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

Answer 1

Answer: (A), (B)

$P\left(B_1\right)=\frac{3}{10}P\left(B_2\right)=\frac{3}{10}P\left(B_3\right)=\frac{4}{10}$
(A) Probability that selected bag is $B_3$ and the chosen ball is green
$=P\left(B_3\right)\times P\left(\frac{G}{B_3}\right)$
$=\frac{4}{10}\times \frac{3}{8}=\frac{3}{20}$
(B) Probability that the selected bag is$B_3$ given that the chosen ball is green $P\left(\frac{B_3}{G}\right)$
$P\left(\frac{B_3}{G}\right)=\frac{P\left(\frac{G}{B_3}\right)P\left(B_3\right)}{P\left(\frac{G}{B_1}\right)P\left(B_1\right)+P\left(\frac{G}{B_2}\right)P\left(B_2\right)+P\left(\frac{G}{B_3}\right)P\left(B_3\right)}$
$=\frac{\frac{4}{10}\times \frac{3}{8}}{\frac{3}{10}\times \frac{5}{10}+\frac{3}{10}\times \frac{5}{8}+\frac{4}{10}\times \frac{3}{8}}=\frac{4}{13}$
(C) Probability that the chosen ball is green, given that the selected bag is $B_3$
$P\left(\frac{G}{B_3}\right)=\frac{3}{8}$
(D) Probability that the chosen ball is green
$P\left(G\right)=P\left(B_1\right)P\left(\frac{G}{B_1}\right)+P\left(B_2\right)P\left(\frac{G}{B_2}\right)+P\left(B_3\right)P\left(\frac{G}{B_3}\right)$
$=\frac{3}{10}\times \frac{5}{10}+\frac{3}{10}\times \frac{5}{8}+\frac{4}{10}\times \frac{3}{8}$
$=\frac{39}{80}$