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Maths Solve the Following Question.(3 Marks)

Probability · Maharashtra Board STD 12 Science (MSBSHSE - English Medium). 61 questions.

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Solve the Following Question.(3 Marks)

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61 questions · self-marked practice — reveal the answer and mark yourself.

Question 13 Marks
A dice is thrown twice and the sum of the numbers appearing is observed to be 6. What is the conditional probability that the number 4 has appeared at least once?
Answer
The sample space of the experiment is {(1, 1), (1, 2), (1, 3), ....., (6, 6)} conisting of 36 outcomes.
$\text{P(A)}=\text{P}(\text{Sum}=6)=\frac{5}{36}$
$\text{P(B)}=\text{P}(4\text{ appears at least once})=\frac{11}{36}$
Now, $\text{P}\Big(\frac{\text{B}}{\text{A}}\Big)=\frac{\text{P}(\text{A and B})}{\text{P(A)}}$
$=\frac{\text{P}(\text{Sum is } 6 \text{ and }4 \text{ has appeard at least once})}{\text{P(A)}}$
$=\frac{\frac{2}{36}}{\frac{5}{36}}$
$=\frac{2}{5}$
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Question 23 Marks
From a pack of 52 cards, 4 are drawn one by one without replacement. Find the probability that all are aces (or kings).
Answer
A = First card Ace
B = Second card Ace
C = Third card Ace
D = Fourth card Ace
P (All four drawn are Ace, without replacement)
$=\text{P}(\text{A})\ \text{P}\Big(\frac{\text{B}}{\text{A}}\Big)\ \text{P}\Big(\frac{\text{C}}{\text{A}\cap\text{B}}\Big)\ \text{P}\Big(\frac{\text{D}}{\text{A}\cap\text{B}\cap\text{C}}\Big)$
$=\frac{4}{52}\times\frac{3}{51}\times\frac{2}{50}\times\frac{1}{49}$ [Since, there are four Ace in 52 cards]
$=\frac{1}{270725}$
Required Probabilty $=\frac{1}{270725}$
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Question 33 Marks
Two coins are tossed once. Find $\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)$ in each of the following:
A = No tail appears,
B = No head appears.
Answer
Sample space of two coins
{HH, HT, TH, TT}
A = No tail appears
A = {HH}
B = No head appears
B = {TT}
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{\text{P}(\text{A}\cap\text{B})}{\text{P(B)}}$
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{0}{1}$
$=0$
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=0$
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Question 43 Marks
Find the chance of drawing 2 white balls in succession from a bag containing 5 red and 7 white balls, the ball first drawn not being replaced.
Answer
Bag contains 5 red and 7 white balls
A = First ball white
B = Second ball white
P (2 white balls drawn without replacement)
$=\text{P}(\text{A})\text{ P}\Big(\frac{\text{B}}{\text{A}}\Big)$
$=\frac{7}{12}\times\frac{6}{11}$
$=\frac{7}{22}$
Required probability $=\frac{7}{22}$
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Question 53 Marks
Ten cards numbered 1 through 10 are placed in a box, mixed up thoroughly and then one card is drawn randomly. If it is known that the number on the drawn card is more than 3, what is the probability that it is an even number?
Answer
Total no. of cards = 10
Let A = drawn number is more than 3
B = drawn number is even
$\text{P}\Big(\frac{\text{B}}{\text{A}}\Big)=\frac{\text{P}(\text{B}\cap\text{A})}{\text{P}(\text{A})}$
Now, $\text{P}(\text{A})=\frac{7}{10}$
$\text{P}(\text{A}\cap\text{B})=\frac{4}{10}$
$\therefore\ \text{P}\Big(\frac{\text{B}}{\text{A}}\Big)=\frac{4}{7}$
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Question 63 Marks
Two coins are tossed once. Find $\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)$ in each of the following:
A = Tail appears on one coin,
B = One coin shows head.
Answer
Sample space of two coins
{HH, HT, TH, TT}
A = Tail appears on one coin
A = {HT, TH}
B = One coin shows head
B = {HT, TH}
$(\text{A}\cap\text{B})=\{{\text{HT}, \text{TH}}\}$
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{\text{P}(\text{A}\cap\text{B})}{\text{P}(\text{B})}$
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{2}{2}$
Hence, $\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=1$
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Question 73 Marks
A die is rolled. If the outcome is an odd number, what is the probability that it is prime?
Answer
A die is rolled.
A = A prime number on die
A = {2, 3, 5}
B = An odd number on die
B = {1, 3, 5}
$(\text{A}\cap\text{B})=\{3,5\}$
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{\text{n}(\text{A}\cap\text{B})}{\text{n}(\text{B})}$
$=\frac{2}{3}$
Required probability $=\frac{2}{3}$
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Question 83 Marks
A speaks the truth 8 times out of 10 times. A die is tossed. He reports that it was 5. What is the probability that it was actually 5?
Answer
Let A denote the event that man reports that 5 occurs and E the event that 5 actually turns up.
$\therefore\text{P(E)}=\frac{1}{6}$ and $\text{P}(\overline{\text{E}})=1-\frac{1}{6}=\frac{5}{6}$
Also, $\text{P}\Big(\frac{\text{A}}{\text{E}}\Big)=$ Probability that man reports that 5 occurs given thet 5 actually turns up = Probability mab speaking the truth $=\frac{8}{10}=\frac{4}{5}$
$\text{P}\Big(\frac{\text{A}}{\overline{\text{E}}}\Big)=$ Probability thet man reports that 5 occurs given that 5 does not turns up = Probability not speaking the truth $=1-\frac{4}{5}=\frac{1}{5}$
$\therefore$ Required probability $\text{P}\Big(\frac{\text{E}}{\text{A}}\Big)=\frac{\text{P}(\text{E})\text{P}\Big(\frac{\text{A}}{\text{E}}\Big)}{\text{P}(\text{E})\text{P}\Big(\frac{\text{A}}{\text{E}}\Big)+\text{P}(\overline{\text{E}})\text{P}\Big(\frac{\text{A}}{\overline{\text{E}}}\Big)}$
$=\frac{\frac{1}{6}\times\frac{4}{5}}{\frac{1}{6}\times\frac{4}{5}+\frac{5}{6}\times\frac{1}{5}}=\frac{4}{9}$
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Question 93 Marks
A speaks truth in 75% and B in 80% of the cases. In what percentage of cases are they likely to contradict each other in narrating the same incident?
Answer
Given
A speaks truth in 75% cases.
B speaks truth in 80% cases.
$\text{P(A)}=\frac{75}{100}\Rightarrow\text{P}(\overline{\text{A}})=\frac{25}{100}$
$\text{P(B)}=\frac{80}{100}\Rightarrow\text{P}(\overline{\text{B}})=\frac{20}{100}$
P(A and B contradict each other)
$=\text{P}\big[(\text{A}\cap\overline{\text{B}})\cup(\overline{\text{A}}\cap\text{B})\big]$
$=\text{P}(\text{A}\cap\overline{\text{B}}) +\text{P}(\overline{\text{A}}\cap\text{B})$
$=\text{P(A)}\text{ P}(\overline{\text{B}})+\text{P}(\overline{\text{A}})\text{ P(B)}$
$=\frac{75}{100}\times\frac{20}{100}+\frac{25}{100}\times\frac{80}{100}$
$=\frac{1500}{1000}+\frac{2000}{10000}$
$=\frac{3500}{10000}$
$=35\%$
Required probability $=35\%$
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Question 103 Marks
A coin is tossed three times. Find $\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)$ in each of the following:
A = At most two tails,
B = At least one tail.
Answer
Sample space for three coins is given by
{HHH, HTH, THH, TTH, HHT, HTT, THT, TTT}
A = At most two tails
A = {HHH, HTH, THT, TTH, HHT, THT, HTT}
B = At least one tail
B = {HTH, THH, TTH, HHT, HTT, THT, TTT}
$(\text{A}\cap\text{B})=\{\text{HTH, THT, TTH, HHT, THT, HTT}\}$
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{\text{P}(\text{A}\cap\text{B})}{\text{P}(\text{B})}$
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{6}{7}$
Hence, $\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{6}{7}$
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Question 113 Marks
Let A and B be two independent events such that $P(A) = p_1$ and $P(B) = p_2$. Describe in words the events whose probabilities are:
$p_1 + p_2 = 2p_1p_2.$
Answer
As $p_1 + p_2 - 2p_1p_2 = (p_1 - p_1p_2) + (p_2 - p_1p_2)$
$= [P(A) - P(A) × P(B)] + [P(B) - P(A) × P(B)]$
And, A and B are indepepndet events.
i.e., $\text{P(A)}\times\text{P(B)}=\text{P}(\text{A}\cap\text{B})$
$\Rightarrow \ \text{p}_1+\text{p}_2-2\text{p}_1\text{p}_2=\big[\text{P(A)}-\text{P}(\text{A}\cap\text{B})\big]+\big[\text{P(B)}-\text{P}(\text{A}\cap\text{B})\big]\\=\text{P(only A)}+\text{P(only B)}$
So, P(only A) + P(only B) $= p_1 + p_2 - 2p_1p_2$
Hence, $p_1 + p_2 - 2p_1p_2 = P$(Exactly one of A and B occurs).
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Question 123 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 133 Marks
A laboratory blood test is $99\%$ effective in detecting a certain disease when its infection is present. However, the test also yields a false positive result for $0.5\%$ of the healthy person tested (i.e. if a healthy person is tested, then, with probability $0.005,$ the test will imply he has the disease). If $0.1\%$ of the population actually has the disease, what is the probability that a person has the disease given that his test result is positive?
Answer
Let $E_1 $ and $E_2 $ denote the events that a person has a disease and a person has no disease, respectively.
$E_1 $ and $E_2 $ are complimentary to each other.
$\therefore P(E_1) + P(E_2) = 1$
$\Rightarrow P(E_2) = 1 - P(E_1) = 1 - 0.001 = 0999$
let A denote the event that the blood test result is positive.
$\therefore P(E_1) = 0.1\% = 0.001$
Now,
$\text{P}\Big(\frac{\text{A}}{\text{E}_1}\Big)=90\%=0.99$
$\text{P}\Big(\frac{\text{A}}{\text{E}_2}\big)=0.5\%=0.005$
Using Baye's theorem, we get
Required probability $\text{P}\Big(\frac{\text{E}_1}{\text{A}}\Big)=\frac{\text{P}(\text{E}_1)\text{P}\Big(\frac{\text{A}}{\text{E}_1}\Big)}{\text{P}(\text{E}_1)\text{P}\Big(\frac{\text{A}}{\text{E}_1}\Big)+\text{P}(\text{E}_2)\text{P}\Big(\frac{\text{A}}{\text{E}_2}\Big)}$
$=\frac{0.001\times0.99}{0.001\times0.99+0.999\times0.005}$
$=\frac{990}{5985}=\frac{22}{133}$
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Question 143 Marks
A box of oranges is inspected by examining three randomly selected oranges drawn without replacement. If all the three oranges are good, the box is approved for sale otherwise it is rejected. Find the probability that a box containing 15 oranges out of which 12 are good and 3 are bad ones will be approved for sale.
Answer
There are 15 organes out of which 12 are good and 3 are bad.
Three oranges selected without replacement are drawn and if they found good the box is approved for salw.
A = First orange good
B = Second orange good
B = Third orange good
P (All three oranges are good)
$=\text{P(A)}\text{ P}\Big(\frac{\text{B}}{\text{A}}\Big)\text{ P}\Big(\frac{\text{C}}{\text{A}\cap\text{B}}\Big)$
$=\frac{12}{15}\times\frac{11}{14}\times\frac{10}{13}$
$=\frac{44}{91}$
Required probability $=\frac{44}{91}$
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Question 153 Marks
An urn contains 3 white, 4 red and 5 black balls. Two balls are drawn one by one without replacement. What is the probability that at least one ball is black?
Answer
Urn contains 3 white, 4 red and 5 black balls. Total balls = 12
Two balls are drawn without replacement
A = First ball is black
B = Second ball is black
P (Atleast one ball is black)
$=\text{P}(\overline{\text{A}}\cup\overline{\text{B}})$
$=1-\text{P}(\overline{\text{A}}\cup\overline{\text{B}})$
$=1-\text{P}(\overline{\text{A}})\text{P }\Big(\overline{\frac{\text{B}}{\text{A}}}\Big)$
$=1-\Big(\frac{7}{12}\times\frac{6}{12}\Big)$
$=1-\frac{7}{22}$
Required probability $=\frac{15}{22}$
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Question 163 Marks
Out of 100 students, two sections of 40 and 60 are formed. If you and your friend are among 100 students, what is the probability that:
You both enter the same section?
Answer
Out of 100 students two friends can enter sections in 100C2 ways.
Let
A = Event both enter in section A (40 students)
B = Event both enter in section B (60 students)
$\text{P(A)}=\frac{^{40}\text{C}_2}{^{100}\text{C}_2},\text{P(B)}=\frac{^{60}\text{C}_2}{^{100}\text{C}_2}$
$=\text{P}(\text{A}\cup\text{B})=\text{P(A)}+\text{P(B)}$
$=\frac{^{40}\text{C}_2+^{60}\text{C}_2}{^{100}\text{C}_2}$
$=\frac{\frac{40\times39}{2}+\frac{60+59}{2}}{\frac{100\times99}{2}}$
$=\frac{780+1770}{4950}$
$=\frac{2550}{4950}$
$=\frac{17}{33}$
P (Both enter same section) $=\frac{17}{33}$
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Question 173 Marks
A bag contains $6$ red and $8$ black balls and another bag contains $8$ red and $6$ black balls. A ball is drawn from the first bag and without noticing its colour is put in the second bag. A ball is drawn from the second bag. Find the probability that the ball drawn is red in colour.
Answer
Given,
Bag $(1)$ contains $6$ red $(R_1)$ and 8 black $(B_1)$ balls
Bag $(2)$ contains $8$ red $(R_2)$ and 6 black $(B_2)$ balls
$A$ ball is drawn from hte first bag and without noticing is colour is pur in the bag $(2)$. Then a ball is drawn from second bag and it is red.
$P($One red ball from bag $2)$
$=\text{P}\big((\text{B}_1\cap\text{P}_2)\cup(\text{R}_1\cap\text{R}_2)\big)$
$=\text{P}(\text{B}_1\cap\text{P}_2)+\text{P}(\text{R}_1\cap\text{R}_2)$
$=\text{P}(\text{B}_1)\text{ P}\Big(\frac{\text{R}_2}{\text{B}_1}\Big)+\text{P}(\text{R}_1)\text{ P}\Big(\frac{\text{R}_2}{\text{R}_1}\Big)$
$=\frac{8}{14}\times\frac{8}{15}+\frac{6}{14}\times\frac{9}{15}$
$=\frac{64+54}{210}$
$=\frac{118}{210}$
$=\frac{59}{105}$
Required probability $=\frac{59}{105}$
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Question 183 Marks
Let $A$ and $B$ be two independent events such that $P(A)=p_1$ and $P(B)=p_2$. Describe in words the events whose probabilities are:
$1-\left(1-p_1\right)\left(1-p_2\right)$.
Answer
As, $1 − (1 − p_1)(1 − p_2) = 1 - [1 - P(A)] \times [1 - P(B)]$
$=1-\text{P}(\overline{\text{A}})\times\text{P}(\overline{\text{B}})$
And, A and B are independent events.
i.e., $\text{P}(\overline{\text{A}})\times\text{P}(\overline{\text{B}})=\text{P}(\overline{\text{A}}\cap\overline{\text{B}})$
$\Rightarrow\ 1-(1-\text{p}_1)(1-\text{p}_2)=1-\text{P}(\overline{\text{A}}\cap\overline{\text{B}})\\=1-\text{P}(\overline{\text{A}}\cap\overline{\text{B}})=\text{P}(\text{A}\cup\text{B})$
So, $\text{P}(\text{A}\cup\text{B})=1-(1-\text{p}_1)(1-\text{p}_2)$
Hence, $1 - (1 - p_1) (1 - p_2) = P$ (At least one of A and B occurs).
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Question 193 Marks
Of the students in a college, it is known that $60 \%$ reside in a hostel and $40 \%$ do not reside in hostel. Previous year results report that $30\%$ of students residing in hostel attain A grade and $20\%$ of ones not residing in hostel attain A grade in their annual examination. At the end of the year, one students is chosen at random from the college and he has an A grade. What is the probability that the selected student is a hosteler?
Answer
Let A, $E_1$ and $E_2$ denote the events that the selected student attains grade A, resedes in a hostel and does not reside in a hostel, respectively.
$\therefore\ \text{P}(\text{E}_1)=\frac{60}{100}$
$\text{P}(\text{E}_1)=\frac{40}{100}$
Now,
$\text{P}\Big(\frac{\text{A}}{\text{E}_1}\Big)=\frac{30}{100}$
$\text{P}\Big(\frac{\text{A}}{\text{E}_2}\Big)=\frac{20}{100}$
Using Baye's theorem,
Required probability $\text{P}\Big(\frac{\text{E}_1}{\text{A}}\Big)=\frac{\text{P}(\text{E}_1)\text{P}\Big(\frac{\text{A}}{\text{E}_1}\Big)}{\text{P}(\text{E}_1)\text{P}\Big(\frac{\text{A}}{\text{E}_1}\Big)+\text{P}(\text{E}_2)\text{P}\Big(\frac{\text{A}}{\text{E}_2}\Big)}$
$=\frac{\frac{60}{100}\times\frac{30}{100}}{\frac{60}{100}\times\frac{30}{100}+\frac{40}{100}\times\frac{20}{100}}$
$=\frac{18}{18+8}=\frac{9}{13}$
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Question 203 Marks
Assume that the chances of a patient having heart attack is $40 \%$. It is also assumed that meditation and yoga course reduces 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 and patient selected at random suffers a heart attack. Find the probability that the patient followed a course of meditation and yoga?
Answer
Let A, $E_1$, and $E_2$ respectively denote the events that a person has heart attack, the selected person followed the course of yoga and meditation, and the person adopted the drug prescription.
$\therefore\ \text{P(A)}=0.40$
$\text{P}(\text{E}_1)=\text{P}(\text{E}_2)=\frac{1}{2}$
$\text{P}(\text{A}|\text{E}_1)=04.\times0.70=0.28$
$\text{P}(\text{A}|\text{E}_2)=0.40\times0.75=0.30$
Probability that the patient suffering a heart attack followe a course of meditation and yoga is given by $P(E_1|A)$.
$\text{P}(\text{E}_1|\text{A})=\frac{\text{P}(\text{E}_1)\times\text{P}(\text{A}|\text{E}_1)}{\text{P}(\text{E}_1)\times\text{P}(\text{A}|\text{E}_1)+\text{P}(\text{E}_2)\times\text{P}(\text{A}|\text{E}_2)}$
$=\frac{\frac{1}{2}\times0.28}{\frac{1}{2}\times0.28+\frac{1}{2}\times0.30}$
$=\frac{14}{29}$
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Question 213 Marks
From a pack of 52 cards, two are drawn one by one without replacement. Find the probability that both of them are kings.
Answer
A = first card is king
B = Second card is also king
Probability of getting two kings (Without replacement)
$=\text{P(A) }\text{P}\Big(\frac{\text{B}}{\text{A}}\Big)$
$=\frac{4}{52}\times\frac{3}{51}$ [Since, 4 kings out of 52 cards.]
$=\frac{1}{13}\times\frac{1}{17}$
$=\frac{1}{221}$
Required probability $=\frac{1}{221}$
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Question 223 Marks
In answering a question on a multiple choice test a student either knows the answer or guesses. Let $\frac{3}{4}$ be the probability that he knows the answer and $\frac{1}{4}$ be the probability that he guesses. Assuming that a student who guesses at the answer will be correct with probability $\frac{1}{4}$. What is the probability that a student knows the answer given that he answered it correctly?
Answer
Let A, $E_1$ and $E_2$ denote the events that the answer is correct, the student knows the answer and the student guesses the answer, respectively.
$\therefore\ \text{P}(\text{E}_1)=\frac{3}{4}$
$\text{P}(\text{E}_2)=\frac{1}{4}$
Now,
$\text{P}\Big(\frac{\text{A}}{\text{E}_1}\Big)=1$
$\text{P}\Big(\frac{\text{A}}{\text{E}_2}\Big)=\frac{1}{4}$
Using Baye's theorem, we get
Required probability $\text{P}\Big(\frac{\text{E}_1}{\text{A}}\Big)=\frac{\text{P}(\text{E}_1)\text{P}\Big(\frac{\text{A}}{\text{E}_1}\Big)}{\text{P}(\text{E}_1)\text{P}\Big(\frac{\text{A}}{\text{E}_1}\Big)+\text{P}(\text{E}_2)\text{P}\Big(\frac{\text{A}}{\text{E}_2}\Big)}$
$=\frac{\frac{3}{4}\times1}{\frac{3}{4}\times1+\frac{1}{4}\times\frac{1}{4}}$
$=\frac{3}{3+\frac{1}{4}}=\frac{12}{13}$
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Question 233 Marks
A bag contains 8 red and 6 green balls. Three balls are drawn one after another without replacement. Find the probability that at least two balls drawn are green.
Answer
A bag contains 8 red and 6 green balls.
Three balls are drawn without replacement
P (At least 2 balls are green)
$=\text{P}\big[(\text{G}_1\cap\text{G}_2\cap\text{R}_1)\cup(\text{G}_1\cap\text{R}_1\cap\text{G}_2)\\\cup(\text{R}_1\cap\text{G}_1\cap\text{G}_2)\cup(\text{G}_1\cap\text{G}_2\cap\text{G}_3)\big]$
$=\text{P}\big(\text{G}_1\cap\text{G}_2\cap\text{R}_1)+\text{ P}(\text{G}_1\cap\text{R}_1\cap\text{G}_2)\\+\text{ P}(\text{R}_1\cap\text{G}_1\cap\text{G}_2)+\text{ P}(\text{G}_1\cap\text{G}_2\cap\text{G}_3)$
$=\text{P}\big(\text{G}_1)\text{ P}\Big(\frac{\text{G}_2}{\text{G}_1}\Big)\text{P}\Big(\frac{\text{R}_1}{\text{G}_1\cap\text{G}_2}\Big)\\+\text{P}(\text{G}_1)\text{ P}\Big(\frac{\text{G}_2}{\text{G}_1}\Big)\text{ P}\Big(\frac{\text{G}_3}{\text{G}_1\cap\text{G}_2}\Big)$
$=\frac{6}{14}\times\frac{5}{13}\times\frac{8}{12}+\frac{6}{14}\times\frac{8}{13}\times\frac{5}{12}\\+\frac{8}{14}\times\frac{6}{13}\times\frac{5}{12}+\frac{6}{14}\times\frac{5}{13}\times\frac{4}{12}$
$=\frac{1}{14}\times\frac{1}{13}\times\frac{1}{12}\times(240+240+240+120)$
$=\frac{840}{14\times13\times12}$
$=\frac{5}{13}$
Required probability $=\frac{5}{13}$
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Question 243 Marks
Ten cards numbered 1 through 10 are placed in a box, mixed up thoroughly and then one card is drawn randomly. If it is known that the number on the drawn card is more than 3, what is the probability that it is an even number?
Answer
Sample space, S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
Consider the given events.
A = An even number on the card
B = A number more than 3 on the card
Clearly,
A = {2, 4, 6, 8, 10}
B = {4, 5, 6, 7, 8, 9, 10}
Now,
$\text{A}\cap\text{B}=\{4, 6, 8, 10\}$
$\therefore \text{Required probability} =\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{\text{n}(\text{A}\cap\text{B})}{\text{n}(\text{B})}=\frac{4}{7}$
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Question 253 Marks
If P(A) = 0.4, P(B) = 0.3 and $\text{P}\Big(\frac{\text{B}}{\text{A}}\Big)=0.5$ find $\text{P}(\text{A}\cap\text{B})$ and $\text{P}\Big(\frac{\text{A}}{\text{B}}\Big).$
Answer
Given,
P(A) = 0.4, P(B) = 0.3 and $\text{P}\Big(\frac{\text{B}}{\text{A}}\Big)=0.5$
We know that,
$\text{P}\Big(\frac{\text{B}}{\text{A}}\Big)=\frac{\text{P}(\text{A}\cap\text{B})}{\text{P(A)}}$
$=0.5=\frac{\text{P}(\text{A}\cap\text{B})}{0.4}$
$\text{P}(\text{A}\cap\text{B})=0.2$
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{\text{P}(\text{A}\cap\text{B})}{\text{P(B)}}$
$=\frac{0.2}{0.3}$
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{2}{3}$
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Question 263 Marks
A, B, and C are independent witness of an event which is known to have occurred. A speaks the truth three times out of four, B four times out of five and C five times out of six. What is the probability that the occurrence will be reported truthfully by majority of three witnesses?
Answer
P(A speaks truth) $=\frac{3}{4}$
P(B speaks truth) $=\frac{4}{5}$
P(C speaks truth) $=\frac{5}{6}$
P(majority speaks truth) = P(two speaks truth) + P(all speak truth)
= P(A) × P(B)[1 - P(C)] + P(A) × P(C)[1 - P(B)] + P(C) × P(B)[1 - P(A)] + P(A) × P(B) × P(C)
$=\frac{3}{4}\times\frac{4}{5}\Big(1-\frac{5}{6}\Big)+\frac{3}{4}\times\frac{5}{6}\Big(1-\frac{4}{5}\Big) \\ +\frac{4}{5}\times\frac{5}{6}\Big(1-\frac{3}{4}\Big)+\frac{3}{4}\times\frac{4}{5}\times\frac{5}{6}$
$=\frac{12}{120}+\frac{15}{120}+\frac{20}{120}+\frac{60}{120}$
$=\frac{107}{120}$
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Question 273 Marks
A pair of dice is thrown. Find the probability of getting 7 as the sum if it is known that the second die always exhibits a prime number.
Answer
Consider the given events
A = A prime number appears on second die.
B = The sum of the numbers on two dice is 7.
Clearly,
A = {(1, 2), (1, 3), (1, 5),
(2, 2), (2, 3), (2, 5),
(3, 2), (3, 3), (3, 5),
(4, 2), (4, 3), (4, 5),
(5, 2), (5, 3), (5, 5),
(6, 2), (6, 3), (6, 5)}
B = {(2, 5), (5, 2), (3, 4), (4, 3), (1, 6), (6, 1)}
Now,
$\text{A}\cap\text{B}=\{(2, 5), (5, 2), (4, 3)\}$
$\therefore\ \text{Required probability}=\text{P}\Big(\frac{\text{B}}{\text{A}}\Big)=\frac{\text{n}(\text{A}\cap\text{B})}{\text{n}(\text{A})}=\frac{3}{18}=\frac{1}{6}$
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Question 283 Marks
A pair of dice is thrown. Find the probability of getting 7 as the sum, if it is known that the second die always exhibits an odd number.
Answer
Here two dice are thrown
A = Getting 7 as sum on two dice
A {(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)}
B = Second die exhibits an odd number
B = {(1,1), (2, 1), (3, 1), (4, 1), (5, 1), (6, 1)
(1, 3), (2, 3), (3, 3), (4, 3), (5, 3), (6, 3)
(1, 5), (2, 5), (3, 5), (4, 5), (5, 5), (5, 5)}
$(\text{A}\cap\text{B})=\{(2, 5), (4, 3), (6, 1)\}$
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{\text{n}(\text{A}\cap\text{B})}{\text{n}(\text{B})}$
$=\frac{3}{18}$
$=\frac{1}{6}$
Hence, Required probability $=\frac{1}{6}$
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Question 293 Marks
X is taking up subjects - Mathematics, Physics and Chemistry in the examination. His probabilities of getting grade A in these subjects are 0.2, 0.3 and 0.5 respectively. Find the probability that he gets,
Grade A in two subject.
Answer
Given,
Probability of getting A grade in mathematics (m) = 0.2
⇒ P(m) = 0.2
Probability of getting A grade in physics (p) = 0.3
⇒ P(p) = 0.3
Probability of getting A grade in chemistry (P) = 0.5
⇒ P(C) = 0.5
P(Getting A grade in two subjects)
$=\text{P}\big((\text{m}\cap\text{p}\cap\overline{\text{c}})\cup(\text{m}\cap\overline{\text{p}}\cap\text{c})\cup(\overline{\text{m}}\cap\text{p}\cap\text{c})\big)$
$=\text{P}(\text{m})\text{P}(\text{p})\text{P}(\overline{\text{c}})+\text{P}(\text{m})\text{P}(\overline{\text{p}})\text{P}(\text{c})+\text{P}(\overline{\text{m}})\text{P}(\text{p})\text{P}(\text{c})$
$=\text{P}(\text{m})\text{P}(\text{p})(1-\text{P(c)})+\text{P(m)}(1-\text{P(p)})\text{p(c)}+(1-\text{p(m)})\text{P(p)}\text{P(c)}$
$=(0.2)(0.3)(1-0.5)+(0.2)(1-0.3)(0.5)+(1-0.2)(0.3)(0.5)$
$=(0.2)(0.3)(0.5) + (0.2)(0.7)(0.5)+(0.8)(0.3)(0.5)$
$=0.03+0.07+0.12$
$=0.22$
Required probability = 0.22
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Question 303 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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Question 313 Marks
If A and B are two independent events such that $\text{P}(\text{A}\cap\text{B})=0.60$ and P(A) = 0.2, find P(B).
Answer
$\text{P}(\text{A}\cup\text{B})=\text{P(A)}+\text{P(B)}-\text{P}(\text{A}\cap\text{B})$
$\text{P}(\text{A}\cup\text{B})=\text{P(A)}+\text{P(B)}-\text{P}(\text{A})\times(\text{B})$
[$\because$ A and B are independent events]
$\Rightarrow\ 0.6=0.2+\text{P(B)}=0.2\times\text{P(B)}$
$\Rightarrow\ 0.6-0.2=\text{P(B)}(1-0.2)$
$\Rightarrow\ \text{P(B)}=\frac{0.6-0.2}{1-0.2}$
$\Rightarrow\ \text{P(B)}=\frac{0.4}{0.8}$
$\Rightarrow\ \text{P(B)}=\frac{1}{2}$
$\Rightarrow\ \text{P(B)}=0.5$
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Question 323 Marks
From a deck of cards, three cards are drawn on by one without replacement. Find the probability that each time it is a card of spade.
Answer
Consider the events,
A = An ace in the first draw
B = An ace in the second draw
C = Getting an ace in the third draw
Now,
$\text{P(A)}=\frac{13}{52}=\frac{1}{4}$
$\text{P}\Big(\frac{\text{B}}{\text{A}}\Big)=\frac{12}{51}=\frac{4}{17}$
$\text{P}\Big(\frac{\text{C}}{\text{A}\cap\text{B}}\Big)=\frac{11}{50}$
$\therefore\ \text{Required probability} = \text{P}(\text{A}\cap\text{B}\cap\text{C})$
$=\text{P(A)}\times\text{P}\Big(\frac{\text{B}}{\text{A}}\Big)\times\text{P}\Big(\frac{\text{C}}{\text{A}\cap\text{B}}\Big)$
$=\frac{1}{4}\times\frac{4}{17}\times\frac{11}{50}$
$=\frac{11}{850}$
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Question 333 Marks
A can hit a target 3 times in 6 shots, B : 2 times in 6 shots and C : 4 times in 4 shots. They fix a volley. What is the probability that at least 2 shots hit?
Answer
P (A hits the target) $=\frac{3}{6}$
P (B hits the target) $=\frac{2}{6}$
P (C hits the target) $=\frac{3}{6}=1$
P (At least 2 shots hit) = P(Exactly 2 shots hit) + P(all 3 shots hit)
$=\frac{3}{6}\Big(1-\frac{2}{6}\Big)+\frac{2}{6}\Big(1-\frac{3}{6}\Big)+\frac{3}{6}\times\frac{2}{6}\times1$
(Here, the probability of C hitting the target is 1. So, it will always hit. When exaxtly 2 shots are hit, then either A hits or B hits.)
$=\frac{3}{6}\times\frac{4}{6}+\frac{2}{6}\times\frac{3}{6}+\frac{6}{36}$
$=\frac{12+6+6}{36}$
$=\frac{24}{36}$
$=\frac{2}{3}$
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Question 343 Marks
In a family, the husband tells a lie in 30% cases and the wife in 35% cases. Find the probability that both contradict each other on the same fact.
Answer
Given,
In a family Husband (H) tells a lie in 30% cases and wife (W) tells a lie in 35%
$\text{P(H)}= 30\%, \text{P}(\overline{\text{H}})=70\%$
$\text{P(W)}= 35\%, \text{P}(\overline{\text{W}})=65\%$
P(Both contradict each other)
$=\text{P}\big[(\text{H}\cap\overline{\text{W}})\cup(\overline{\text{H}}\cap\text{W})\big]$
$=\text{P}(\text{H}\cap\overline{\text{W}})+\text{P}(\overline{\text{H}}\cap\text{W})$
$=\text{P(H)}\text{ P}(\overline{\text{W}})+\text{P}(\overline{\text{H}})\text{ P(W)}$
$=\frac{30}{100}\times\frac{65}{100}\times\frac{70}{100}\times\frac{35}{100}$
$=\frac{1950+2450}{10000}$
$=\frac{4400}{10000}$
$=0.44$
Required probability = 0.44
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Question 353 Marks
Two dice are thrown and it is known that the first die shows a 6. Find the probability that the sum of the numbers showing on two dice is 7.
Answer
Two dice are thrown
A = Sun of the numbers showing on the dice is 7
A = {(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)}
B = First dies shows a 6
= {(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)
$(\text{A}\cap\text{B})=\{(6, 1)\}$
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{\text{n}(\text{A}\cap\text{B})}{\text{n}(\text{B})}$
$=\frac{1}{6}$
Required probability $=\frac{1}{6}$
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Question 363 Marks
If a machine is correctly set up it produces $90\%$ acceptable items. If it is incorrectly set up it produces only $40%$ acceptable item. Past experience shows that $80\%$ of the setups are correctly done. If after a certain set up, the machine produces $2$ acceptable items, find the probability that the machine is correctly set up.
Answer
Let A be the event that the machine produces two acceptable items.
Also, let $E_1$ represent the event that the machine is correctly set up and $E_2$ represent the event that the machine is incorrectly set up
$\therefore\ \text{P}(\text{E}_1)=0.8$
$\text{P}(\text{E}_2)=0.2$
Now,
$\text{P}\Big(\frac{\text{A}}{\text{E}_1}\Big)=0.9\times0.9=0.81$
$\text{P}\Big(\frac{\text{A}}{\text{E}_2}\Big)=0.40\times0.40=0.16$
Using Bayes, theorem, we get
Required probability $=\text{P}\Big(\frac{\text{E}_1}{\text{A}}\Big)=\frac{\text{P}(\text{E}_1)\text{P}\Big(\frac{\text{A}}{\text{E}_1}\Big)}{\text{P}(\text{E}_1)\text{P}\Big(\frac{\text{A}}{\text{E}_1}\Big)+\text{P}(\text{E}_2)\text{P}\Big(\frac{\text{A}}{\text{E}_2}\Big)}$
$=\frac{0.8\times0.81}{0.8\times0.81+0.2\times0.61}$
$=\frac{81}{85}$
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Question 373 Marks
Tickets are numbered from 1 to 10. Two tickets are drawn one after the other at random. Find the probability that the number on one of the tickets is a multiple of 5 and on the other a multiple of 4.
Answer
Tickers are numbered from 1 to 10.
Two tickets are drawn.
Consider, A = Multiple of 5
B = Multiple of 4
$\text{P(A)}=\frac{2}{10}$
[Since 5, 10 are multiple of 5]
$\text{P(A)}=\frac{1}{5}$
$\text{P(B)}=\frac{2}{10}$
$\text{P(B)}=\frac{1}{5}$
[Since 4, 8 are multiple of 4]
P (One number multiple of 5 and other multiple of 4)
$=\text{P}\big[(\text{A}\cap\text{B})\cup(\text{B}\cap\text{A)}\big]$
$=\text{P}(\text{A}\cap\text{B})+\text{P}(\text{B}\cap\text{A)}$
$=\text{P(A) }\text{P}\Big(\frac{\text{B}}{\text{A}}\Big)+\text{P(B) P}\Big(\frac{\text{A}}{\text{B}}\Big)$
$=\frac{1}{5}\times\frac{2}{9}+\frac{1}{5}\times\frac{2}{9}$
$=\frac{4}{45}$
Required probability $=\frac{4}{45}$
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Question 383 Marks
In a school there are 1000 students, out of which 430 are girls. It is known that out of 430, 10% of the girls study in class XII. What is the probability that a student chosen randomly studies in class XII given that the chosen student is a girl?
Answer
Suppose S represents a student chosen randomlu studying in class XII and G represents a female student chosen randomly.
We have.
$\text{P}(\text{G})=\frac{430}{1000}$
$\text{P}\Big(\frac{\text{S}}{\text{G}}\Big)=\frac{43}{1000}$
Now,
$\text{P}\Big(\frac{\text{S}}{\text{G}}\Big)=\frac{\frac{43}{1000}}{\frac{430}{1000}}=\frac{1}{10}$
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Question 393 Marks
A card is drawn from a well-shuffled deck of 52 cards and then a second card is drawn. Find the probability that the first card is a heart and the second card is a diamond if the first card is not replaced.
Answer
Consider the given events.
A = A heart in the first draw
B = A diamond in the second draw
Now,
$\text{P(A)}=\frac{13}{52}=\frac{1}{4}$
$\text{P}\Big(\frac{\text{B}}{\text{A}}\Big)=\frac{13}{51}$
$\therefore\text{Required probability}=\text{P}(\text{A}\cap\text{B})=\text{P(A)}\times\text{P}\Big(\frac{\text{B}}{\text{A}}\Big)$
$=\frac{1}{4}\times\frac{13}{51}=\frac{13}{204}$
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Question 403 Marks
An urn contains $10$ white and $3$ black balls. Another urn contains $3$ white and $5$ black balls. Two are drawn from first urn and put into the second urn and then a ball is drawn from the latter. Find the probability that its is a white ball.
Answer
A white ball can be drawn in three mutually exclusive ways:
  1. By transferring two black balls from first to second urn, then drawing a white ball.
  2. By transferring two white balls from first to second urn, then drawing a white ball.
  3. By transferring a white and a black ball from first to second urn, then drawing a white ball.
Let $E_1, E_2, E_3$ and A be the events as defined below:
$E_1$ = Two black balls are transferred from first to second bag
$E_2$ = Two white balls are transferred from first to second bag
$E_2$ = A white and a black ball is transferred from first to second bag
A = A white ball is drawn
$\therefore\ \text{P}(\text{E}_1)=\frac{^{3}\text{C}_2}{^{13}\text{C}_2}=\frac{3}{78}$
$\text{P}(\text{E}_2)=\frac{^{10}\text{C}_2}{^{13}\text{C}_2}=\frac{45}{78}$
$\text{P}(\text{E}_3)=\frac{^{10}\text{C}_2\times^{3}\text{C}_1}{^{13}\text{C}_2}=\frac{30}{78}$
Now,
$\text{P}\Big(\frac{\text{A}}{\text{E}_1}\Big)=\frac{3}{10}$
$\text{P}\Big(\frac{\text{A}}{\text{E}_2}\Big)=\frac{5}{10}$
$\text{P}\Big(\frac{\text{A}}{\text{E}_3}\Big)=\frac{4}{10}$
Using the law of total probability, we get
Required probability $=\text{P(A)}=\text{P}(\text{E}_1)\text{ P}\Big(\frac{\text{A}}{\text{E}_1}\Big)+\text{P}(\text{E}_2)\text{ P}\Big(\frac{\text{A}}{\text{E}_2}\Big)+\text{P}(\text{E}_3)\text{ P}\Big(\frac{\text{A}}{\text{E}_3}\Big)$
$=\frac{3}{78}\times\frac{3}{10}+\frac{45}{78}\times\frac{5}{10}+\frac{30}{78}\times\frac{4}{10}$
$=\frac{9}{780}+\frac{225}{780}+\frac{120}{780}$
$=\frac{354}{780}=\frac{59}{130}$
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Question 413 Marks
A bag contains $4$ white, $7$ black and $5$ red balls. $4$ balls are drawn with replacement. What is the probability that at least two are white?
Answer
Number of white balls $= 4$
Number of black balls $= 7$
Number of red balls $= 5$
Total balls $= 16$
Number of ways in which $4$ balls can be drawn from $16$ balls $=\ ^{16}C_4$
Let $A =$ getting at least two white ball = Getting $2, 3, 4$ white balls
Number of ways of choosing $2$ white balls $=\ ^4C_2 \times\ ^{12}C_2$
Number of ways of chossing $3$ white balls $=\ ^4C_3 \times\ ^{12}C_1$
Number of ways of choosing $4$ white balls $=\ ^4C_4 \times\ ^{12}C_0$
$\therefore\ \text{P(A)}=\frac{^{4}\text{C}_2\times ^{12}\text{C}_2 + ^{4}\text{C}_3\times ^{12}\text{C}_1+^{4}\text{C}_1\times ^{12}\text{C}_0}{^{16}\text{C}_4}=\frac{67}{256}$
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Question 423 Marks
The probability of student A passing an examination is $\frac{2}{9}$ and of student B passing is $\frac{5}{9}$. Assuming the two events: 'A passes', 'B passes' as independent, find the probability of:
Only one of them passing the examination.
Answer
Given,
The probability of A passing exam $=\frac{2}{9}$
The probability of B passing exam $=\frac{5}{9}$
$\Rightarrow\ \text{P(A)}=\frac{2}{9},\text{P(B)}=\frac{5}{9}$
P (Only one of them passing exam)
$=\text{P}\big((\text{A}\cap\overline{\text{B}})\cup(\overline{\text{A}}\cap\text{B})\big)$
$=\text{P}(\text{A}\cap\overline{\text{B}})+\text{P}(\overline{\text{A}}\cap\text{B})$
$=\text{P}(\text{A})\text{ P}(\overline{\text{B}})+\text{P}(\overline{\text{A}})\text{ P}(\text{B})$
$=\text{P(A)}(1-\text{P(B)})+(1-\text{P(A)})\text{P(B)}$
$=\frac{2}{9}\Big(1-\frac{5}{9}\Big)+\Big(1-\frac{2}{9}\Big)\frac{5}{9}$
$=\frac{2}{9}\times\frac{4}{9}+\frac{7}{9}\times\frac{5}{9}$
$=\frac{8}{81}+\frac{35}{81}$
$=\frac{43}{81}$
Required probability $=\frac{43}{81}$
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Question 433 Marks
A factory has two machines A and B. Past records show that the machine A produced $60 \%$ of the items of output and machine B produced $40 \%$ of the items. Further $2 \%$ of the items produced by machine A were defective and $1 \%$ produced by machine B were defective. If an item is drawn at random, what is the probability that it is defective?
Answer
Let $E_1, E_2$ A be defined as,
$E_1$ = Item produced by machine A
$E_2$ = Item produced by machine B
A = The item drawn is defective
$\text{P}(\text{E}_1)=60\%$
$=\frac{60}{100}$
$\text{P}(\text{E}_2)=40\%$
$=\frac{40}{100}$
$\text{P}(\text{A}|\text{E}_1)=\text{P}$ [Defective item from machine A]
$=2\%$
$=\frac{2}{100}$
$\text{P}\Big(\frac{\text{A}}{\text{E}_2}\Big)=\text{P}$ [Defective item from machine B]
$=1\%$
$=\frac{1}{100}$
BY law of total probability,
$\text{P(A)}=\text{P}(\text{E}_1)\text{ P}\Big({\text{A}}|{\text{E}_1}\Big)+\text{P}(\text{E}_2)\text{ P}\Big(\frac{\text{A}}{\text{E}_2}\Big)$
$=\frac{60}{100}\times\frac{2}{100}+\frac{40}{100}\times\frac{1}{100}$
$=\frac{120+40}{10000}$
$=\frac{160}{10000}$
$=0.016$
Required probability = 0.016
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Question 443 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 453 Marks
A die is tossed twice. Find the probability of getting a number greater than 3 on each toss.
Answer
Given, A and B are independent events and $\text{P}(\text{A}\cap\text{B})=0.60,\text{P(A)}=0.2$
A and B are independent events,
So, $\text{P}(\text{A}\cap\text{B})=\text{P(A) }\text{P(B)}$
We know that,
$\text{P}(\text{A}\cup\text{B})=\text{P(A)}+\text{P(A)}-\text{P}(\text{A}\cap\text{B})$
$0.6=0.2+\text{P(B)}-\text{P(B) }\text{P(B)}$
$0.6-0.2=\text{P(B)}-0.2\text{P(B)}$
$0.4=0.8\text{P(B)}$
$\text{P(B)}=\frac{0.4}{0.8}$
$\text{P(B)}=0.5$
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Question 463 Marks
A die is thrown twice and the sum of the numbers appearing is observed to be 8. What is the conditional probability that the number 5 has appeared at least once?
Answer
Consider the given events.
A = 5 appears on the die at least once
B = The sum of the numbers on two dice is 8.
Clearly,
A = {(1, 5), (2, 5), (3, 5), (4, 5), (5, 5), (6, 5), (5, 1), (5, 2), (5, 3), (5, 4), (5, 6)}
B = {(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)}
Now,
$\text{A}\cap\text{B}=\{(3, 5), (5, 3)\}$
$\therefore\ \text{Required probability}=\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{\text{n}(\text{A}\cap\text{B})}{\text{n(B)}}=\frac{2}{5}$
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Question 473 Marks
Compute $\text{P}\Big(\frac{\text{A}}{\text{B}}\Big),$ if P(B) = 0.5 and $\text{P}(\text{A}\cap\text{B})=0.32$
Answer
Given,
$\text{P(B)}=0.5,\text{P}(\text{A}\cap\text{B})=0.32$
We know that,
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{\text{n}(\text{A}\cap\text{B})}{\text{n}(\text{B})}$
$=\frac{0.32}{0.5}$
$=\frac{32}{50}$
$=\frac{16}{25}$
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{16}{25}$
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Question 483 Marks
Two cards are drawn successively without replacement from a well-shuffled deck of 52 cards. Find the probability of exactly one ace.
Answer
Two cards are drawn without replacement.
There are total 4 ace.
A = Getting Ace
P (Exactly one ace out of 2 cards)
$=\text{P}\big((\text{A}\cap\overline{\text{A}})\cup(\overline{\text{A}}\cap\text{A})\big)$
$=\text{P(A)P}\Big(\frac{\overline{\text{A}}}{\text{A}}\big)=\text{P}(\overline{\text{A}})\text{P}\Big(\frac{\text{A}}{\overline{\text{A}}}\Big)$
$=\frac{4}{52}\times\frac{48}{51}+\frac{48}{52}\times\frac{4}{51}$
$=\frac{96}{663}$
$=\frac{32}{221}$
required probability $=\frac{32}{221}$
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Question 493 Marks
A coin is tossed three times. Find $\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)$ in each of the following:
A = Heads on third toss,
B = Heads on first two tosses.
Answer
Sample space for three coins is given by
{HHH, HTH, THH, TTH, HHT, HTT, THT, TTT}
A = Head on third toss
A = {HHH, HTH, THH, TTH}
B = Head on first two toss
B = {HHH, HHT}
$(\text{A}\cap\text{B})=\{\text{HHH}\}$
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{\text{n}(\text{A}\cap\text{B})}{\text{n}(\text{B})}$
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{1}{2}$
Hence, $\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{1}{2}$
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Question 503 Marks
A bag contains $1$ white and $6$ red balls, and a second bag contains $4$ white and $3$ red balls. One of the bags is picked up at random and a ball is randomly drawn from it, and is found to be white in colour. Find the probability that the drawn ball was from the first bag.
Answer
Let A, $E_1$ and $E_2$ denote the events that the ball is white, bag I is chosen and bag II is chosen, respectively.
$\therefore\text{P}(\text{E}_1)=\frac{1}{2}$
$\text{P}(\text{E}_2)=\frac{1}{2}$
Now,
$\text{P}\Big(\frac{\text{A}}{\text{E}_1}\Big)=\frac{1}{7}$
$\text{P}\Big(\frac{\text{A}}{\text{E}_2}\Big)=\frac{4}{7}$
Using Baye's therorem, we get
Required probability $=\text{P}\Big(\frac{\text{E}_1}{\text{A}}\Big)=\frac{\text{P}(\text{E}_1)\text{P}\Big(\frac{\text{A}}{\text{E}_1}\Big)}{\text{P}(\text{E}_1)\text{P}\Big(\frac{\text{A}}{\text{E}_1}\Big)+\text{P}(\text{E}_2)\text{P}\Big(\frac{\text{A}}{\text{E}_2}\Big)}$
$=\frac{\frac{1}{2}\times\frac{1}{7}}{\frac{1}{2}\times\frac{1}{7}+\frac{1}{2}\times\frac{4}{7}}$
$=\frac{1}{1+4}=\frac{1}{5}$
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Question 513 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 523 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 533 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 543 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 553 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 563 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 573 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 583 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 593 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 603 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 613 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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