Question

There are three bags ${{B}_{1}}$, ${{B}_{2}}$ and ${{B}_{3}}$. The bag ${{B}_{1}}$ contains 5 red and 5 green balls. ${{B}_{2}}$ contains 3 red and 5 green balls and ${{B}_{3}}$ contains 5 red and 3 green balls. Bags ${{B}_{1}}$, ${{B}_{2}}$ and ${{B}_{3}}$ have probabilities \[\dfrac{3}{10}\], \[\dfrac{3}{10}\] and \[\dfrac{4}{10}\] respectively of being chosen. A bag is selected at random and a ball is chosen at random from the bag. Then which of the following options is/are correct?
(This question has multiple correct options)
(a) Probability that the chosen ball is green given that the selected bag is ${{B}_{3}}$ is $\dfrac{3}{8}$.
(b) Probability that the selected bag is ${{B}_{3}}$, given that the chosen ball is green is $\dfrac{5}{13}$.
(c) Probability that the chosen ball is green equals $\dfrac{39}{80}$.
(d) Probability that the selected bag is ${{B}_{3}}$, given that the chosen ball is green, equals $\dfrac{3}{10}$

Answer 1

Hint: To solve this question, we will require the concepts of Bayes’ Theorem and Total Probability. The probabilities of the three bags are given to us and we can find the probability of getting a green or a red ball from any of the bags as we are given the number of green and red balls in each bag. To understand the data better, we will tabulate the given data. It will give a clear insight. We will consider each option and find the probability of the event given in that option and find whether the option matches the probability we got or not.

Complete step-by-step answer:
It is given to us that there are three bags ${{B}_{1}}$, ${{B}_{2}}$ and ${{B}_{3}}$ and bag ${{B}_{1}}$ contains 5 red and 5 green balls. ${{B}_{2}}$ contains 3 red and 5 green balls and ${{B}_{3}}$ contains 5 red and 3 green balls. Bags ${{B}_{1}}$, ${{B}_{2}}$ and ${{B}_{3}}$ have probabilities \[\dfrac{3}{10}\], \[\dfrac{3}{10}\] and \[\dfrac{4}{10}\] respectively of being chosen.
We will tabulate this data.

	Probability	Red balls	Green balls	Total balls
${{B}_{1}}$	\[\dfrac{3}{10}\]	5	5	10
${{B}_{2}}$	\[\dfrac{3}{10}\]	3	5	8
${{B}_{3}}$	\[\dfrac{4}{10}\]	5	3	8

Now, we will consider the event given in option (a).
We need to find the probability that the chosen ball is green, given that the bag is ${{B}_{3}}$.
If it is given that the bag is ${{B}_{3}}$, this means the bag is already chosen. We only need to find the probability of green balls from all the balls in the bag ${{B}_{3}}$.
The number of green balls in the third bag is 3, whereas the total number of balls is 8.
$\Rightarrow P(G\ \text{from}\ {{B}_{3}})=\dfrac{3}{8}$
Hence, option (a) is correct.
Events in option (b) and (d) are the same.
We have to find the probability that the selected bag is ${{B}_{3}}$, given that the chosen ball is green.
To find the probability of this event, we need to use the Bayes’ Theorem.
According to Bayes’ Theorem $P\left( {{B}_{3}}\left| G \right. \right)=\dfrac{P\left( G|{{B}_{3}} \right)P\left( {{B}_{3}} \right)}{P\left( G|{{B}_{1}} \right)P\left( {{B}_{1}} \right)+P\left( G|{{B}_{2}} \right)P\left( {{B}_{2}} \right)+P\left( G|{{B}_{3}} \right)P\left( {{B}_{3}} \right)}$
The probability of green ball from ${{B}_{1}}$ will be $P\left( G|{{B}_{1}} \right)=\dfrac{5}{10}$.
The probability of green ball from ${{B}_{2}}$ will be $P\left( G|{{B}_{2}} \right)=\dfrac{5}{8}$.
The probability of green ball from ${{B}_{3}}$ will be $P\left( G|{{B}_{3}} \right)=\dfrac{3}{8}$.
Thus, we will substitute all the probabilities in the Bayes’ Theorem.
$\begin{align}
  & \Rightarrow P\left( {{B}_{3}}\left| G \right. \right)=\dfrac{\dfrac{3}{8}\times \dfrac{4}{10}}{\dfrac{5}{10}\times \dfrac{3}{10}+\dfrac{5}{8}\times \dfrac{3}{10}+\dfrac{3}{8}\times \dfrac{4}{10}} \\
 & \Rightarrow P\left( {{B}_{3}}\left| G \right. \right)=\dfrac{\dfrac{3}{8}\times \dfrac{4}{10}}{\dfrac{4}{8}\times \dfrac{3}{10}+\dfrac{5}{8}\times \dfrac{3}{10}+\dfrac{3}{8}\times \dfrac{4}{10}} \;\;\;\; [\because \dfrac{5}{10} = \dfrac{4}{8} =\dfrac{1}{2} ] \\
\end{align}$
$\begin{align}
  & \Rightarrow P\left( {{B}_{3}}\left| G \right. \right)=\dfrac{12}{12+15+12} \\
 & \Rightarrow P\left( {{B}_{3}}\left| G \right. \right)=\dfrac{12}{39} \\
 & \Rightarrow P\left( {{B}_{3}}\left| G \right. \right)=\dfrac{4}{13} \\
\end{align}$
Hence, neither of option (b) or (d) is correct.
Now, we shall proceed to the event in option (c).
We need to find the probability of choosing a green ball from any bag. So it can be done as follows, we choose first bag and green ball from it or choose second bag and a green ball from it or the third bag and green ball form it.
$\begin{align}
  & \Rightarrow P\left( G \right)=P\left( {{B}_{1}} \right)P\left( G|{{B}_{1}} \right)+P\left( {{B}_{2}} \right)P\left( G|{{B}_{2}} \right)+P\left( {{B}_{3}} \right)P\left( G|{{B}_{3}} \right) \\
 & \Rightarrow P\left( G \right)=\dfrac{3}{10}\times \dfrac{5}{10}+\dfrac{3}{10}\times \dfrac{5}{8}+\dfrac{4}{10}\times \dfrac{3}{8} \\
 & \Rightarrow P\left( G \right)=\dfrac{3}{10}\times \dfrac{4}{8}+\dfrac{3}{10}\times \dfrac{5}{8}+\dfrac{4}{10}\times \dfrac{3}{8}\;\;\;\ [\because \dfrac{5}{10} = \dfrac{4}{8} =\dfrac{1}{2} ] \\
 & \Rightarrow P\left( G \right)=\dfrac{12+15+12}{80} \\
 & \Rightarrow P\left( G \right)=\dfrac{39}{80} \\
\end{align}$
Hence, option (c) is also correct.

Note: The problem is an example of conditional probability. Conditional probability is calculated for such events whose outcomes are dependent on the occurrence of another event. Bayes’ Theorem of conditional probability is used to calculate probabilities of such events. It is to be noted that Bayes’ theorem can only be applied when all the events are mutually exclusive and independent events.