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, $B_3$ 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 option 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)

Probability that the chosen ball is green equals $\frac{39}{80}$

Given, $Bag{B}_1\to$ (5 Red, 5 Green),$Bag{B}_2\to$ (3 Red, 5 Green), $Bag{B}_3\to$(5 Red, 3 Green)
Probability of chosing bag 1, bag 2 and bag 3: $P\left(B_1\right)=\frac{3}{10},P\left(B_2\right)=\frac{3}{10},P\left(B_3\right)=\frac{4}{10}$
Probability of chosing green ball
$P\left(G\right)=P\left(B_1\right)\times P\left(\frac{G}{B_1}\right)+P\left(B_2\right)\times P\left(\frac{G}{B_2}\right)+P\left(B_3\right)\times P\left(\frac{G}{B_3}\right)$
$=\frac{3}{10}\times \frac{5}{10}+\frac{3}{10}\times \frac{5}{8}+\frac{4}{10}\times \frac{3}{8}=\frac{39}{80}$

Probability that the chosen ball is green, given that the selected bag is $B_3,=P\left(\frac{G}{B_3}\right)=\frac{3}{8}$

Probability that the selected bag is $B_3,$ and the chosen ball is green $=P\left(\frac{B_3}{G}\right)$
Using Baye's Theorem
$=P\left(\frac{B_3}{G}\right)=\frac{P\left(B_3\right).P\left(\frac{G}{B_3}\right)}{P\left(G\right)}=\frac{\frac{4}{10}\times \frac{3}{8}}{\frac{39}{80}}=\frac{4}{13}$