
Probability that B can solve the same problem $=\frac{3}{5}$
$\Rightarrow\ \text{P(B)}=\frac{3}{5}$
$\Rightarrow\ \text{P}(\overline{\text{B}})=1-\frac{3}{5}$
$\text{P}(\overline{\text{B}})=\frac{2}{5}$
P(None of them solve the problem)
$=\text{P}(\overline{\text{A}}\cap\overline{\text{B}})$
$=\text{P}(\overline{\text{A}})\text{ P}(\overline{\text{B}})$
$=\frac{1}{2}\times\frac{2}{5}$
$=\frac{2}{15}$
Required probability $=\frac{2}{15}$
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Question 1154 Marks
A bag contains 7 white, 5 black and 4 red balls. Four balls are drawn without replacement. Find the probability that at least three balls are black.
Answer
A bag contains 7 white, 5 black and 4 red balls.
Four balls are drawn without replacement
P (At least three balls are black)
= P(3 black balls and one not black or 4 black balls)
= P(3 black and one not black) + P(4 black balls)
$=\frac{{^5}\text{C}_3\times{^{11}}\text{C}_1}{{^{16}}\text{C}_4}+\frac{{^{5}}\text{C}_4}{^{16}\text{C}_4}$
$=\frac{\frac{5!}{3!2!}\times11+\frac{5!}{4!1!}}{\frac{16!}{4!12!}}\ \Big[\text{Since } ^\text{n}\text{C}_\text{r}=\frac{\text{n}!}{\text{r}!(\text{n}-\text{t})!}\Big]$
$=\frac{\frac{5.4}{2}\times11+5}{\frac{16\times15\times14\times13}{4\times3\times2}}$
$=\frac{(110+5)}{1820}$
$=\frac{115}{1820}$
$=\frac{23}{364}$
Required probability $=\frac{23}{364}$
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Question 1164 Marks
An item is manufactured by three machines A, B and C. Out of the total number of items manufactured during a specified period, 50% are manufactured on machine A, 30% on B and 20% on C, 2% of the items produced on A and 2% of items produced on B are defective and 3% of these produced on C are defective. All the items stored at one godown. One item is drawn at random and is found to be defective. What is the probability that it was manufactured on machine A?
Answer
Consider the following events:
E1 = Item is produced by machine A,
E2 = Item is produced by machine B,
E3 = Item is produced by machine C,
A = Item is defective
Clearly,
$\text{P}(\text{E}_1)=\frac{50}{100}=\frac{1}{2},\text{P}(\text{E}_2)=\frac{30}{100}=\frac{3}{10},\text{P}(\text{E}_3)=\frac{20}{100}=\frac{1}{5}$
$\text{P}\Big(\frac{\text{A}}{\text{E}_1}\Big)=\frac{2}{100},\text{P}\Big(\frac{\text{A}}{\text{E}_2}\Big)=\frac{2}{100},\text{P}\Big(\frac{\text{A}}{\text{E}_3}\Big)=\frac{3}{100}$
Required probability $=\text{P}\Big(\frac{\text{E}_1}{\text{A}}\Big)$
$=\frac{\text{P}(\text{E}_1)\text{ P}\Big(\frac{\text{A}}{\text{E}_1}\Big)}{\text{P}(\text{E}_1)\text{ P}\Big(\frac{\text{A}}{\text{E}_1}\Big)+\text{P}(\text{E}_2)\text{ P}\Big(\frac{\text{A}}{\text{E}_2}\Big)+\text{P}(\text{E}_3)\text{ P}\Big(\frac{\text{A}}{\text{E}_3}\Big)}$
$=\frac{\frac{1}{2}\times\frac{2}{100}}{\frac{1}{2}\times\frac{2}{100}+\frac{3}{10}\times\frac{2}{100}+\frac{1}{5}\times\frac{3}{100}}$
$=\frac{5}{11}$
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Question 1174 Marks
A box contains 100 tickets, each bearing one of the numbers from 1 to 100. If 5 tickets are drawn successively with replacement from the box, find the probability that all the tickets bear numbers divisible by 10.
Answer
Let P denote the probability of getting a ticket bearing number divisibal by 10, So
$\text{p}=\frac{10}{100}$ [Since there are 10, 20, 30, 40, 50, 60, 70, 80, 90, 100 which are divisible by 10] 
$\text{p}=\frac{1}{10}$
$\text{q}=1-\frac{1}{10}$ [Since p + q = 1]
$\text{q}=\frac{9}{10}$
Let X denote the variable representing the number of tickets bearing a number divisible by 10 out of 5 tickets. probability of getting r tickets bearing a number divisible by 10 out ot n tickets is given by
$\text{P}(\text{X = r})=\text{ }^\text{n}\text{c}_{\text{r}}\text{p}^{\text{r}}\text{q}^{\text{n}-\text{r}}$
$=\text{ }^{\text{5}}\text{c}_{\text{r}}\big(\frac{1}{10}\big)^{\text{r}}\big(\frac{9}{10}\big)^{5-\text{r}}\dots(1)$
Probability of getting all the tickets bearing a number divisible by 10
$=\text{ }^5\text{c}_5\big(\frac{1}{10}\big)^5\big(\frac{9}{10}\big)^{5-5}$ [Using (1)]
$=1.\big(\frac{1}{10}\big)^5\big(\frac{9}{10}\big)^0$
$=\big(\frac{1}{10}\big)^5$
Required probability $=\big(\frac{1}{10}\big)^5$
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Question 1184 Marks
In a game, a man wins a rupee for a six and loses a rupee for any other number when a fair die is thrown. The man decided to throw a die thrice but to quit as and when he gets a six. Find the expected value of the amount he wins/loses.
Answer
When a die is thrown, probability of getting a six (p) $=\frac{1}{6}$

Therefore, $=\text{q}=1-\text{p}=1-\frac{1}{6}=\frac{5}{6}$

  1. If he gets a six in first throw, then,

Probability of getting a six $=\frac{1}{6}$

  1. If he does not get a six, in first throw, but he gets a six in the second throw, then

Its probability $=\frac{5}{6}\times \frac{1}{6}=\frac{5}{36}$

Probability that he does not get a six in first two throws and he gets a six in thied throw

$=\frac{5}{6}\times\frac{5}{6}\times\frac{1}{6}=\frac{25}{216}$

Probability that he does not get a six in any of the three throws $=\bigg(\frac{5}{6}\bigg)^3=\frac{125}{216}$

In first throw he gets a six, will receive Rs.1

If he gets a six in second throw, he will receive Rs.(1 - 1) = 0

If he gets a six in third throw, he will receive Rs.(- 1 - 1 + 1)=Rs. -1 =he will loss Rs. 1

Exapected value $=\frac{1}{6}\times1+\bigg(\frac{5}{6}\times\frac{1}{6}\bigg)\times0+\bigg(\frac{5}{6}\times\frac{5}{6}\times\frac{1}{6}\bigg)\times(-1)$

$=\frac{1}{6}-\frac{25}{216}=\frac{11}{216}\ (\text{loss})$

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Question 1194 Marks
Given that the two numbers appearing on throwing two dice are different. Find the probability of the event 'the sum of numbers on the dice is 4'.
Answer
A = Two numbers on two dice are different
= {(1, 2), (1, 3), (1, 4), (1, 5), (1, 6)
(2, 1), (2, 3), (2, 4), (2, 5), (2, 6)
(3, 1), (3, 2), (3, 4), (3, 5), (3, 6)
(4, 1), (4, 2), (4, 3), (4, 5), (4, 6)
(5, 1), (5, 2), (5, 3), (5, 4), (5, 6)
(6, 1), (6, 2), (6, 3), (6, 4), (6, 5)}
B = Sum of numbers on the dice is 4
B = {(1, 3), (2, 2), (3, 1)}
$\text{A}\cap\text{B}=\{(1,3),(3,1)\}$
Required probability $=\text{P}\Big(\frac{\text{B}}{\text{A}}\Big)$
$=\frac{\text{n}(\text{A}\cap\text{B})}{\text{n}(\text{A})}$
$=\frac{2}{30}$
Required probability $=\frac{1}{15}$
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Question 1204 Marks
The random variable X can take only the values 0, 1, 2. Given that P(X = 0) = P (X = 1) = p and that E(X2) = E[X], find the value of p.
Answer
Since, X = 0, 1, 2 and P (X) at X = 0 and 1 is p, let at X = 2, P (X) is x.
⇒ p + p + x = 1
⇒ x = 1 – 2p
We get, the following distribution
X
0
1
2
P(X)
p
q
1 - 2p
 $\therefore\text{E}[\text{X}]=\sum\text{XP}(\text{X})$
$=0\cdot\text{p}+1\cdot\text{p}+2(1-2\text{p})$
$=\text{p}+2-4\text{p}$
$=2-3\text{p}$
And $\text{E}(\text{X}^2)=\sum\text{X}^2\text{P}(\text{X})$
$=0\cdot\text{p}+1\cdot\text{p}+4\cdot(1-2\text{p})$
$=\text{p}+4-8\text{p}$
$=2-7\text{p}$
Also, given that $\text{E}(\text{X}^2)=\text{E}[\text{X}]$
$\Rightarrow4-7\text{p}=2-3\text{p}$
$\Rightarrow4\text{p}=2$
$\Rightarrow\text{p}=\frac{1}{2}$
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Question 1214 Marks
For A, B and C the chances of being selected as the manager of a firm are in the ratio 4:1:2 respectively. The respective probabilities for them to introduce a radical change in marketing strategy are 0.3, 0.8 and 0.5. If the change does take place, find the probability that it is due to the appointment of B or C.
Answer
Let E1, E2, E3 and A be event as:
E1 = A is appointed
E2 = B is appointed
E3 = C is appointed
A = A change does take place
$\text{P}(\text{E}_1)=\frac{4}{7}$
$\text{P}(\text{E}_2)=\frac{1}{7}$
$\text{P}(\text{E}_3)=\frac{2}{7}$
P(A|E1) = P(Changes tale place by A)
= 0.3
$\text{P}\Big(\frac{\text{A}}{\text{E}_2}\Big)=\text{P}$ (Changes take place by B)
= 0.8
$\text{P}\Big(\frac{\text{A}}{\text{E}_3}\Big)=\text{P}$ (Changes take place by C)
= 0.5
To find, P(Changes were taken place by B or C) $=\text{P}\Big(\frac{\text{E}_2}{\text{A}}\Big)+\text{P}\Big(\frac{\text{E}_3}{\text{A}}\Big)$
By baye's theorem,
$=\text{P}\Big(\frac{\text{E}_2}{\text{A}}\Big)+\text{P}\Big(\frac{\text{E}_3}{\text{A}}\Big)=\frac{\text{P}(\text{E}_2)\text{P}\Big(\frac{\text{A}}{\text{E}_2}\Big)+\text{P}(\text{E}_3)\text{P}\Big(\frac{\text{A}}{\text{E}_3}\Big)}{\text{P}(\text{E}_1)\text{P}\Big(\frac{\text{A}}{\text{E}_1}\Big)+\text{P}(\text{E}_2)\text{P}\Big(\frac{\text{A}}{\text{E}_2}\Big)+\text{P}(\text{E}_3)\text{P}\Big(\frac{\text{A}}{\text{E}_3}\Big)}$
$=\frac{\frac{1}{7}\times\frac{8}{10}+\frac{2}{7}\times\frac{5}{10}}{\frac{4}{7}\times\frac{3}{10}+\frac{1}{7}\times\frac{8}{10}\times\frac{2}{7}\times\frac{5}{10}}$
$=\frac{\frac{18}{70}}{\frac{30}{70}}$
$=\frac{18}{30}$
$=\frac{3}{5}$
Required probability $=\frac{3}{5}$
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Question 1224 Marks
A pair of dice is thrown 4 times. If getting a doublet is considered a success, find the probability distribution of the number of successes.
Answer
Let p be the probability of getting a doublet in a throw of a pair of dice, so
$\text{p}=\frac{6}{36}$ [Since (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)]
$=\frac{1}{6}$
$\text{q}=1-\frac{1}{6}$ [Since p + q = 1]
$=\frac{5}{6}$
Let X denote the number of grtting doublets i.e. success out of 4 times. So, probility distribution is given by
$\text{X}$
 $\text{P(X)}$
$0$
 $\text{ }^4\text{C}_0\big(\frac{1}{6}\big)^0\big(\frac{5}{6}\big)^{4-0}=\big(\frac{5}{6}\big)^4$
$1$
 $\text{ }^4\text{C}_1\big(\frac{1}{6}\big)^1\big(\frac{5}{6}\big)^{4-1}=4\big(\frac{1}{6}\big)\big(\frac{5}{6}\big)^3=\frac{2}{3}\big(\frac{5}{6}\big)^3$
$2$
 $\text{ }^4\text{C}_2\big(\frac{1}{6}\big)^2\big(\frac{5}{6}\big)^{4-2}=\frac{4.3}{2}\big(\frac{1}{6}\big)^2\big(\frac{5}{6}\big)^2=\frac{25}{216}$
$3$
 $\text{ }^4\text{C}_3\big(\frac{1}{6}\big)^3\big(\frac{5}{6}\big)^{4-3}=\frac{4.3}{2}\big(\frac{1}{6}\big)^3\big(\frac{5}{6}\big)=\frac{5}{324}$
$4$
 $\text{ }^4\text{C}_4\big(\frac{1}{6}\big)^4\big(\frac{5}{6}\big)^{4-4}=\big(\frac{1}{6}\big)^4=\frac{1}{1296}$
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Question 1234 Marks
Two cards are drawn simultaneosly from a well shuffled deck of 52 cards. Find the probability distribution of the number of the successes, when getting a spade is considered a success.
Answer
Let X denote the numbers of spades in a sample of two cards drawn from a well shuffled pack of 52 playing cards. Then, X can take the values 0, 1 and 2.
Now,
P(X = 0)
= P(no spade)
$=\frac{\text{}^{39}\text{C}_2}{\text{}^{52}\text{C}_2}$
$=\frac{741}{1326}$
$=\frac{19}{34}$
P(X = 1)
= P(1 spade)
$=\frac{\text{}^{13}\text{C}_1\times\text{}^{39}\text{C}_1}{\text{}^{52}\text{C}_2}$
$=\frac{507}{1326}$
$=\frac{13}{34}$
P(X = 2)
= P(2 spade)
$=\frac{\text{}^{13}\text{C}_2}{\text{}^{52}\text{C}_2}$
$=\frac{78}{1326}$
$=\frac{1}{17}$
Thus, the probability distribution of X is given by
$\text{X}$
$0$
$1$
$2$
$\text{P}(\text{X})$
$\frac{19}{34}$
$\frac{13}{34}$
$\frac{1}{17}$
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Question 1244 Marks
A coin is tossed thrice and all the eight outcomes are assumed equally likely. In which of the following cases are the following events A and B are independent?
A = the first throw results in head,
B = the last throw results in tail.
Answer
A coin is tossed thrice
Samplw space = {HHT, HTT, THT, TTT, HHH, HTH, THH, TTH}
A = The first throw results in head
A = {HHT, HTH, HHH, HTT}
B = The last throw in tail
B = {HHT, HTT, THT, TTT}
$\text{A}\cap\text{B}=\{\text{HHT, HTT}\}$
$\text{P}(\text{A}\cap\text{B})=\frac{2}{8}=\frac{1}{4}$
$\text{P(A) }\text{P(B)}=\frac{1}{2},\frac{1}{2}$
$\text{P(A) }\text{P(B)}=\frac{1}{4}$
$\text{P(A)}.\text{P(B)}=\text{P}(\text{A}\cap\text{B})$
So, A and B are independent events.
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Question 1254 Marks
Three dice are thrown at the sametime. Find the probability of getting three two’s, if it is known that the sum of the numbers on the dice was six.
Answer
As three dice are thrown at the same time, so we have sample space [n(S)] = 63 = 216
Let E1 is the event when the sum of numbers on the dice was six and E2 is the event when three two’s occurs.
⇒ E1 = {(1, 1, 4), (1, 2, 3), (1, 3, 2), (1, 4, 1), (2, 1, 3) (2, 2, 2), (2, 3, 1), (3, 1, 2), (3, 2, 1), (4, 1)}
⇒ n(E1) = 10 and E2 ={2, 2, 2}
⇒ n(E1) = 1
Also, $(\text{E}_1\cap\text{E}_2)=1$
$\therefore\text{P}\Big(\frac{\text{E}_2}{\text{E}_1}\Big)=\frac{\text{P}(\text{E}_1\cap\text{E}_2)}{\text{P}(\text{E}_2)}$
$=\frac{\frac{1}{216}}{\frac{10}{216}}=\frac{1}{10}$
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Question 1264 Marks
A die is tossed twice. A 'success' is getting an odd number on a toss. Find the variance of the number of successes.
Answer
It is given that "success" denotes the event of getting the number 1, 3 or 5. Then,

P(success) $=\frac{1}{2}$

Also, "failure" denotes the event of getting the numbers 2, 4, or 6. Then,

P(failure) $=\frac{1}{2}$

Let X denote the event of getting success. Then, X can take the values 0, 1 and 2.

Now,

P(X = 0) = P(no success) $=\frac{1}{2}\times\frac{1}{2}=\frac{1}{4}$

P(X = 1) = P(1 success) $=\Big(\frac{1}{2}\times\frac{1}{2}\Big)+\Big(\frac{1}{2}\times\frac{1}{2}\Big)=\frac{1}{2}$

P(X = 2) = P(2 success) $=\frac{1}{2}\times\frac{1}{2}=\frac{1}{4}$

Thus, the probability distribution of X is given by

$\text{X}$

 $\text{P}(\text{X})$
$0$

 $\frac{1}{4}$
$1$

 $\frac{1}{2}$
$2$

 $\frac{1}{4}$
Computation of mean and variance

$\text{X}_\text{i}$

 $\text{P}_\text{i}$

 $\text{P}_\text{i}\text{X}_\text{i}$

 $\text{P}_\text{i}\text{X}_\text{i}^2$
$0$

 $\frac{1}{4}$

 $0$

 $0$
$1$

 $\frac{1}{2}$

 $\frac{1}{2}$

 $\frac{1}{2}$
$2$

 $\frac{1}{4}$

 $\frac{1}{2}$

 $1$

 

 

 $\sum\text{p}_\text{i}\text{x}_\text{i}=1$

 $\sum\text{p}_\text{i}\text{x}_\text{i}^2=\frac{3}{2}$
Mean $=\sum\text{p}_\text{i}\text{x}_\text{i}=1$

Variance $=\sum\text{p}_\text{i}\text{x}_\text{i}^2-({\text{Mean}})^2=\frac{3}{2}-1=\frac{1}{2}$

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Question 1274 Marks
A pair of dice is thrown 4 times. If getting a doublet is considered a success, find the probability of two successes.
Answer
The repeated tosses of a pair of dice are Bernoulli trials. Let X denote the number of times of getting doublets in an experiment of throwing two dice simultaneously four times.
Probability of getting doublets in a single throw of the pair of dice is
$\text{p}=\frac{6}{36}=\frac{1}{6}$
$\therefore\ \text{q}=1-\text{p}=1-\frac{1}{6}=\frac{5}{6}$
Clearly, X has the binomial distribution with n = 4, $\text{p}=\frac{1}{6},\ \text{and}\ \text{q}=\frac{5}{6}$
$\therefore\ \text{P}(\text{X}=\text{x})=\ ^\text{n}\text{C}_\text{x}\text{q}^\text{n-x}\text{p}^\text{x},\ \text{where}\ \text{x}=0,\ 1,\ 2,\ 3...\text{n}$
$=\ ^4\text{C}_\text{x}\Bigg(\frac{5}{6}\Bigg)^{4-\text{x}}. \Bigg(\frac{1}{6}\Bigg)^\text{x}$
$=\ ^4\text{C}_\text{x}\cdot\frac{5}{6^4}^{4-\text{x}}$
∴ P(2 successes) = P(X = 2)
$=\ ^4\text{C}_\text{2}\cdot\frac{5}{6^4}^{4-\text{2}}$
$=6.\frac{25}{1296}$
$=\frac{25}{216}$
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Question 1284 Marks
In an examination, 20 questions of true-false type are asked. Suppose a student tosses a fair coin to determine his answer to each question. If the coin falls heads, he answers 'true'; if it falls tails, he answers 'false'. Find the probability that he answers at least 12 questions correctly.
Answer
Let X represent the number of correctly answered questions out of 20 questions.
The repeated tosses of a coin are Bernoulli trails. Since “head” on a coin represents the true answer and “tail” represents the false answer, the correctly answered questions are Bernoulli trials.
$\therefore\ \text{p}=\frac{1}{2}$
$\therefore\ \text{q}=1-\text{p}=1-\frac{1}{2}=\frac{1}{2}$
X has a binomial distribution with n = 20 and $\text{p}=\frac{1}{2}$
$\therefore\ \text{P}(\text{X=x})=\ ^\text{n}\text{C}_\text{x}\text{q}^\text{n-x}\text{p}^\text{x},\ \text{where x}=1,\ 2,\ ...\text{n}$
$=\ ^{20}\text{C}_\text{x}\bigg(\frac{1}{2}\bigg)^{20-\text{x}}.\bigg(\frac{1}{2}\bigg)^\text{x}$
$=\ ^{20}\text{C}_\text{x}\bigg(\frac{1}{2}\bigg)^{20}$
P(at least 12 questions answered correctly) = P(X ≥ 12)
$=\text{P}(\text{X}=12)+\text{P}(\text{X}=13)+...+\text{P}(\text{X}=20)$
$=\ ^{20}\text{C}_{12}\bigg(\frac{1}{2}\bigg)^{20}+\ ^{20}\text{C}_{13}\bigg(\frac{1}{2}\bigg)^{20}+...+\ ^{20}\text{C}_{20}\bigg(\frac{1}{2}\bigg)^{20}$
$=\bigg(\frac{1}{2}\bigg)^{20}.\bigg[\ ^{20}\text{C}_{12}+\ ^{20}\text{C}_{13}+...+\ ^{20}\text{C}_{20}\bigg]$
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Question 1294 Marks
A fair coin is tossed four times. Let X denote the longest string of heads accuring. Find the probability distribution mean and variance of X.
Answer
Let the event of getting a head = H and getting a tail = T
Let X denote the variable longest consecutive heads accuring in 4 tosses. The possible values are
X = 0 (no heads) {T, T, T, T}
X = 1 (1 head) {H, T, T, T}
X = 2 (2 heads) {H, H, T, T}
X = 3 (3 heads) {H, H, H, T}
X = 4 (4 heads) {H, H, H, H}
n(S) = {(HHHH), (HHHT), (HHTT), (HTHH), (HTHT), (HTTH), (HTTT), (THHH), (THTH), (THHT), (THTT), (THTT), (TTHH), (TTHT), (TTTH), (TTTT)}
$\text{P}(\text{X}=0)=\frac{1}{16}$
$\text{P}(\text{X}=1)=\frac{7}{16}$
$\text{P}(\text{X}=2)=\frac{5}{16}$
$\text{P}(\text{X}=3)=\frac{2}{16}$
$\text{P}(\text{X}=4)=\frac{1}{16}$
Thus, the probability distribution is
$\text{X}$
 $0$
 $1$
 $2$
 $3$
 $4$
$\text{p}_\text{i}=\text{P}(\text{X})$
 $\frac{1}{16}$
 $\frac{7}{16}$
 $\frac{5}{16}$
 $\frac{2}{16}$
 $\frac{1}{16}$
$\text{p}_\text{i}\text{x}_\text{i}^2$
 $0$
 $\frac{7}{16}$
 $\frac{20}{16}$
 $\frac{18}{16}$
 $1$
Mean $=\sum_\text{i=1 to n}\text{X}_\text{i}\times\text{P}(\text{X}_\text{i})$
Mean, $\mu=0\times\frac{1}{16}+1\times\frac{7}{16}+2\times\frac{5}{16}+3\times\frac{2}{16}+4\times\frac{1}{16}$
$=0+\frac{7}{16}+\frac{10}{16}+\frac{6}{16}+\frac{4}{16}$
$=\frac{27}{16}=1.7$
Variance Var(X) $=\sum\text{p}_\text{i}\text{x}_\text{i}^2-\big(\sum\text{p}_\text{i}\text{x}_\text{i}\big)^2$
$=\frac{61}{16}-(1.7)^2$
$=3.825-2.89$
$=0.935$
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Question 1304 Marks
A bag contains 4 white, 7 black and 5 red balls. Three balls are drawn one after the other without replacement. Find the probability that the balls drawn are white, black and red respectively.
Answer
Given bag contains 4 white, 7 black and 5 red balls.
Total number of balls = 16
Three balls are drawn without replacement
A = First ball is white
B = Second ball is black
C = Third balls is red
P (Three balls drawn are white, black, red respectively)
$=\text{P}(\text{A}) \text{ P}\Big(\frac{\text{B}}{\text{A}}\Big) \text{ P}\Big(\frac{\text{C}}{\text{A}\cap\text{B}}\Big)$
$=\frac{4}{16}\times\frac{7}{15}\times\frac{5}{14}$
$=\frac{1}{24}$
Required probability $=\frac{1}{24}$
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Question 1314 Marks
A coin is tossed 5 times. What is the probability of getting at least 3 heads?
Answer
Probability of getting head on one throw of coin $=\frac{1}{2}$
So, $\text{p}=\frac{1}{2}$
$\text{q}=1-\frac{1}{2}$
$\text{q}=\frac{1}{2}$ [Since p + q = 1]
The coin is tossed 5 times. Let x denote the number of getting head as 5 tosses of coins.
So probability of getting r head in n tosses of coin is given by
$\text{P}(\text{x = r})=\text{ }^{\text{n}}\text{c}_{\text{r}}\text{p}^{\text{r}}\text{q}^{\text{n - r}}$
$\text{P}(\text{x = r})=\text{ }^5\text{c}_{\text{r}}\big(\frac{1}{2}\big)^{\text{r}}\big(\frac{1}{2}\big)^{5-\text{r}}\dots(1)$
probability of getting at least 3 heads
$=\text{P}(\text{X}=3)+\text{P}(\text{x}=4)+\text{p}(\text{x}=5) $
$=\text{ }^5\text{c}_3\big(\frac{1}{2}\big)^3.\big(\frac{1}{2}\big)^{5-3}+\text{ }^5\text{c}_4\big(\frac{1}{2}\big)^4\big(\frac{1}{2}\big)^{5-4}+\text{ }^5\text{c}_5\big(\frac{1}{2}\big)^5\big(\frac{1}{2}\big)^0$ [Using (1)]
$=\text{ }^5\text{c}_3\big(\frac{1}{2}\big)^3\big(\frac{1}{2}\big)^2+\text{ }^5\text{c}_4\big(\frac{1}{2}\big)^4\big(\frac{1}{2}\big)+\text{ }^5\text{c}_5\big(\frac{1}{2}\big)^5.1$
$=\frac{5.4}{2}.\big(\frac{1}{2}\big)^5+5\big(\frac{1}{2}\big)^5+1.\big(\frac{1}{2}\big)^5$
$=\big(\frac{1}{2}\big)^5[10+5+1]$
$=16.\frac{1}{32}$
$=\frac{1}{2}$
The required probability is $=\frac{1}{2}$
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Question 1324 Marks
Five cards are drawn successively with replacement from a well-shuffled deck of 52 cards. What is the probability that:
  1. All the five cards are spades?
  2. Only 3 cards are spades?
  3. None is a spade?
Answer
Let X represent the number of spade cards among the five cards drawn. Since the drawing of card is with replacement, the trials are Bernoulli trials.

In a well shuffled deck of 52 cards, there are 13 spade cards.

$\Rightarrow\ \text{p}=\frac{13}{52}=\frac{1}{4}$

$\therefore\ \text{q}=1-\frac{1}{4}=\frac{3}{4}$

X has a binomial distribution with n = 5 and $\text{p}=\frac{1}{4}$

$\text{p}(\text{X=x})=\ ^\text{n}\text{C}_\text{x}\text{q}^\text{n-x}\text{p}^\text{x},\ \text{where x}=0,\ 1,\ ...\text{n}$

$=\ ^5\text{C}_\text{x}\Big(\frac{3}{4}\Big)^{5-\text{x}}\Big(\frac{1}{4}\Big)^\text{x}$

  1. P(all five cards are spades) = P(X = 5)

$=\ ^5\text{C}_{5}\Big(\frac{3}{4}\Big)^0.\Big(\frac{1}{4}\Big)^{5}$

$=1\cdot\frac{1}{1024}$

$=\frac{1}{1024}$

  1. P(only 3 cards are spades) = P(X = 3)

$=\ ^5\text{C}_{3}\cdot\Big(\frac{3}{4}\Big)^2.\Big(\frac{1}{4}\Big)^{3}$

$=10\cdot\frac{9}{16}\cdot\frac{1}{64}$

$=\frac{45}{512}$

  1. P(none is a spade) = P(X = 0)

$=\ ^5\text{C}_{0}\cdot\Big(\frac{3}{4}\Big)^5.\Big(\frac{1}{4}\Big)^{0}$

$=1\cdot\frac{243}{1024}$

$=\frac{243}{1024}$

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Question 1334 Marks
A coin is tossed thrice and all the eight outcomes are assumed equally likely. In which of the following cases are the following events A and B are independent?
A = the number of heads is two,
B = the last throw results in head.
Answer
Sample space for throwing a coin thrice
= {HHT, HTT, THT, TTT, HHH, HTH, THH, TTH}
A = The number of heads is two
A = {HHT, THH, HTH}
B = The Last throw results in head
B = {HHH, HTH, THH, TTH}
$\text{A}\cap\text{B}=\{\text{THH, HTH}\}$
$\text{P(A)}=\frac{3}{8}$
$\text{P(B)}=\frac{4}{8}=\frac{1}{2}$
$\text{P}(\text{A}\cap\text{B})=\frac{2}{8}=\frac{1}{4}$
$\text{P(A)}.\text{P}(\text{B})=\frac{3}{8}\times\frac{1}{2}$
$=\frac{3}{16}$
$\text{P(A)}.\text{P(B)}\neq\text{P}(\text{A}\cap\text{B})$
So, A and B are not independent events.
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Question 1344 Marks
A card from a pack of 52 cards is lost. From the remaining cards of the pack, two cards are drawn and are found to be both diamonds. Find the probability of the lost card being a diamond.
Answer
Let E1 = the missing card is a diamond, E2 = the missing card is a spade, E3 = the missing card is a club, E4 = the missing card is a heart and A = drawing of two heart cards from the remaining cards.
$\text{P}(\text{E}_1)=\frac{13}{52}=\frac{1}{4},\ \text{P}(\text{E}_2)=\frac{13}{52}=\frac{1}{4},$
$\text{P}(\text{E}_3)=\frac{13}{52}=\frac{1}{4},\ \text{P}(\text{E}_4)=\frac{13}{52}=\frac{1}{4}$
$\text{P}(\text{A}|\text{E}_1)$ = P(drawing 2 heart cards given that one diamond card is missing) $=\frac{\text{C}(12,\ 2)}{\text{C}(51,\ 2)}$
Similarly, $ \text{P}(\text{A}|\text{E}_2)=\frac{\text{C}(13,\ 2)}{\text{C}(51,\ 2)},\ \text{P}(\text{A}|\text{E}_3)=\frac{\text{C}(13,\ 2)}{\text{C}(51,\ 2)}\ \text{and}\ \text{P}(\text{A}|\text{E}_4)=\frac{\text{C}(13,\ 2)}{\text{C}(51,\ 2)}$
By Bayes’ theorem,
$\text{P}(\text{E}_1|\text{A})=\frac{\text{P}(\text{E}_1)\text{P}(\text{A}|\text{E}_1)}{\text{P}(\text{E}_1)\text{P}(\text{A}|\text{E}_1)+\text{P}(\text{E}_2)\text{P}(\text{A}|\text{E}_2)+\text{P}(\text{E}_3)\text{P}(\text{A}|\text{E}_3)+\text{P}(\text{E}_4)\text{P}(\text{A}|\text{E}_4)}$
$=\frac{\frac{1}{4}\times\frac{\text{C}(12,\ 2)}{\text{C}(51,\ 2)}}{\frac{1}{4}\times\frac{\text{C}(12,\ 2)}{\text{C}(51,\ 2)}+\frac{1}{4}\times\frac{\text{C}(13,\ 2)}{\text{C}(51,\ 2)}+\frac{1}{4}\times\frac{\text{C}(13,\ 2)}{\text{C}(51,\ 2)}+\frac{1}{4}\times\frac{\text{C}(13,\ 2)}{\text{C}(51,\ 2)}}$
$=\frac{66}{66+78+78+78}=\frac{11}{50}$
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Question 1354 Marks
Explain why the experiment of tossing a coin three times is said to have binomial distribution.
Answer
We know that, in a Bindmial distribution,

  1. There are 2 outcomes for each trial.
  2. There is a fixed number of trials.
  3. The probability of success must be the same for all the trials.

When coin is tossed, possible outcomes are Hesd and Tail.

Since coin is tossed three times, we have fixed number of trials.

Also probility of Head and Tail in each trial is $\frac{1}{2}.$

Thus given experiment is said to have binomial distribution.

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Question 1364 Marks
Find the probability distribution of the number of doublets in three throws of a pair of dice and find its mean.
Answer
Throwing a doublet i.e. $\big\{(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)\big\}$
Total number of outcomes = 36
Let p be the probability of success therefore
$\text{p}=\frac{6}{3}6=\frac{1}{6}$
Let q be the probability of failure therefore $\text{q}=1-\text{p}=1-\frac{1}{6}=\frac{5}{6}$
Since there is three rows of dise so n = 3
Let X be the random variable for getting doublet, therefore X can take at max 3 values.
$\text{P(X}=0)=\text{ }^3\text{c}_0\text{p}^0\text{q}^{3}=\big(\frac{5}{6}\big)^{3}=\frac{125}{216}$
$\text{P(X}=1)=\text{ }^3\text{c}_1\text{p}^1\text{q}^2=3.\frac{1}{6}.\big(\frac{5}{6}\big)^2=\frac{75}{216}$
$\text{P(X}=2)=\text{ }^3\text{c}_2\text{p}^2\text{q}^1=3.\big(\frac{1}{6}\big)^2.\big(\frac{5}{6}\big)=\frac{15}{216}$
$\text{P(X}=3)=\text{ }^3\text{c}_3\text{p}^3\text{q}^0=\big(\frac{1}{6}\big)^3=\frac{1}{216}$
Mean
$\mu=\sum^3_{\text{i}-1}\text{X}_{\text{i}}\text{P(X}_{\text{i}})=0.\frac{125}{216}+1.\frac{75}{216}+2.\frac{15}{216}+3.\frac{1}{216}$
$=\frac{75+30+3}{216}=\frac{108}{216}=\frac{1}{2}$
Hence the mean is $=\frac{1}{2}$
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Question 1374 Marks
An unbiased die is tossed twice. Find the probability of getting 4, 5, or 6 on the first toss and 1, 2, 3 or 4 on the second toss.
Answer
Given an unbiased die is tossed twise
A = Getting 4, 5 or 6 on the first toss
B = 1, 2, 3 or 4 on second toss
$\Rightarrow\ \text{P(B)}=\frac{3}{6}$
$\text{P(A)}=\frac{1}{2}$
and, $\text{P(B)}=\frac{4}{6}$
$\text{P(B)}=\frac{2}{3}$
P (Getting 4, 5 or 6 on the first toss and 1, 2, 3 or 4 on second toss)
$=\text{P}(\text{A}\cap\text{B})$
$=\text{P(A) }\text{P(B)}$
$=\frac{1}{2}\times\frac{2}{3}$
$=\frac{1}{3}$
Required probability $=\frac{1}{3}$
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Question 1384 Marks
If A and B are two events such that $\text{P}(\text{A})=\frac{1}{3},\text{P(B)}=\frac{1}{5}$ and $\text{P}(\text{A}\cup\text{B})=\frac{11}{30}$ find $\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)$ and $\text{P}\Big(\frac{\text{B}}{\text{A}}\Big).$
Answer
Given,
$\text{P}(\text{A})=\frac{1}{3},\text{P(B)}=\frac{1}{5}$ and $\text{P}(\text{A}\cup\text{B})=\frac{11}{30}$
$\text{P}(\text{A}\cup\text{B})=\text{P}(\text{A})+\text{P(B)}-\text{P}(\text{A}\cap\text{B})$
$\text{P}(\text{A}\cap\text{B})=\frac{1}{3}+\frac{1}{5}-\frac{11}{30}$
$=\frac{10+6-11}{30}$
$=\frac{5}{30}$
$\text{P}(\text{A}\cap\text{B})=\frac{1}{6}$
We know that,
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{\text{P}(\text{A}\cap\text{B})}{\text{P}(\text{B})}$
$=\frac{\frac{1}{6}}{\frac{1}{5}}$
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{5}{6}$
$\text{P}\Big(\frac{\text{B}}{\text{A}}\Big)=\frac{\text{P}(\text{B}\cap\text{A})}{\text{P}(\text{A})}$
$\text{P}\Big(\frac{\text{B}}{\text{A}}\Big)=\frac{\frac{1}{6}}{\frac{1}{3}}$
$=\frac{1}{6}\times\frac{3}{1}$
$=\frac{1}{2}$
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{5}{6},\text{P}\Big(\frac{\text{B}}{\text{A}}\Big)=\frac{1}{2}$
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Question 1394 Marks
A couple has two children. Find the probability that both the children are,
  1. Males, if it is known that at least one of the children is male.
  2. Females, if it is known that the elder child is a female.
Answer
Consider the given events.
A = Both the children are female.
B = The elder child is a female.
C = At least one child is a male.
D = Both children are male.
Clearly,
S = {M1M2, M1F2, F1M2, F1F2}
A = {F1F2}
B = {F1M2, F1F2}
C = {M1F2, F1M2, M1M2}
D = {M1M2}
[Here, first child is elder and second is younger]
$\text{D}\cap\text{C}=\big\{\text{M}_1\text{M}_2\big\}$ and $\text{A}\cap\text{B}=\big\{\text{F}_1\text{F}_2\big\}$
  1. Required probability $=\text{P}\Big(\frac{\text{D}}{\text{C}}\Big)=\frac{\text{n}(\text{D}\cap\text{C})}{\text{n}(\text{C})}=\frac{1}{3}$
  2. Required Probability $= \text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{\text{n}(\text{A}\cap\text{B})}{\text{n}(\text{B})}=\frac{1}{2}$
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Question 1404 Marks
In a certain college, 4% of boys and 1% of girls are taller than 1.75 metres. Further more, 60% of the students in the colleges are girls. A student selected at random from the college is found to be taller than 1.75 metres. Find the probability that the selected students is girl.
Answer
Consider the following events
E1 = The selected student is a girl
E2 = The selected student is not a girl
A = The student is taller than 1.75 meters
We have,
$\text{P}(\text{E}_1)=60\%=\frac{60}{100}=0.6$
$\text{P}(\text{E}_2)=1-\text{P}(\text{E}_1)=1-0.6=0.4$
$\text{P}\Big(\frac{\text{A}}{\text{E}_1}\Big)=$ Probability that the student is taller than 1.75 meters given that the student is a girl
$\text{P}\Big(\frac{\text{A}}{\text{E}_1}\Big)=\frac{1}{100}=0.01$
And
$\text{P}\Big(\frac{\text{A}}{\text{E}_2}\Big)=$ Probability that the student is taller than 1.75 meters given that the student is not a girl
$\text{P}\Big(\frac{\text{A}}{\text{E}_2}\Big)=\frac{4}{100}=0.04$
Now,
Required probability $\text{P}\Big(\frac{\text{E}_1}{\text{A}}\Big)=\frac{\text{P}(\text{E}_1)\text{P}\Big(\frac{\text{A}}{\text{E}_1}\Big)}{\text{P}(\text{E}_1)\text{P}\Big(\frac{\text{A}}{\text{E}_1}\Big)+\text{P}(\text{E}_2)\text{P}\Big(\frac{\text{A}}{\text{E}_2}\Big)}$
$=\frac{0.6\times0.01}{0.6\times0.01+0.4\times0.04}$
$=\frac{\frac{6}{1000}}{\frac{22}{1000}}$
$=\frac{3}{11}$
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Question 1414 Marks
By examining the chest X-ray, probability that T.B. is detected when a person is actually suffering is 0.99. The probability that the doctor diagnoses incorrectly that a person has T.B. on the basic of X-ray is 0.001. In a certain city 1 in 1000 persons suffers from T.B. A person is selected at random is diagnosed to have T.B. what is the chance that he actually has T.B.?
Answer
Consider events E1, E2 and A as
E1 = The person selected is actually having T.B.
E2 = The person selected not having T.B.
E3 = The person diagnosed to have T.B.
Given,
$\text{P}(\text{E}_1)=\frac{1}{1000}$
$\text{P}(\text{E}_2)=\frac{999}{1000}$
P(A|E1) = P(Person diagnosed to have T.B. and he is actually having T.B.)
$=0.99$
$\text{P}\Big(\frac{\text{A}}{\text{E}_2}\Big)=$ P(Person diagnossed to have a T.B. and he is not a actually having T.B.)
$0.001$
To find, P(Person diagnosed to have T.B. is actually having T.B.) $=\text{P}\Big(\frac{\text{E}_1}{\text{A}}\Big)$
By baye's theorem,
$\text{P}\Big(\frac{\text{E}_1}{\text{A}}\Big)=\frac{\text{P}(\text{E}_1)\text{P}\Big(\frac{\text{A}}{\text{E}_1}\Big)}{\text{P}(\text{E}_1)\text{P}\Big(\frac{\text{A}}{\text{E}_1}\Big)+\text{P}(\text{E}_2)\text{P}\Big(\frac{\text{A}}{\text{E}_2}\Big)}$
$=\frac{\frac{1}{1000}\times0.99}{\frac{1}{1000}\times0.99+\frac{999}{1000}\times0.001}$
$=\frac{990}{990+999}$
$=\frac{990}{1989}$
$=\frac{110}{221}$
Required probability $=\frac{110}{221}$
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Question 1424 Marks
An experiment succeeds twice as often as it fails. Find the probability that in the next 6 trials there will be at least 4 successes.
Answer
Let X denote the number of successes in 6 trials.
It is given that successes are twice the failures.
$\Rightarrow\text{p}=2\text{q}$
$\text{p + q}=1$
$\Rightarrow3\text{q}=1$
$\Rightarrow\text{q}=\frac{1}{3}$
$\therefore\text{p}=1-\frac{1}{3}=\frac{2}{3}$
$\text{ n}=6$
Hence, the distribution is given by
$\text{P(X = r})=\text{ }^6\text{C}_{\text{r}}\big(\frac{2}{3}\big)^{\text{r}}\big(\frac{1}{3}\big)^{6-\text{r}},\text{r}=0,1,2,\dots6$
$\text{P(atleast 4 successes})=\text{P(X}\geq4)$
$=\text{P(X}=4)+\text{P(X}=5)+\text{P(X}=6)$
$\text{ }^6\text{C}_4\big(\frac{2}{3}\big)^4\big(\frac{1}{3}\big)^{6-4}+\text{ }^6\text{C}_5\big(\frac{2}{3}\big)^5\big(\frac{1}{3})^{6-5}+\text{ }^6\text{C}_6\big(\frac{2}{3}\big)^6\big(\frac{1}{3}\big)^{6-6}$
$=\frac{15(2^4)+6(32)+64}{3^6}$
$=\frac{240+192+64}{729}$
$=\frac{496}{729}$
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Question 1434 Marks
A bag contains 20 tickets, numbered from 1 to 20. Two tickets are drawn without replacement. What is the probability that the first ticket has an even number and the second an odd number.
Answer
Total number of tickets are 20 numbered from 1, 2, 3, ..... 20.
Number of tickets with even numbers,
= 10 [Since, even numbers are 2, 4, 6, 8, 10, 12, 14, 16, 18, 20]
Number of tickets with odd numbers,
= 10 [Since, odd numbers are 1, 3, 5, 7, 9, 11, 13, 15, 17,19]
Two cards are drawn without replacement.
A = tickets with even numbers
B = tickets with odd numbers
P (first ticket has even number and second has odd number)
$=\text{P(A) }\text{P}\Big(\frac{\text{B}}{\text{A}}\Big)$
$=\frac{10}{20}\times\frac{10}{19}$
$=\frac{5}{19}$
Required probability $=\frac{5}{19}$
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Question 1444 Marks
A die is thrown three times, find the probability that 4 appears on the third toss if it is given that 6 and 5 appear respectively on first two tosses.
Answer
A = 4 appears on third toss, if a die is thrown three times
= {(1, 1, 4), (1, 2, 4), (1, 3, 4), (1, 4, 4), (1, 5, 4), (1, 6, 4)
(2, 1, 4), (2, 2, 4), (2, 3, 4), (2, 4, 4), (2, 5, 4), (2, 6, 4)
(3, 1, 4), (3, 2, 4), (3, 3, 4), (3, 4, 4), (3, 5, 4), (3, 6, 4)
(4, 1, 4), (4, 2, 4), (4, 3, 4), (4, 4, 4), (4, 5, 4), (4, 6, 4)
(5, 1, 4), (5, 2, 4), (5, 3, 4), (5, 4, 4), (5, 5, 4), (5, 6, 4)
(6, 1, 4), (6, 2, 4), (6, 3, 4), (6, 4, 4), (6, 5, 4), (6, 6, 4)}
B = 6 and 5 appears respectively on first two tosses, if die is tosses three times
B = {(6, 5, 1), (6, 5, 2), (6, 5, 3), (6, 5, 4), (6, 5, 5), (6, 5, 6)}
$\text{A}\cap\text{B}=\{(6,5,4)\}$
Required probability $=\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)$
$=\frac{\text{n}(\text{A}\cap\text{B})}{\text{n}(\text{B})}$
$=\frac{1}{6}$
Required Probability $=\frac{1}{6}$
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Question 1454 Marks
Five defective mangoes are acciedently mixed with 15 good ones. Four mangoes are drawn at random from this lot. Find the probability distribution of the number of defective mangoes.
Answer
Let X denote number of defective mangoes in a sample of 4 mangoes drawn from a bag containing 5 defective mangoes and 15 good mangoes. Then, X can take the values 0, 1, 2, 3 and 4.

Now,

P(X = 0)

P(no defective mango)

$=\frac{\text{}^{15}\text{C}_4}{\text{}^{20}\text{C}_4}$

$=\frac{1365}{4845}$

$=\frac{91}{323}$

P(X = 1)

= P(1 defective mango)

$=\frac{\text{}^5\text{C}_1\times\text{}^{15}\text{C}_3}{\text{}^{20}\text{C}_4}$

$=\frac{2275}{4845}$

$=\frac{455}{969}$

P(X = 2)

= P(2 defective mangoes)

$=\frac{\text{}^5\text{C}_2\times\text{}^{15}\text{C}_2}{\text{}^{20}\text{C}_4}$

$=\frac{1050}{4845}$

$=\frac{70}{323}$

P(X = 2)

= P(3 defective mangoes)

$=\frac{\text{}^5\text{C}_3\times\text{}^{15}\text{C}_1}{\text{}^{20}\text{C}_4}$

$=\frac{150}{4845}$

$=\frac{10}{323}$

P(X = 3)

= P(4 defective mangoes)

$=\frac{\text{}^5\text{C}_4}{\text{}^{20}\text{C}_4}$

$=\frac{5}{4845}$

$=\frac{1}{969}$

The required probability distribution of X is given by

$\text{X}$
$0$
$1$
$2$
$3$
$4$
$\text{P}(\text{X})$
$\frac{91}{323}$
$\frac{455}{969}$
$\frac{70}{323}$
$\frac{10}{323}$
$\frac{1}{969}$
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Question 1464 Marks
An urn contains 4 red and 3 blue balls. Find the probability distribution of the number of blue balls in a random draw of 3 balls with replacement.
Answer
Urn has 4 red and 3 blue balls. 3 balls are drawn with replacement. Let X denote the numbers of blue balls drawn out of 3 drawn.
So, X has values 0, 1, 2, 3
$\text{P}(\text{X}=0)=\text{P}\big(\overline{\text{B}}_1\big)\times\text{P}\big(\overline{\text{B}}_2\big)\times\text{P}\big(\overline{\text{B}}_3\big)$
$=\frac{4}{7}\times\frac{4}{7}\times\frac{4}{7}$
$=\frac{144}{343}$
$\text{P}(\text{X}=1)=\text{P}(\text{B}_1)\text{P}\big(\overline{\text{B}}_2\big)\text{P}\big(\overline{\text{B}}_3\big)+\text{P}\big(\overline{\text{B}}_1\big)\text{P}(\text{B}_2)\text{P}\big(\overline{\text{B}}_3\big)+\text{P}\big(\overline{\text{B}}_1\big)\text{P}\big(\overline{\text{B}}_2\big)\text{P}(\text{B}_3)$
$=\frac{3}{7}\times\frac{4}{7}\times\frac{4}{7}+\frac{4}{7}\times\frac{3}{7}\times\frac{4}{7}+\frac{4}{7}\times\frac{4}{7}\times\frac{3}{7}$
$=\frac{144}{343}$
$\text{P}(\text{X}=2)=\text{P}(\text{B}_1)\text{P}(\text{B}_2)\text{P}\big(\overline{\text{B}}_3\big)+\text{P}(\text{B}_1)\text{P}\big(\overline{\text{B}}_2\big)\text{P}(\text{B}_3)+\text{P}\big(\overline{\text{B}}_1\big)\text{P}(\text{B}_2)\text{P}(\text{B}_3)$
$=\frac{3}{7}\times\frac{3}{7}\times\frac{4}{7}+\frac{3}{7}\times\frac{4}{7}\times\frac{3}{7}+\frac{4}{7}\times\frac{3}{7}\times\frac{3}{7}$
$=\frac{108}{343}$
$\text{P}(\text{X}=3)=\text{P}(\text{B}_1)\text{P}(\text{B}_2)\text{P}(\text{B}_3)$
$=\frac{3}{7}\times\frac{3}{7}\times\frac{3}{7}$
$=\frac{27}{343}$
So,
Required probability distribution is
$\text{X}:$
$0$
$1$
$2$
$3$
$\text{P}(\text{X}):$
$\frac{64}{343}$
$\frac{144}{343}$
$\frac{108}{343}$
$\frac{27}{343}$
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Question 1474 Marks
If the mean and variance of a random variable X having  a binomial distribution are 4 and 2 respectively, find P (X = 1).
Answer
Let n and p be the parmeters of binomial distribution,
Given,
$\text{Mean = np}=4$
$\text{Variance = npq}=2$
Dividing equation (2) by (1),
$\frac{\text{npq}}{\text{np}}=\frac{2}{4}$
$\text{q}=\frac{1}{2}$.
$\text{p}=1-\frac{1}{2}$
$\text{p}=\frac{1}{2}$
Put the value of p in equation (1),
$\text{np}=4$
$\text{n}\big(\frac{1}{2}\big)=4$
$\text{n}=8$
Hence, binomial distribution is given by
$\text{P(X = r})=\text{ }^{\text{n}}\text{c}_{\text{r}}\text{p}^{\text{r}}\text{q}^{\text{n}-\text{r}}$
$\text{P(X= r})\text{ }^8\text{c}_{\text{r}}\big(\frac{1}{2}\big)^{\text{r}}\big(\frac{1}{2}\big)^{8-\text{r}}$
$\text{P(X=}1)$
$=\text{ }^8\text{c}_{1}\big(\frac{1}{2}\big)^{1}\big(\frac{1}{2}\big)^{8-1}$
$=8\big(\frac{1}{2}\big)^8$
$=\big(\frac{1}{2}\big)^5$
$=\frac{1}{32}$
$\text{P(X}=1)=\frac{1}{32}$
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Question 1484 Marks
Find the variance of the distribution:
$\text{x}$
 $0$
 $1$
 $2$
 $3$
 $4$
 $5$
$\text{P}(\text{x})$
 $\frac{1}{6}$
 $\frac{5}{18}$
 $\frac{2}{9}$
 $\frac{1}{6}$
 $\frac{1}{9}$
 $\frac{1}{18}$
Answer
We have,
$\text{X}$
 $0$
 $1$
 $2$
 $3$
 $4$
 $5$
$\text{P}(\text{X})$
 $\frac{1}{6}$
 $\frac{5}{18}$
 $\frac{2}{9}$
 $\frac{1}{6}$
 $\frac{1}{9}$
 $\frac{1}{18}$
$\text{X}\text{P}(\text{X})$
 $0$
 $\frac{5}{18}$
 $\frac{4}{9}$
 $\frac{1}{2}$
 $\frac{4}{9}$
 $\frac{5}{18}$
$\text{X}^2\text{P}(\text{X})$
 $0$
 $\frac{5}{18}$
 $​​\frac{8}{9}$
 $\frac{3}{2}$
 $\frac{16}{9}$
 $\frac{25}{18}$
$\therefore\text{Variance}=\text{E}(\text{X}^2)-\big[\text{E}(\text{X})\big]^2$
$=\sum\text{X}^2\text{P}(\text{X})-\big[\sum\text{X}\text{P}(\text{X})\big]^2$
$=\Big[0+\frac{5}{18}+\frac{8}{9}+\frac{3}{2}+\frac{16}{9}+\frac{25}{18}\Big]-\Big[0+\frac{5}{18}+\frac{4}{9}+\frac{1}{2}+\frac{4}{9}+\frac{5}{18}\Big]^2$
$=\frac{105}{18}-\frac{35^2}{18^2}$
$=\frac{18\times105-35\times35}{18^2}$
$=\frac{19\times35}{324}$
$=\frac{665}{324}$
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Question 1494 Marks
If A and B are two events such that,
$\text{P(A)}=\frac{1}{2},\text{P(B)}=\frac{1}{3}$ and $\text{P}(\text{A}\cap\text{B})=\frac{1}{4},$ then find $\text{P}(\text{A}|\text{B}), \text{ P}(\text{B}|\text{A}), \text{ P}(\overline{\text{A}}|\text{B})$ and $\text{P}(\overline{\text{A}}|\overline{\text{B}}).$
Answer
We have,
$\text{P(A)}=\frac{1}{2},\text{P(B)}=\frac{1}{3}$ and $\text{P}(\text{A}\cap\text{B})=\frac{1}{4},$
Also, $\text{P}(\overline{\text{B}})=1-\text{P(B)}=1-\frac{1}{3}=\frac{2}{3}$
As, $\text{P}(\text{A}\cup\text{B})=\text{P(A)}+\text{P(B)}-\text{P}(\text{A}\cap\text{B})$
$=\frac{1}{2}+\frac{1}{3}-\frac{1}{4}$
$=\frac{6+4-3}{12}$
$\Rightarrow\ \text{P}(\text{A}\cup\text{B})=\frac{7}{12}$
Also, $\text{P}(\overline{\text{A}}\cap\text{B})=\text{P(B)}-\text{P}(\text{A}\cap\text{B})$
$\Rightarrow\ \text{P}(\overline{\text{A}}\cap\text{B})=\frac{1}{3}-\frac{1}{4}$
$\Rightarrow\ \text{P}(\overline{\text{A}}\cap\text{B})=\frac{4-3}{12}$
$\Rightarrow\ \text{P}(\overline{\text{A}}\cap\text{B})=\frac{1}{12}$
And, $\Rightarrow\ \text{P}(\overline{\text{A}}\cap\overline{\text{B}})=\text{P}(\overline{\text{A}\cup\text{B}})$
$=1-\text{P}({\text{A}\cup\text{B}})$
$=1-\frac{7}{12}$
$=\frac{5}{12}$
Now,
$\text{P}(\text{A}|\text{B})=\frac{\text{P}(\text{A}\cap\text{B})}{\text{P}(\text{B})}=\frac{\Big(\frac{1}{4}\Big)}{\Big(\frac{1}{3}\Big)}=\frac{3}{4},$
$\text{P}(\text{B}|\text{A})=\frac{\text{P}(\text{A}\cap\text{B})}{\text{P}(\text{A})}=\frac{\Big(\frac{1}{4}\Big)}{\Big(\frac{1}{2}\Big)}=\frac{2}{4}=\frac{1}{2},$
$\text{P}(\overline{\text{A}}|\text{B})=\frac{\text{P}(\overline{\text{A}}\cap\text{B})}{\text{P(B)}}=\frac{\Big(\frac{1}{12}\Big)}{\Big(\frac{1}{3}\Big)}=\frac{3}{12}=\frac{1}{4}$ and
$\text{P}(\overline{\text{A}}|\overline{\text{B}})=\frac{\text{P}(\overline{\text{A}}\cap\overline{\text{B}})}{\text{P}(\overline{\text{B}})}=\frac{\Big(\frac{5}{12}\Big)}{\Big(\frac{2}{3}\Big)}=\frac{15}{24}=\frac{5}{8}$
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Question 1504 Marks
An article manufactured by a company consists of two parts X and Y. In the process of manufacture of the part X, 9 out of 100 parts may be defective. Similarly, 5 out of 100 are likely to be defective in the manufacture of part Y. Calculate the probability that the assembled product will not be defective.
Answer
Given,
Prat X has 9 out of 100 defective
⇒ Part X has 91 out of 100 non defective
Part Y has out of 100 defective
⇒ Part Y has 95 out of 100 non defective
Consider,
X = A non defective part X
Y = A non defective Part Y
$\Rightarrow\ \text{P(x)}=\frac{91}{100}$ and $\text{P(Y)}=\frac{95}{100}$
= P(Assembled product will bot be defective)
= P(Niether X defective nor Y defective)
$=\text{P}(\text{X}\cap\text{Y})$
$=\text{P(X) }\text{P(Y)}$
$=\frac{91}{100}\times\frac{95}{100}$
$=0.8645$
Required probability = 0.8645
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