Solution

The urn contains 5 red and 5 black balls.

Let a red ball be drawn in the first attempt.

$\therefore P($ drawing a red ball $)=\frac{5}{10}=\frac{1}{2}$

If two red balls are added to the urn, then the urn contains 7 red and 5 black balls.

$P(.$ drawing a red ball) $=\frac{7}{12}$

Let a black ball be drawn in the first attempt.

$\therefore P$ (drawing a black ball in the first attempt) $=\frac{5}{10}=\frac{1}{2}$

If two black balls are added to the urn, then the urn contains 5 red and 7 black balls.

$P($ drawing a red ball $)=\frac{5}{12}$

Therefore, probability of drawing second ball as red is

$\frac{1}{2} \times \frac{7}{12}+\frac{1}{2} \times \frac{5}{12}=\frac{1}{2}(\frac{7}{12}+\frac{5}{12})=\frac{1}{2} \times 1=\frac{1}{2}$

Solution

Let $E_1$ and $E_2$ be the events of selecting first bag and second bag respectively. $\therefore P(E_1)=P(E_2)=\frac{1}{2}$

Let $A$ be the event of getting a red ball.

$\Rightarrow P(A \mid E_1)=P($ drawing a red ball from first bag $)=\frac{4}{8}=\frac{1}{2}$

$\Rightarrow P(A \mid E_2)=P($ drawing a red ball from second bag $)=\frac{2}{8}=\frac{1}{4}$

The probability of drawing a ball from the first bag, given that it is red, is given by $P$ $(E_2 \mid A)$.

By using Bayes’ theorem, we obtain

$ \begin{aligned} P(E_1 \mid A) & =\frac{P(E_1) \cdot P(A \mid E_1)}{P(E_1) \cdot P(A \mid E_1)+P(E_2) \cdot P(A \mid E_2)} \\ & =\frac{\frac{1}{2} \cdot \frac{1}{2}}{\frac{1}{2} \cdot \frac{1}{2}+\frac{1}{2} \cdot \frac{1}{4}} \\ & =\frac{\frac{1}{4}}{\frac{1}{4}+\frac{1}{8}} \\ & =\frac{\frac{1}{3}}{\frac{3}{8}} \\ & =\frac{2}{3} \end{aligned} $

Solution

Let $E_1$ and $E_2$ be the events that the student is a hostler and a day scholar respectively and $A$ be the event that the chosen student gets grade $A$.

$\therefore P(E_1)=60 %=\frac{60}{100}=0.6$

$P(E_2)=40 %=\frac{40}{100}=0.4$

$P(A \mid E_1)=P$ (student getting an A grade is a hostler) $=30 %=0.3$

$P(A \mid E_2)=P($ student getting an A grade is a day scholar $)=20 %=0.2$

The probability that a randomly chosen student is a hostler, given that he has an A grade, is given by $P(E_1 \mid A)$.

By using Bayes’ theorem, we obtain

$ \begin{aligned} P(E_1 \mid A) & =\frac{P(E_1) \cdot P(A \mid E_1)}{P(E_1) \cdot P(A \mid E_1)+P(E_2) \cdot P(A \mid E_2)} \\ & =\frac{0.6 \times 0.3}{0.6 \times 0.3+0.4 \times 0.2} \\ & =\frac{0.18}{0.26} \\ & =\frac{18}{26} \\ & =\frac{9}{13} \end{aligned} $

Solution

Let $E_1$ and $E_2$ be the respective events that the student knows the answer and he guesses the answer.

Let $A$ be the event that the answer is correct.

$\therefore P(E_1)=\frac{3}{4}$

$P(E_2)=\frac{1}{4}$

The probability that the student answered correctly, given that he knows the answer, is 1.

$\therefore P(A \mid E_1)=1$

Probability that the student answered correctly, given that he guessed, is $\frac{1}{4}$.

$\therefore P(A \mid E_2)=\frac{1}{4}$

The probability that the student knows the answer, given that he answered it correctly, is given by $P(E_1 \mid A)$

By using Bayes’ theorem, we obtain

$ \begin{aligned} P(E_1 \mid A) & =\frac{P(E_1) \cdot P(A \mid E_1)}{P(E_1) \cdot P(A \mid E_1)+P(E_2) \cdot P(A \mid E_2)} \\ & =\frac{\frac{3}{4} \cdot 1}{\frac{3}{4} \cdot 1+\frac{1}{4} \cdot \frac{1}{4}} \\ & =\frac{\frac{3}{4}}{\frac{3}{4}+\frac{1}{16}} \\ & =\frac{\frac{3}{13}}{\frac{16}{16}} \\ & =\frac{12}{13} \end{aligned} $

Solution

Let $E_1$ and $E_2$ be the respective events that a person has a disease and a person has no disease.

Since $E_1$ and $E_2$ are events complimentary to each other,

$\therefore P(E_1)+P(E_2)=1$

$\Rightarrow P(E_2)=1-P(E_1)=1-0.001=0.999$

Let $A$ be the event that the blood test result is positive.

$P(E_1)=0.1 %=\frac{0.1}{100}=0.001$

$P(A \mid E_1)=P($ result is positive given the person has disease $)=99 %=0.99$

$P(A \mid E_2)=P($ result is positive given that the person has no disease $)=0.5 %=0.005$

Probability that a person has a disease, given that his test result is positive, is given by P $(E_1 \mid A)$.

By using Bayes’ theorem, we obtain

$ \begin{aligned} P(E_1 \mid A) & =\frac{P(E_1) \cdot P(A \mid E_1)}{P(E_1) \cdot P(A \mid E_1)+P(E_2) \cdot P(A \mid E_2)} \\ & =\frac{0.001 \times 0.99}{0.001 \times 0.99+0.999 \times 0.005} \\ & =\frac{0.00099}{0.00099+0.004995} \\ & =\frac{0.00099}{0.005985} \\ & =\frac{990}{5985} \\ & =\frac{110}{665} \\ & =\frac{22}{133} \end{aligned} $

Solution

Let $E_1, E_2$, and $E_3$ be the respective events of choosing a two headed coin, a biased coin, and an unbiased coin.

$\therefore P(E_1)=P(E_2)=P(E_3)=\frac{1}{3}$

Let $A$ be the event that the coin shows heads.

A two-headed coin will always show heads.

$\therefore P(A \mid E_1)=P($ coin showing heads, given that it is a two-headed coin $)=1$

Probability of heads coming up, given that it is a biased coin $=75 %$

$\therefore P(A \mid E_2)=P($ coin showing heads, given that it is a biased coin $)=\frac{75}{100}=\frac{3}{4}$

Since the third coin is unbiased, the probability that it shows heads is always $\frac{1}{2}$.

$\therefore P(A \mid E_3)=P($ coin showing heads, given that it is an unbiased coin $)=\frac{1}{2}$

The probability that the coin is two-headed, given that it shows heads, is given by $P(E_1 \mid A)$.

By using Bayes’ theorem, we obtain

$ \begin{aligned} P(E_1 \mid A) & =\frac{P(E_1) \cdot P(A \mid E_1)}{P(E_1) \cdot P(A \mid E_1)+P(E_2) \cdot P(A \mid E_2)+P(E_3) \cdot P(A \mid E_3)} \\ & =\frac{\frac{1}{3} \cdot 1}{\frac{1}{3} \cdot 1+\frac{1}{3} \cdot \frac{3}{4}+\frac{1}{3} \cdot \frac{1}{2}} \\ & =\frac{\frac{1}{3}}{\frac{1}{3}(1+\frac{3}{4}+\frac{1}{2})} \\ & =\frac{1}{\frac{9}{4}} \\ & =\frac{4}{9} \end{aligned} $

Solution

Let $E_1, E_2$, and $E_3$ be the respective events that the driver is a scooter driver, a car driver, and a truck driver.

Let $A$ be the event that the person meets with an accident.

There are 2000 scooter drivers, 4000 car drivers, and 6000 truck drivers.

Total number of drivers $=2000+4000+6000=12000$

$P(E_1)=P(.$ driver is a scooter driver) $=\frac{2000}{12000}=\frac{1}{6}$

$P(E_2)=P$ (driver is a car driver) $=\frac{4000}{12000}=\frac{1}{3}$

$P(E_3)=P(.$ driver is a truck driver) $=\frac{6000}{12000}=\frac{1}{2}$

$P(A \mid E_1)=P($ scooter driver met with an accident $)=0.01=\frac{1}{100}$

$P(A \mid E_2)=P($ car driver met with an accident $)=0.03=\frac{3}{100}$

$P(A \mid E_3)=P($ truck driver met with an accident $)=0.15=\frac{15}{100}$

The probability that the driver is a scooter driver, given that he met with an accident, is given by $P(E_1 \mid A)$.

By using Bayes’ theorem, we obtain

$ \begin{aligned} P(E_1 \mid A) & =\frac{P(E_1) \cdot P(A \mid E_1)}{P(E_1) \cdot P(A \mid E_1)+P(E_2) \cdot P(A \mid E_2)+P(E_3) \cdot P(A \mid E_3)} \\ & =\frac{\frac{1}{6} \cdot \frac{1}{100}}{\frac{1}{6} \cdot \frac{1}{100}+\frac{1}{3} \cdot \frac{3}{100}+\frac{1}{2} \cdot \frac{15}{100}} \\ & =\frac{\frac{1}{6} \cdot \frac{1}{100}}{\frac{1}{100}(\frac{1}{6}+1+\frac{15}{2})} \end{aligned} $

$ \begin{aligned} & =\frac{\frac{1}{6}}{\frac{104}{12}} \\ & =\frac{1}{6} \times \frac{12}{104} \\ & =\frac{1}{52} \end{aligned} $

Solution

Let $E_1$ and $E_2$ be the respective events of items produced by machines $A$ and $B$. Let $X$ be the event that the produced item was found to be defective.

$\therefore$ Probability of items produced by machine $A, P(E_1)=60 %=\frac{3}{5}$

Probability of items produced by machine $B, P(E_2)=40 %=\frac{2}{5}$

Probability that machine A produced defective items, $P(X \mid E_1)=2 %=\frac{2}{100}$

Probability that machine $B$ produced defective items, $P(X \mid E_2)=1 %=\frac{1}{100}$

The probability that the randomly selected item was from machine $B$, given that it is defective, is given by $P(E_2 \mid X)$.

By using Bayes’ theorem, we obtain

$ \begin{aligned} P(E_2 \mid X) & =\frac{P(E_2) \cdot P(X \mid E_2)}{P(E_1) \cdot P(X \mid E_1)+P(E_2) \cdot P(X \mid E_2)} \\ & =\frac{\frac{2}{5} \cdot \frac{1}{100}}{\frac{3}{5} \cdot \frac{2}{100}+\frac{2}{5} \cdot \frac{1}{100}} \\ & =\frac{\frac{2}{500}}{\frac{6}{500}+\frac{2}{500}} \\ & =\frac{2}{8} \\ & =\frac{1}{4} \end{aligned} $

Solution

Let $E_1$ and $E_2$ be the respective events that the first group and the second group win the competition. Let $A$ be the event of introducing a new product.

$P(E_1)=$ Probability that the first group wins the competition $=0.6$

$P(E_2)=$ Probability that the second group wins the competition $=0.4$

$P(A \mid E_1)=$ Probability of introducing a new product if the first group wins $=0.7$

$P(A \mid E_2)=$ Probability of introducing a new product if the second group wins $=0.3$

The probability that the new product is introduced by the second group is given by $P(E_2 \mid A)$.

By using Bayes’ theorem, we obtain

$ \begin{aligned} P(E_2 \mid A) & =\frac{P(E_2) \cdot P(A \mid E_2)}{P(E_1) \cdot P(A \mid E_1)+P(E_2) \cdot P(A \mid E_2)} \\ & =\frac{0.4 \times 0.3}{0.6 \times 0.7+0.4 \times 0.3} \\ & =\frac{0.12}{0.42+0.12} \\ & =\frac{0.12}{0.54} \\ & =\frac{12}{54} \\ & =\frac{2}{9} \end{aligned} $

Solution

Let $E_1$ be the event that the outcome on the die is 5 or 6 and $E_2$ be the event that the outcome on the die is $1,2,3$, or 4 .

$\therefore P(E_1)=\frac{2}{6}=\frac{1}{3}$ and $P(E_2)=\frac{4}{6}=\frac{2}{3}$

Let $A$ be the event of getting exactly one head.

$P(A \mid E_1)=$ Probability of getting exactly one head by tossing the coin three times if she gets 5 or $6=\frac{3}{8}$

$P(A \mid E_2)=$ Probability of getting exactly one head in a single throw of coin if she gets 1 ,

2,3, or $4=\frac{1}{2}$

The probability that the girl threw $1,2,3$, or 4 with the die, if she obtained exactly one head, is given by $P(E_2 \mid A)$.

By using Bayes’ theorem, we obtain

$ \begin{aligned} P(E_2 \mid A) & =\frac{P(E_2) \cdot P(A \mid E_2)}{P(E_1) \cdot P(A \mid E_1)+P(E_2) \cdot P(A \mid E_2)} \\ & =\frac{\frac{2}{3} \cdot \frac{1}{2}}{\frac{1}{3} \cdot \frac{3}{8}+\frac{2}{3} \cdot \frac{1}{2}} \\ & =\frac{\frac{1}{3}}{\frac{1}{3}(\frac{3}{8}+1)} \\ & =\frac{1}{\frac{11}{8}} \\ & =\frac{8}{11} \end{aligned} $

Solution

Let $E_1, E_2$ and $E_3$ be the respective events of the time consumed by machines $A, B$, and C for the job.

$ \begin{aligned} & P(E_1)=50 %=\frac{50}{100}=\frac{1}{2} \\ & P(E_2)=30 %=\frac{30}{100}=\frac{3}{10} \\ & P(E_3)=20 %=\frac{20}{100}=\frac{1}{5} \end{aligned} $

Let $X$ be the event of producing defective items.

$ \begin{aligned} & P(X \mid E_1)=1 %=\frac{1}{100} \\ & P(X \mid E_2)=5 %=\frac{5}{100} \\ & P(X \mid E_3)=7 %=\frac{7}{100} \end{aligned} $

The probability that the defective item was produced by $A$ is given by $P(E_1 \mid A)$.

By using Bayes’ theorem, we obtain

$ \begin{aligned} P(E_1 \mid X) & =\frac{P(E_1) \cdot P(X \mid E_1)}{P(E_1) \cdot P(X \mid E_1)+P(E_2) \cdot P(X \mid E_2)+P(E_3) \cdot P(X \mid E_3)} \\ & =\frac{\frac{1}{2} \cdot \frac{1}{100}}{\frac{1}{2} \cdot \frac{1}{100}+\frac{3}{10} \cdot \frac{5}{100}+\frac{1}{5} \cdot \frac{7}{100}} \\ & =\frac{\frac{1}{100} \cdot \frac{1}{2}}{\frac{1}{100}(\frac{1}{2}+\frac{3}{2}+\frac{7}{5})} \\ & =\frac{\frac{1}{17}}{\frac{5}{5}} \\ & =\frac{5}{34} \end{aligned} $

Solution

Let $E_1$ and $E_2$ be the respective events of choosing a diamond card and a card which is not diamond.

Let $A$ denote the lost card.

Out of 52 cards, 13 cards are diamond and 39 cards are not diamond. $\therefore P(E_1)=\frac{13}{52}=\frac{1}{4}$

$P(E_2)=\frac{39}{52}=\frac{3}{4}$

When one diamond card is lost, there are 12 diamond cards out of 51 cards.

Two cards can be drawn out of 12 diamond cards in ${ }^{12} C_2$ ways.

Similarly, 2 diamond cards can be drawn out of 51 cards in ${ }^{51} C_2$ ways. The probability of getting two cards, when one diamond card is lost, is given by $P(A \mid E_1)$.

$P(A \mid E_1)=\frac{{ }^{12} C_2}{{ }^{51} C_2}=\frac{12 !}{2 ! 10 !} \times \frac{21 \times 49 !}{51 !}=\frac{11 \times 12}{50 \times 51}=\frac{22}{425}$

When the lost card is not a diamond, there are 13 diamond cards out of 51 cards.

Two cards can be drawn out of 13 diamond cards in ${ }^{13} C_2$ ways whereas 2 cards can be drawn out of 51 cards in ${ }^{51} C_2$ ways.

The probability of getting two cards, when one card is lost which is not diamond, is given by $P(A \mid E_2)$.

$ P(A \mid E_2)=\frac{{ }^{13} C_2}{{ }^{51} C_2}=\frac{13 !}{2 ! \times 11 !} \times \frac{2 ! \times 49 !}{51 !}=\frac{12 \times 13}{50 \times 51}=\frac{26}{425} $

The probability that the lost card is diamond is given by $P(E_1 \mid A)$.

By using Bayes’ theorem, we obtain

$ \begin{aligned} P(E_1 \mid A) & =\frac{P(E_1) \cdot P(A \mid E_1)}{P(E_1) \cdot P(A \mid E_1)+P(E_2) \cdot P(A \mid E_2)} \\ & =\frac{\frac{1}{4} \cdot \frac{22}{425}}{\frac{1}{4} \cdot \frac{22}{425}+\frac{3}{4} \cdot \frac{26}{425}} \\ & =\frac{\frac{1}{425}(\frac{22}{4})}{\frac{1}{425}(\frac{22}{4}+\frac{26 \times 3}{4})} \end{aligned} $

$ \begin{matrix} =\frac{\frac{11}{2}}{25} \\ =\frac{11}{50} \end{matrix} $

Solution

Let $E_1$ and $E_2$ be the events such that

$E_1:$ A speaks truth

$E_2$ : A speaks false

Let $X$ be the event that a head appears.

$P(E_1)=\frac{4}{5}$

$\therefore P(E_2)=1-P(E_1)=1-\frac{4}{5}=\frac{1}{15}$

If a coin is tossed, then it may result in either head $(H)$ or tail $(T)$.

The probability of getting a head is $\frac{1}{2}$ whether A speaks truth or not.

$\therefore P(X \mid E_1)=P(X \mid E_2)=\frac{1}{2}$

The probability that there is actually a head is given by $P(E_1 \mid X)$.

$ \begin{aligned} P(E_1 \mid X) & =\frac{P(E_1) \cdot P(X \mid E_1)}{P(E_1) \cdot P(X \mid E_1)+P(E_2) \cdot P(X \mid E_2)} \\ & =\frac{\frac{4}{5} \cdot \frac{1}{2}}{\frac{4}{5} \cdot \frac{1}{2}+\frac{1}{5} \cdot \frac{1}{2}} \\ & =\frac{\frac{1}{2} \cdot \frac{4}{5}}{\frac{1}{2}(\frac{4}{5}+\frac{1}{5})} \\ & =\frac{4}{\frac{5}{1}} \\ & =\frac{4}{5} \end{aligned} $

Therefore, the correct answer is A.

Solution

If $A \subset B$, then $A \cap B=A$ $\Rightarrow P(A \cap B)=P(A)$

Also, $P(A)<P(B)$

Consider $P(A \mid B)=\frac{P(A \cap B)}{P(B)}=\frac{P(A)}{P(B)} \neq \frac{P(B)}{P(A)} \ldots$ (1)

Consider $P(A \mid B)=\frac{P(A \cap B)}{P(B)}=\frac{P(A)}{P(B)} \ldots$ (2)

It is known that, $P(B) \leq 1$

$\Rightarrow \frac{1}{P(B)} \geq 1$

$\Rightarrow \frac{P(A)}{P(B)} \geq P(A)$

From (2), we obtain

$\Rightarrow P(A \mid B) \geq P(A)$

$\therefore P(A \mid B)$ is not less than $P(A)$.

Thus, from (3), it can be concluded that the relation given in alternative $C$ is correct.