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MATHS 3 Marks Question

Probability · Rajasthan Board (RBSE) STD 12 Science (Rajasthan - English Medium). 183 questions.

Questions · Page 4 of 4

3 Marks Question

Question 1513 Marks
Two dice are thrown. Find the probability that the numbers appeared has the sum 8, if it is known that the second die always exhibits 4.
Answer
Two dice are thrown.
A = Sum on the dice is 8
A = {(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)}
B = Second die always exhibits 4
B = {(1, 4), (2, 4), (3, 4), (4, 4), (5, 4), (6, 4)}
$(\text{A}\cap\text{B})=\{(4, 4)\}$
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{\text{n}(\text{A}\cap\text{B})}{\text{n}(\text{B})}$
$=\frac{1}{6}$
Required probability $=\frac{1}{6}$
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Question 1523 Marks
If A and B be two events such that $\text{P(A)}=\frac{1}{4},\text{P(B)}=\frac{1}{3}$ and $\text{P}(\text{A}\cap\text{B})=\frac{1}{2},$ show that A and B are independent events.
Answer
$\text{P}(\text{A}\cup\text{B})=\text{P(A)}+\text{P(B)}-\text{P}(\text{A}\cap\text{B})$
$\Rightarrow\ \text{P}(\text{A}\cap\text{B})=\text{P(A)}+\text{P(B)}-\text{P}(\text{A}\cup\text{B})$
$\Rightarrow\ \text{P}(\text{A}\cap\text{B})=\frac{1}{4}+\frac{1}{3}-\frac{1}{2}$
$\Rightarrow\ \text{P}(\text{A}\cap\text{B})=\frac{3+4-6}{12}$
$\Rightarrow\ \text{P}(\text{A}\cap\text{B})=\frac{1}{12}=\frac{1}{4}\times\frac{1}{3}=\text{P(A)}\text{ P(B)}$
Thus, A and B are independent events.
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Question 1533 Marks
Determine P(E|F) : A coin is tossed three times.
E : at most two tails, F : at least one tail.
Answer
E : at most two tails
$\text{E}=(\text{HTT, THT, TTH, HHT, HTH, THH, HHH})$$\ \ \ \ \ \ \ \ \Rightarrow\ \ \ \ \text{n}(\text{E})=7$
$\text{P}\left(\text{E}\right)=\frac{\text{n}\left(\text{E}\right)}{\text{n}\left(\text{S}\right)}=\frac{7}{8}$
F : at least one tail
$\text{F}=(\text{TTT, HTT, THT, TTH, HHT, HTH, THH})\ \ \ \ \ \ \ \ \Rightarrow\ \ \ \ \ \text{n}(\text{F})=7$
$\text{P}\left(\text{F}\right)=\frac{\text{n}\left(\text{F}\right)}{\text{n}\left(\text{S}\right)}=\frac{7}{8}$
$\therefore\ \ \ \ \ \ \ \text{E}\cap\text{F}=\left(\text{HTT, THT, TTH, HHT, HTH, THH}\right)\Rightarrow\ \ \ \text{n}\left(\text{E}\cap\text{F}\right)=6$
$ \therefore\ \ \ \ \ \ \ \text{P}\left(\text{E}\cap\text{F}\right)=\frac{\text{n}\left(\text{E}\cap\text{F}\right)}{\text{n}\left(\text{S}\right)}=\frac{6}{8}$
$\text{And}\ \ \ \ \text{P}\left(\text{E}|\text{F}\right)=\frac{\text{P}\left(\text{E}\cap\text{F}\right)}{\text{P}\left(\text{F}\right)}=\frac{\frac{6}{8}}{\frac{7}{8}}=\frac{6}{7}$
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Question 1543 Marks
A coin is tossed three times. Find $\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)$ in each of the following:
A = At least two heads,
B = At most two heads.
Answer
Sample space for three coins is given by {HHH, HTH, THH, TTH, HHT, HTT, THT, TTT} A = At least two heads A = {HHT, HHT, HTH, THH} B = At most two heads B = {HHT, HTT, THT, TTT, HTH, THH, TTH} $(\text{A}\cap\text{B})=\{\text{HHT, HTH, THH}\}$ $\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{\text{P}(\text{A}\cap\text{B})}{\text{P}(\text{B})}$ $\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{3}{7}$Hence, $\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{3}{7}$
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Question 1553 Marks
An unbiased coin is tossed. If the result is a head, a pair of unbiased dice is rolled and the sum of the numbers obtained is noted. If the result is a tail, a card from a well shuffled pack of eleven cards numbered $2, 3, 4, ....., 12$ is picked and the number on the card is noted. What is the probability that the noted number is either $7$ or $8$?
Answer
Let $E_1, E_2$ and $A$ be the events as defined below:
$E_1 =$ The coin shows a head
$E_2 =$ The coin shows a head
$A =$ The noted number is $7$ or $8$
$\therefore\ \text{P}(\text{E})_1=\frac{1}{2}$
$\text{P}(\text{E})_2=\frac{1}{2}$
Now,
$\text{P}\Big(\frac{\text{A}}{\text{E}_1}\Big)=\frac{11}{36}$
$\text{P}\Big(\frac{\text{A}}{\text{E}_2}\Big)=\frac{2}{11}$
Using the law of total probability, we get
Required probability $=\text{P(A)}=\text{P}(\text{E}_1)\text{ P}\Big(\frac{\text{A}}{\text{E}_1}\Big)+\text{P}(\text{E}_2)\text{ P}\Big(\frac{\text{A}}{\text{E}_2}\Big)$
$=\frac{1}{2}\times\frac{11}{36}+\frac{1}{2}\times\frac{2}{11}$
$=\frac{11}{72}+\frac{1}{11}$
$=\frac{121+72}{792}$
$=\frac{193}{792}$
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Question 1563 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 and, hence, find its mean.
Answer
Let n and p be the parameters of binomial distribution,
Given, $\text{n}=6$
$\text{Mean + Variance}=\frac{10}{3}$
$\text{np + npq}=\frac{10}{3}$
$\text{6p}+\text{6pq}=\frac{10}{3}$
$\text{6p(1 + q})=\frac{10}{3}$
$6(1-\text{q})(1+\text{q})=\frac{10}{3}$ [Since p + q = 1]
$6(1-\text{q}^2)=\frac{10}{3}$
$1-\text{q}^2=\frac{10}{18}$
$-\text{q}^2=\frac{5}{9}-1$
$-\text{q}^2=-\frac{4}{9}$
$\text{q}^2=\frac{4}{9}$
$\text{q}=\frac{2}{3}$
$\text{p}=1-\text{q}$
$=1-\frac{2}{3}$
$\text{p}=\frac{1}{3}$
Hence, the binomial distribution is given by,
$\text{P(X = r})=\text{ }^6\text{C}_{\text{r}}\big(\frac{1}{3}\big)^{\text{r}}\big(\frac{2}{3}\big)^{6-\text{r}}$
as $\text{r}=0,1,2\dots6$
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Question 1573 Marks
If the mean and variance of a binomial variate X are 2 and 1 respectively, find P (X > 1).
Answer
$\text{Mean}=2,\text{variance}=1$
$\therefore\text{q}=\frac{\text{Variance}}{\text{Mean}}=\frac{1}{2}$
and $\text{p}=1-\frac{1}{2}=\frac{1}{2}$
$\text{n}=\frac{\text{Mean}}{\text{p}}=\frac{2}{\frac{1}{2}}=4$
The binomial distribution is given by
$\text{P(X = r})=\text{ }^4\text{C}_{\text{r}}\big(\frac{1}{2}\big)^{\text{r}}\big(\frac{1}{2}\big)^{4-\text{r}}$
$\therefore\text{P(X}=0)=\text{ }^4\text{C}_0\big(\frac{1}{2}\big)^0\big(\frac{1}{2}\big)^{4-0},\text{r}=0,1,2,3,4$
$=\big(\frac{1}{2}\big)^4$
$\text{P(X}>1)=1-\text{P(X}=0)$
$=1-\big(\frac{1}{2}\big)^4$
$=\frac{15}{16}$
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Question 1583 Marks
A die is thrown three times. Find $\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)$ and $\text{P}\Big(\frac{\text{B}}{\text{A}}\Big)$, if
A = 4 appears on the third toss,
B = 6 and 5 appear respectively on first two tosses.
Answer
Consider the given events.
A = Getting 4 on third throw
B = Getting 6 on first throw and and 5 on second throw
Clearly,
A = {(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), ...... (6, 1, 4), (6, 2 4), (6, 3, 4), (6, 4, 4), (6, 5, 4), (6, 6, 4)}
B = {(6, 5, 1), (6, 5, 2), (6, 5, 3), (6, 5, 4), (6, 5, 5), (6, 5, 6)}
Now,
$(\text{A}\cap\text{B})=\{(6, 5, 4)\}$
$\therefore\ \text{Required probability} = \text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{\text{n}(\text{A}\cap\text{B})}{\text{n}(\text{B})}=\frac{1}{6}$
$\therefore\ \text{Required probability} = \text{P}\Big(\frac{\text{B}}{\text{A}}\Big)=\frac{\text{n}(\text{A}\cap\text{B})}{\text{n}(\text{A})}=\frac{1}{36}$
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Question 1593 Marks
If A and B are events such that P(A) = 0.6, P(B) = 0.3 and $\text{P}(\text{A}\cap\text{B})=0.2$ find $\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)$ and $\text{P}\Big(\frac{\text{A}}{\text{B}}\Big).$
Answer
Given:
P(A) = 0.6
P(B) = 0.3
$\text{P}(\text{A}\cap\text{B})=0.2$
Now,
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{\text{P}(\text{A}\cap\text{B})}{\text{P(B)}}$
$\Rightarrow\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{0.2}{0.3}=\frac{2}{3}$
$\text{P}\Big(\frac{\text{B}}{\text{A}}\Big)=\frac{\text{P}(\text{A}\cap\text{B})}{\text{P(A)}}$
$\Rightarrow\text{P}\Big(\frac{\text{B}}{\text{A}}\Big)=\frac{0.2}{0.6}=\frac{1}{3}$
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Question 1603 Marks
A coin is tossed three times, if head occurs on first two tosses, find the probability of getting head on third toss.
Answer
Consider the given events.
A = Getting head on third toss
B = Getting head on first two tosses
Clearly,
A = {(H, H, H), (H, T, H), (T, H, H), (T, T, H)}
B = {(H, H, H), (H, H, T)}
Now,
$\text{A}\cap\text{B}=\{\text{H},\text{H},\text{H}\}$
$\therefore\text{Required probability}=\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{\text{n}(\text{A}\cap\text{B})}{\text{n}(\text{B})}=\frac{1}{2}$
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Question 1613 Marks
Four cards are successively drawn without replacement from a deck of $52$ playing cards. What is the probability that all the four cards are kings?
Answer
Let $E_1, E_2, E_3$ and $E_4$ are the events that the first, second, third and fourth card is king, respectively.
$\therefore\text{P}(\text{E}_1\cap\text{E}_2\cup\text{E}_3\cap\text{E}_4)$
$=\text{P}(\text{E}_1)\cdot\text{P}\Big(\frac{\text{E}_2}{\text{E}_1}\Big)\cdot\text{P}\Big(\frac{\text{E}_3}{\text{E}_1\cap\text{E}_2}\Big)\cdot\text{P}\Big[\frac{\text{E}_4}{(\text{E}_1\cap\text{E}_2\cup\text{E}_3\cap\text{E}_4)}\Big]$
$=\frac{4}{52}\cdot\frac{3}{51}\cdot\frac{2}{50}\cdot\frac{1}{49}$
$=\frac{1}{13\times17\times25\times49}$
$=\frac{1}{270725}$
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Question 1623 Marks
There are three coins. One is a two headed coin $($having head on both faces$),$ another is a biased coin that comes up heads $75\%$ of the time and third is an unbiased coin. One of the three coins is chosen at random and tossed, it shows heads, what is the probability that it was the two headed coin?
Answer
Let $E_{1 }= a$ two headed coin, $E_{2 }= a$ biased coin, $E_{3 }= an$ unbiased coin and $A = A $ head is shown
$\text{Now}\ \ \text{P}(\text{E}_1)=\frac{1}{3},\ \text{P}(\text{E}_2)=\frac{1}{3},\ \text{P}(\text{E}_3)=\frac{1}{3}$
$\text{P}(\text{A}|\text{E}_1)=1,\ \text{P}(\text{A}|\text{E}_2)=\frac{75}{100}=\frac{3}{4}\ \text{and}\ \text{P}(\text{A}|{\text{E}_3})=\frac{1}{2}$
Therefore, by Bayes’ theorem,
$\text{P}(\text{E}_1|\text{A})=\frac{\text{P}(\text{E}_1)\text{P}(\text{A}|\text{E}_1)}{\text{P}(\text{E}_1)\text{P}(\text{A}|\text{E}_1)+\text{P}(\text{E}_2)\text{P}(\text{A}|\text{E}_2)+\text{P}(\text{E}_3)\text{P}(\text{A}|\text{E}_3)}$
$=\frac{\frac{1}{3}\times1}{\frac{1}{3}\times1+\frac{1}{3}\times\frac{3}{4}+\frac{1}{3}\times\frac{1}{2}}$
$=\frac{{4}}{{1}+\frac{3}{4}+\frac{1}{2}}$
$=\frac{4}{4+3+2}$
$=\frac{4}{9}$
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Question 1633 Marks
There are 3 red and 5 black balls in bag 'A'; and 2 red and 3 black balls in bag 'B'. One ball is drawn from bag 'A' and two from bag 'B'. Find the probability that out of the 3 balls drawn one is red and 2 are black.
Answer
It si givem that bag A contains 3 red and 5 balck balls (3R, 5B) and bag B contains 2 red and 3 black balls (2R, 3B).
Now,
P(One red and 2 black) = P(one red from bag A and two black from bag B) + P(black ball from bag A and remaining balls from bag B)
$=\frac{3}{8}\times\frac{3}{5}\times\frac{2}{4}+\frac{5}{8}\times\frac{2}{5}\times\frac{3}{4}\times2$
$=\frac{9}{80}+\frac{30}{80}$
$=\frac{39}{80}$
Note: 2 is multiplied by second term because there are two ways to select red and black balls from bag B.
While the first way is to pick black ball first, followed by red, the second way is to pick red ball first, followed by black.
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Question 1643 Marks
A is known to speak truth $3$ times out of $5$ times. He throws a die and reports that it is one. Find the probability that it is actually one.
Answer
Let $A \ E_1$ and $E_2$ denote the events that the man reports the appearance of $1$ on throwing a die, $1$ occurs and $1$ does not occur, respectively.
$\therefore\ \text{P}(\text{E}_1)=\frac{1}{6}$
$\text{P}(\text{E}_2)=\frac{5}{6}$
Now,
$\text{P}\Big(\frac{\text{A}}{\text{E}_1}\Big)=\frac{3}{5}$
$\text{P}\Big(\frac{\text{A}}{\text{E}_2}\Big)=\frac{3}{5}$
Using Baye's theorem, we get
Required probability $\text{P}\Big(\frac{\text{E}_1}{\text{A}}\Big)=\frac{\text{P}(\text{E}_1)\text{P}\Big(\frac{\text{A}}{\text{E}_1}\Big)}{\text{P}(\text{E}_1)\text{P}\Big(\frac{\text{A}}{\text{E}_1}\Big)+\text{P}(\text{E}_2)\text{P}\Big(\frac{\text{A}}{\text{E}_2}\Big)}$
$=\frac{\frac{1}{6}\times\frac{3}{5}}{\frac{1}{6}\times\frac{3}{5}+\frac{5}{6}\times\frac{2}{5}}$
$=\frac{3}{3+10}$
$=\frac{3}{13}$
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Question 1653 Marks
A die is tossed twice. A 'success' is getting an even number on a toss. Find the variance of number of successes.
Answer
Let p be the probability of getting an even number on the toss when a dice is thrown.
Let q be the probability of not getting an even number on the toss when a dice is thrown.
Then $\text{p}=\frac{3}{2}=\frac{1}{2}$ and $\text{q}=\frac{1}{2}$
Clearly, X follows binomial distribution with $\text{n}=2,\text{p}=\frac{1}{2}.$
$\therefore\text{Variance = npq}=2\times\frac{1}{2}\times\frac{1}{2}=\frac{1}{2}$
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Question 1663 Marks
Three machines $E_1, E_2, E_3$ in a certain factory produce $50\%, 25\%$ and $25\%,$ respectively, of the total daily output of electric bulbs. It is known that $4\%$ of the tubes produced one each of the machines $E_{1 }$ and $E_2$ are defective, and that $5\%$ of those produced on $E_3$ are defective. If one tube is picked up at random from a day's production, then calculate the probability that it is defective.
Answer
Let $D$ be the event that the picked up tube is defective.
Let $A_1, A_2$ and $A_3$ be the events that the tube is produced on macjines $E_1, E_2$ and $E_3$ respectively.
$P(D) = P(A_1) P(D|A_1) + P(A_2) P(D|A_2) + P(A_3) P(D|A_3) .....(i)$
$\text{P}(\text{A}_1)=\frac{50}{100}=\frac{1}{2},\text{P}(\text{A}_2)=\frac{25}{100}=\frac{1}{4},\text{P}(\text{A}_3)=\frac{25}{100}=\frac{1}{4}$
$\text{P}(\text{D}|\text{A}_1)=\text{P}(\text{D}|\text{A}_2)=\frac{4}{100}=\frac{1}{25}$
$\text{P}(\text{D}|\text{A}_3)=\frac{5}{100}=\frac{1}{20}$
Putting these values in $(i),$ we get
$\text{P(D)}=\frac{1}{2}\times\frac{1}{25}+\frac{1}{4}\times\frac{1}{25}+\frac{1}{4}\times\frac{1}{20}$
$\text{P(D)}=\frac{17}{400}$
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Question 1673 Marks
For a loaded die, the probabilities of outcome are given as under:
$P(1) = P(2) = 0.2, P(3) = P(5) = P(6) = 0.1$ and $P(4) = 0.3.$ the die is thrown two times. If the die were fair, determine whether or not the events $A$ and $B$ are independent.
Answer
We have $A = A = \{(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)\}$
$\therefore n(A) = 6$ and $n(S) = 6^2 = 36$
$\therefore\text{P}(\text{A})=\frac{\text{n}(\text{A})}{\text{n}(\text{S})}$
$=\frac{6}{36}=\frac{1}{6}$
and $B = \{(4, 6), (6, 4), (5, 5), (6, 5), (5, 6), (6, 6)\}$
$\Rightarrow n(B) = 6$
$\therefore\text{P}(\text{B})=\frac{\text{n}(\text{B})}{\text{n}(\text{S})}$
$=\frac{6}{36}=\frac{1}{6}$
Also, $\text{A}\cap\text{B}=\left\{(5,5),(6,6)\right\}$
 $\therefore\text{P}(\text{A}\cap\text{B})=\frac{2}{36}=\frac{1}{8}$
Also, $\text{P}(\text{A})\cdot\text{P}(\text{B})=\frac{1}{36}$
Thus, $\text{P}(\text{A}\cap\text{B})\neq\text{P}(\text{A})\cdot\text{P}(\text{B})$
So, $A$ and $B$ are not independent events.
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Question 1683 Marks
The items produced by a company contain 10% defective items. Show that the probability of getting 2 defective items in a sample of 8 items is $\frac{28\times9^6}{10^8}$.
Answer
Let X denote the number of defective items in the items produced by the company.
Then, X follows binomial distribution with n = 8.
$\text{p}=10\%=\frac{1}{10}$
$\text{q}=1-\text{p}=\frac{9}{10}$
Hence, the distribution is given by
$\text{P(X = r)}=\text{ }^8\text{C}_{\text{r}}\big(\frac{1}{10}\big)^{\text{r}}\big(\frac{9}{10}\big)^{8-\text{r}}$
Prob of getting 2 defective items $=\text{P}(\text{X}=2)$
$=\text{ }^8\text{C}_2\big(\frac{1}{10}\big)^2\big(\frac{9}{10}\big)^{8-2}$
$=\frac{28\text{ x }9^6}{10^8}$
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Question 1693 Marks
A box has $5$ blue and $4$ red balls. One ball is drawn at random and not replaced. Its colour is also not noted. Then another ball is drawn at random. What is the probability of second ball being blue?
Answer
A box $5$ blue and $4$ red balls.
Let $E_1$ is the event that first ball drawn is blue, $E_2$ is the event that first ball.
drawn is red and $E$ is the event that second ball drawn is blue.
$\therefore\text{P}(\text{E})=\text{P}(\text{E}_1)\cdot\text{P}\Big(\frac{\text{E}}{\text{E}_1}\Big)+\text{P}(\text{E}_2)\cdot\text{P}\Big(\frac{\text{E}}{\text{E}_2}\Big)$
$=\frac{5}{9}\cdot\frac{4}{8}+\frac{4}{9}\cdot\frac{5}{8}$
$=\frac{20}{72}+\frac{20}{72}$
$=\frac{40}{72}$
$=\frac{5}{9}$
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Question 1703 Marks
The probability is 0.02 that an item produced by a factory is defective. A shipment of 10,000 items is sent to its warehouse. Find the expected number of defective items and the standard deviation.
Answer
Let p denote the probability of a defective item produced in the factory, so
$\text{p}=0.02$
$=\frac{2}{100}$
$\text{p}=\frac{1}{50}$
$\text{q}=1-\frac{1}{50}$ [Since p + q = 1]
$=\frac{49}{50}$
Given $\text{n}=10,000$
Expected number of defective item $=\text{np}$
$=10000\times\frac{1}{50}$
$=200$
$\text{Standard deviation}=\sqrt{\text{npq}}$
$=\sqrt{10000\times\frac{1}{50}\times\frac{49}{50}}$
$=14$
$\text{Expected No. of defective items}=200$
$\text{Standard deviation}=14$
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Question 1713 Marks
A bag contains 5 white, 7 red and 3 black balls. If three balls are drawn one by one without replacement, find the probability that none is red.
Answer
Bag contains 5 white, 7 red and 3 black balls.
Total number of balls = 15
Three balls are drawn without replacement
A = first ball is red
B = Second ball is red
C = Third balls is red
P (Three balls are drawn, non is red)
$=\text{P}(\overline{\text{A}})\text{P}\Big(\overline{\frac{\text{B}}{\text{A}}}\Big)\text{P}\Big(\frac{\overline{\text{C}}}{\text{A}\cap\text{B}}\Big)$
$=\frac{8}{15}\times\frac{7}{14}\times\frac{6}{13}$
[Since, number of non red balls = 5 + 3 = 8]
$=\frac{8}{65}$
Required probability $=\frac{8}{65}$
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Question 1723 Marks
Ten eggs are drawn successively, with replacement, from a lot containing 10% defective eggs. Find the probability that there is at least one defective egg.
Answer
Let X be the number of defective eggs drawn from 10 eggs.
Then, X follows a binomial distribution with n = 10
Let p be the probability that a drawn egg is defective.
$\therefore\text{p}=10\%=\frac{1}{10},\text{q}=\frac{9}{10}$
Hence, the distribution is given by
$\text{P(X = r})=\text{ }^{10}\text{C}_{\text{r}}\big(\frac{1}{10}\big)^{\text{r}}\big(\frac{9}{10}\big)^{10-\text{r}},\text{r}=0,1,2\dots10$
$\text{P(there is at least one defectiv egg})=\text{P(X}\geq1)$
$=1-\text{P(X}=0)$
$=1-\text{ }^{10}\text{C}_0\big(\frac{1}{10}\big)^0\big(\frac{9}{10}\big)^{10-0}$
$=1-\big(\frac{9}{10}\big)^{10}$
$=1-\frac{9^{10}}{10^{10}}$
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Question 1733 Marks
Find the binomial distribution when the sum of its mean and variance for 5 trials is 4.8.
Answer
Number of trials in the binomial distribution = 5
If p is the probability for success, then
$\text{np + npq}=4.8$
Or $\Rightarrow5\text{p}+5\text{p}(1-\text{p})=4.8$
By factorising, we get
$\big(\text{p}-0.8\big)\big(\text{p}-1.2)=0$
As p cannot exceed 1,
$\text{p}=0.8$ or $\frac{4}{5}$
and $\text{q}=1-\text{p}=\frac{1}{5}$
$\therefore\text{P(X = r})=\text{ }^{5}\text{C}_{\text{r}}\big(\frac{4}{5}\big)^{\text{r}}\big(\frac{1}{5}\big)^{5-\text{r}},\text{r}=0,1,2,\dots5$
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Question 1743 Marks
If each element of a second order determinant is either zero or one, what is the probability that the value of the determinant is positive? $($Assume that the individual entries of the determinant are chosen independently, each value being assumed with probability $\frac{1}{2} ).$
Answer
There are four entries in determinant of $2 \times 2$ order.
Each entry may be filled up in two ways with $0$ or $1.$
$\therefore$ number of determinants that can be formed $= 2^4 = 16$
$\therefore$ total number of cases $= 16$
The value of determinant is positive in the cases
$ \begin{vmatrix} 1\ 0\\0\ 1 \end{vmatrix}, \begin{vmatrix} 1\ 0\\1\ 1 \end{vmatrix}, \begin{vmatrix} 1\ 1\\0\ 1 \end{vmatrix},$
$\therefore$ number of favourable cases $= 3$
$\therefore$ the probability that the determinant is positive $=\frac{3}{16}.$
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Question 1753 Marks
Find the probability distribution of the number of heads, when three coins are tossed.
Answer
Let X denote number of heads in three tosses of a coin. Then, X can take the values 0, 1, 2 and 3.
Now,
$\text{P}(\text{X=0})=\text{P}(\text{TTT})=\frac{1}{8},\text{P}(\text{X}=1)$ $=\text{P}(\text{HTT or TTH or THT})=\frac{3}{8}$
$\text{P}(\text{X=2})\text{P}(\text{HTH or THH or HHT})$ $\\=\frac{3}{8},\text{P}(\text{X}=3)=\text{P}(\text{HHH})=\frac{1}{8}$
Thus, the probability distribution of X is given by
$\text{X}$
$0$
$1$
$2$
$3$
$\text{P}(\text{X})$
$\frac{1}{8}$
$\frac{3}{8}$
$\frac{3}{8}$
$\frac{1}{8}$
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Question 1763 Marks
Two dice are thrown simultaneously. If X denotes the number of sixes, find the expectation of X.
Answer
Two dice thrown simultaneously is the same the die thrown 2 times.
Let S = {1, 2, 3, 4, 5, 6} ⇒ n(S) = 6
Let A denotes the number 6 ⇒ A = {6} ⇒ n(A) = 1
$\text{P}(\text{A})=\frac{\text{n}(\text{A})}{\text{n}(\text{S})}=\frac{1}{6}\ \text{and}\ \text{P}(\overline{\text{A}})=1-\frac{1}{6}=\frac{5}{6}$
n = 2, r = 0, 1, 2
$\text{P}(\text{X}=0)=\text{P}(\overline{\text{A}}).\text{P}(\overline{\text{A}})=\bigg(\frac{5}{6}\bigg)^2=\frac{25}{36}$
$\text{P}(\text{X}=1)=2\text{P}(\text{A}).\text{P}(\overline{\text{A}})=2\times\frac{1}{6}\times\frac{5}{6}=\frac{10}{36}$
$\text{P}(\text{X}=2)=\text{P}(\text{A}).\text{P}(\text{A})=\frac{1}{6}\times\frac{1}{6}=\frac{1}{36}$
$\text{E}(\text{X})=\sum\limits_{i=1}^{2}\text{x}_i\text{p}(\text{x}_i)=0\times\frac{25}{36}+1\times\frac{10}{36}+2\times\frac{1}{36}=\frac{12}{36}=\frac{1}{3}$
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Question 1773 Marks
Suppose that $5\%$ of men and $0.25\%$ of women have grey hair. A grey haired person is selected at random. What is the probability of this person being male? Assume that there are equal number of males and females.
Answer
Let $E_1, E_2, E$ be the events
$E_1 : $'selected person is a male'
$E_2 :$ 'selected person is a female',
$E :$ 'selected person is grey haired'
$\therefore\ \text{P}(\text{E}_1)=\text{P}(\text{E}_2)=\frac{1}{2}.$
$\text{and}\ \ \text{P}(\text{E}|\text{E}_1)=\frac{5}{100}=\frac{1}{20},\ \text{P}(\text{E}|\text{E}_2)=\frac{0.25}{100}=\frac{1}{400}.$
Required probaility = $\text{P}(\text{E}_1|\text{E})$
$=\frac{\text{P}(\text{E}_1)\text{P}(\text{E}|\text{E}_1)}{\text{P}(\text{E}_1)\text{P}(\text{E}|\text{E}_1)+\text{P}(\text{E}_2)\text{P}(\text{E}|\text{E}_2)}$
$=\frac{\frac{1}{2}\times\frac{1}{20}}{\frac{1}{2}\times\frac{1}{20}+\frac{1}{2}\times\frac{1}{400}}=\frac{\frac{1}{20}}{\frac{1}{20}+\frac{1}{400}}=\frac{\frac{1}{20}}{\frac{20+1}{400}}=\frac{\frac{1}{20}}{\frac{21}{400}}$
$=\frac{1}{2}\times\frac{400}{21}=\frac{20}{21}.$
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Question 1783 Marks
Three cards are drawn with replacement from a well shuffled pack of 52 cards. Find the probability that the cards are a king, a queen and a jack.
Answer
Three cards are drawn with replacement fron a pack of cards
There are 4 Kings, 4 Queens, 5 Jacks.
P (1 King, 1 Queen, 1 Jack)
$=\text{P}\big[(\text{K}\cap\text{Q}\cap\text{J})\cup(\text{K}\cap\text{J}\cap\text{Q})\cup(\text{J}\cap\text{K}\cap\text{Q})\\\cup(\text{J}\cap\text{Q}\cap\text{K})\cup(\text{Q}\cap\text{K}\cap\text{L})\cup(\text{Q}\cap\text{J}\cap\text{K})\big]$
$=\text{P}(\text{K}\cap\text{Q}\cap\text{J})+\text{P}(\text{K}\cap\text{J}\cap\text{Q})+\text{P}(\text{J}\cap\text{K}\cap\text{Q})\\+\text{P}(\text{J}\cap\text{Q}\cap\text{K})+\text{P}(\text{Q}\cap\text{K}\cap\text{L})+\text{P}(\text{Q}\cap\text{J}\cap\text{K})$
$=\text{P}(\text{K}) \text{P}(\text{Q}) \text{P}(\text{J})+\text{P}(\text{K}) \text{P}(\text{J}) \text{P}(\text{Q})+\text{P}(\text{J}) \text{P}(\text{K}) \text{P}(\text{Q}) \\ +\text{P}(\text{J}) \text{P}(\text{Q}) \text{P}(\text{K})+\text{P}(\text{Q}) \text{P}(\text{K}) \text{P}(\text{L})+\text{P}(\text{Q}) \text{P}(\text{J}) (\text{K})$
$=\frac{4}{52}\times\frac{4}{52}\times\frac{4}{52}+\frac{4}{52}\times\frac{4}{52}\times\frac{4}{52}+\frac{4}{52}\times\frac{4}{52}\times\frac{4}{52} \\ +\frac{4}{52}\times\frac{4}{52}\times\frac{4}{52}+\frac{4}{52}\times\frac{4}{52}\times\frac{4}{52}+\frac{4}{52}\times\frac{4}{52}\times\frac{4}{52}$
$=\frac{6}{13\times13\times13}$
$=\frac{6}{2197}$
Required probability $=\frac{6}{2197}$
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Question 1793 Marks
A coin is tossed 5 times. What is the probability that tail appears an odd number of times?
Answer
Let X denote the number of tails when a coin is tossed 5 times.
X follows a binomial distribution with $\text{n}=5;\text{p}=\frac{1}{2};\text{q}=1-\text{p}=\frac{1}{2}$
Then $\text{P}(\text{X = r})=\text{ }^5\text{C}_{\text{r}}\big(\frac{1}{2}\big)^{\text{r}}\big(\frac{1}{2}\big)^{\text{n}-\text{r}}=\text{ }^5\text{C}_{\text{r}}\big(\frac{1}{2}\big)^5$
The required probability $=\text{P}(\text{X = odd})$
$=\text{P}(\text{X}=1)+\text{P}(\text{X}=3)+\text{P}(\text{X}=5)$
$=\text{ }^5\text{C}_1\big(\frac{1}{2}\big)^5+\text{ }^5\text{C}_3\big(\frac{1}{2}\big)^5+\text{ }^5\text{C}_5\big(\frac{1}{2}\big)^5$
$=\big(\frac{1}{2}\big)^5\ [5+10+1]$
$=\frac{16}{32}$
$=\frac{1}{2}$
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Question 1803 Marks
Find the probability of getting 5 exactly twice in 7 throws of a die.
Answer
The repeated tossing of a die are Bernoulli trials. Let X represent the number of times of getting 5 in 7 throws of the die.
Probability of getting 5 in a single throw of the die, $\text{p}=\frac{1}{6}$
$\therefore\ \text{q}=1-\text{p}=1-\frac{1}{6}=\frac{5}{6}$
Clearly, X has the probability distribution with n = 7 and $\text{p}=\frac{1}{6}$
$\therefore\ \text{P}(\text{X=x})=\ ^\text{n}\text{C}_\text{x}\text{q}^\text{n-x}\text{p}^\text{x}=\ ^7\text{C}_\text{x}\bigg (\frac{5}{6}\bigg)^{7-\text{x}}.\bigg (\frac{1}{6}\bigg)^\text{x}$
P(getting 5 exactly twice) = P(X = 2)
$=\ ^7\text{C}_2\bigg (\frac{5}{6}\bigg)^5.\bigg (\frac{1}{6}\bigg)^{2}$
$=21\cdot\Big(\frac{5}{6}\Big)^5\cdot\frac{1}{36}$
$=\Big(\frac{7}{12}\Big)\Big(\frac{5}{6}\Big)^5$
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Question 1813 Marks
A bag contains 3 white, 4 red and 5 black balls. Two balls are drawn one after the other, without replacement. What is the probability that one is white and the other is black?
Answer
Bag contain 3 white, 4 red, 5 black balls.
Two balls are drawn without replacemenet,
P (One ball is white and other black)
$=\text{P}\big[(\text{W}\cap\text{B})\cup(\text{B}\cap\text{W})\big]$
$=\text{P}\big[(\text{W}\cap\text{B})+\text{P}(\text{B}\cap\text{W})\big]$
$=\text{P(W)}\text{ P}\Big(\frac{\text{B}}{\text{W}}\Big)+\text{P(B)}\text{ P}\Big(\frac{\text{W}}{\text{B}}\Big)$
$=\frac{3}{12}\times\frac{5}{12}+\frac{5}{12}\times\frac{3}{11}$
$=\frac{15}{132}+\frac{15}{132}$
$=\frac{30}{132}$
$=\frac{5}{22}$
Required probability $=\frac{5}{22}$
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Question 1823 Marks
The mean of a binomial distribution is 20 and the standard deviation 4. Calculate the parameters of the binomial distribution.
Answer
Given that mean, i.e. $\text{np}=20\dots(1)$
and standard deviation, i.e. $\text{npq}=4$
$\sqrt{\text{npq}}=4$
$\Rightarrow\text{npq}=16\dots(2)$
Dividing eq. (2) by eq. (1), we get
$\text{q}=\frac{16}{20}=\frac{4}{5}$
and $\text{p}=\frac{1}{5};$
$\therefore\text{n}=\frac{\text{Mean}}{\text{p}}=100$
$\text{P(X = r})=\text{ }^{100}\text{C}_{\text{r}}\big(\frac{1}5{})^{\text{r}}\big(\frac{4}{5}\big)^{100-\text{r}},\text{r}=0,1,2\dots100$
Therefore, the parameters are $\text{n}=100$ and $\text{p}=\frac{1}{5}$
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Question 1833 Marks
If A, B and C are independent events such that P(A) = P(B) = P(C) = p, then find the probability of occurrence of at least two of A, B and C.
Answer
P(At least two of A, B and C occur) = P(Exactly two of A, B and C occurs) + P(All three occurs)
$=\big[\text{P}(\text{A}\cap\text{B})-\text{P}(\text{A}\cap\text{B}\cap\text{C})\big]+\big[\text{P}(\text{B}\cap\text{C})-\text{P}(\text{A}\cap\text{B}\cap\text{C})\big]\\+\big[\text{P}(\text{A}\cap\text{C})-\text{P}(\text{A}\cap\text{B}\cap\text{C})\big]+\text{P}(\text{A}\cap\text{B}\cap\text{C})$
$=\text{P}(\text{A}\cap\text{B})+\text{P}(\text{B}\cap\text{C})+\text{P}(\text{A}\cap\text{C})-3\text{P}(\text{A}\cap\text{B}\cap\text{C})+\text{P}(\text{A}\cap\text{B}\cap\text{C})$
$=\text{P}(\text{A}\cap\text{B})+\text{P}(\text{B}\cap\text{C})+\text{P}(\text{A}\cap\text{C})-2\text{P}(\text{A}\cap\text{B}\cap\text{C})$
$=\text{P}(\text{A})+\text{P}(\text{B})+\text{P}(\text{B})+\text{P}(\text{C})+\text{P}(\text{A})+\text{P}(\text{C})-2\text{P}(\text{A})+\text{P}(\text{B})+\text{P}(\text{C})$
(As, A, B, and C are independe)
$=\text{P}\times\text{P}+\text{P}\times\text{P}+\text{P}\times\text{P}-2\text{P}\times\text{P}\times\text{P}$
$=\text{P}^2+\text{P}^2+\text{P}^2-2\text{P}^3$
$=3\text{P}^2-2\text{P}^3$
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3 Marks Question - Page 4 - MATHS STD 12 Science Questions - Vidyadip