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 $\frac{3}{10},\frac{3}{10}$ and $\frac{4}{10}$ 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?

• A. Probability that the selected bag is $B_3$ and the chosen ball is green equals $\frac{3}{10}$.
• B. Probability that the chosen ball is green equals $\frac{39}{80}$.
• C. Probability that the chosen ball is green, given that the selected bag is $B_3$, equals $\frac{3}{8}$.
• D. Probability that the selected bag is $B_3$, given that the chosen balls is green, equals $\frac{5}{13}$.

Answer 1

Answer: (B), (C)

Probability that the chosen ball is green, given that the selected bag is $B_3$, equals $\frac{3}{8}$.

Explanation of the correct options

Option $\left(B\right)$: Probability that the chosen ball is green equals $\frac{39}{80}$.

Let's plot the tree diagram,

Thus,

$\begin{array}{rcl}P\left(G\right)& =& \frac{5}{10}\times \frac{3}{10}+\frac{5}{8}\times \frac{3}{10}+\frac{3}{8}\times \frac{4}{10}\\ & =& \frac{60+75+60}{400}\\ & =& \frac{195}{400}\\ & =& \frac{39}{80}\end{array}$

Hence option $\left(B\right)$ is the correct option.

Option $\left(C\right)$: Probability that the chosen ball is green, given that the selected bag is $B_3$, equals $\frac{3}{8}$.

$P\left(\frac{G}{B_3}\right)=\frac{3}{8}$

Hence option $\left(C\right)$ is the correct option.

Explanation of the incorrect options

Option $\left(A\right)$: Probability that the selected bag is $B_3$ and the chosen ball is green equals $\frac{3}{10}$.

$\begin{array}{rcl}P(B_3\cap G)& =& P\left(\frac{G}{B_3}\right).P\left(B_3\right)\\ & =& \frac{3}{8}\times \frac{4}{10}\\ & =& \frac{3}{20}\end{array}$

Hence option $\left(A\right)$ is the incorrect option.

Option $\left(D\right)$: Probability that the selected bag is $B_3$, given that the chosen balls is green, equals $\frac{5}{13}$

$\begin{array}{rcl}P\left(\frac{B_3}{G}\right)& =& \frac{P\left(G\cap B_3\right)}{P\left(G\right)}\\ & =& \frac{{\displaystyle \frac{3}{20}}}{{\displaystyle \frac{39}{80}}}\\ & =& \frac{4}{13}\end{array}$

Hence option $\left(D\right)$ is the incorrect option.

Hence, option $\left(B\right)$ and $\left(C\right)$ are the correct options.