Question

A bag A contains 2 white and 3 red balls, and another bag B contains 4 white and 5 red balls. One ball is drawn at random from one of the bags and it is found to be red. Find the probability that it was drawn from bag B.
$\begin{aligned} & \left[ a \right]\ P\left( {{E}_{2}}|E \right)=\dfrac{25}{52} \\\ & \left[ b \right]\ P\left( {{E}_{2}}|E \right)=\dfrac{24}{52} \\\ & \left[ c \right]\ P\left( {{E}_{2}}|E \right)=\dfrac{26}{52} \\\ & \left[ d \right]\ P\left( {{E}_{2}}|E \right)=\dfrac{23}{52} \\\ \end{aligned}$

Answer 1

Hint: Assume that E is the event of drawing out a red ball, ${{E}_{1}}$ the event of drawing a ball from bag A and ${{E}_{2}}$ the event of drawing out a ball from B. Use law of total probability, which states that if ${{E}_{1}},{{E}_{2}},\cdots ,{{E}_{n}}$ are mutually exclusive and exhaustive events then for any event A, $P\left( A \right)=\sum\limits_{r=1}^{n}{P\left( A|{{E}_{r}} \right)P\left( {{E}_{r}} \right)}$, to determine the probability of event E. Use the fact that $P\left( A\bigcap E \right)=P\left( E \right)P\left( A|E \right)=P\left( A \right)P\left( E|A \right)$. Hence determine $P\left( {{E}_{2}}|E \right)$.

Complete step by step solution:
Let E be the event of drawing a red ball
Let ${{E}_{1}}$ be the event of drawing out a ball from bag A
Let ${{E}_{2}}$ be the event of drawing out a ball from the bag B.
Hence we need to determine the probability of event ${{E}_{2}}$ given that E has occurred.
Now, we have $P\left( {{E}_{1}} \right)=\dfrac{1}{2}$ and $P\left( {{E}_{2}} \right)=\dfrac{1}{2}$(Since the bags are chosen at random)
Also, we have
$P\left( E|{{E}_{1}} \right)=\dfrac{3}{5}$(Since there are three red balls and two white balls in bag A) and $P\left( E|{{E}_{2}} \right)=\dfrac{5}{9}$(Since there are five red balls and 4 white balls in bag B )
Since ${{E}_{1}}$ and ${{E}_{2}}$ are mutually exclusive and exhaustive events, we get
$P\left( E \right)=P\left( {{E}_{1}} \right)P\left( E|{{E}_{1}} \right)+P\left( {{E}_{2}} \right)P\left( E|{{E}_{2}} \right)$(By law of total probability)
Hence, we have
$P\left( E \right)=\dfrac{1}{2}\times \dfrac{3}{5}+\dfrac{1}{2}\times \dfrac{5}{9}=\dfrac{26}{45}$
Also, we know that $P\left( A\bigcap B \right)=P\left( A \right)P\left( B|A \right)=P\left( B \right)P\left( A|B \right)$
Hence, we have
$P\left( {{E}_{2}}\bigcap E \right)=P\left( {{E}_{2}} \right)P\left( E|{{E}_{2}} \right)=P\left( E \right)P\left( {{E}_{2}}|E \right)$
Hence, we have
$\begin{aligned} & \dfrac{1}{2}\times \dfrac{5}{9}=\dfrac{26}{45}\times P\left( {{E}_{2}}|E \right) \\\ & \Rightarrow P\left( {{E}_{2}}|E \right)=\dfrac{5\times 45}{2\times 9\times 26}=\dfrac{25}{52} \\\ \end{aligned}$
Hence the probability that the ball is drawn from bag B given that it is red is $\dfrac{25}{52}$
Hence option [a] is correct.

Note: The above result can directly be achieved using Bayes theorem, which says that if ${{E}_{1}},{{E}_{2}},\cdots ,{{E}_{n}}$ describe a partition on sample space of a random experiment and E is any event, then $P\left( {{E}_{k}}|E \right)=\dfrac{P\left( {{E}_{k}} \right)P\left( E|{{E}_{k}} \right)}{\sum\limits_{r=1}^{n}{P\left( {{E}_{r}} \right)P\left( E|{{E}_{r}} \right)}}$
Substituting the values, we get
$P\left( {{E}_{2}}|E \right)=\dfrac{\dfrac{1}{2}\times \dfrac{5}{9}}{\dfrac{1}{2}\times \dfrac{3}{5}+\dfrac{1}{2}\times \dfrac{5}{9}}=\dfrac{\dfrac{5}{9}}{\dfrac{3}{5}+\dfrac{5}{9}}=\dfrac{25}{52}$, which is the same as obtained above.