Bag I contains 3 red and 4 black balls and Bag II contains 4 red and 5 black balls. One ball is transferred from Bag I to Bag II and then a ball is drawn from Bag II. The ball so drawn is found to be red in colour. Find the probability that the transferred ball is black.

Home » Bag I contains 3 red and 4 black balls and Bag II contains 4 red and 5 black balls. One ball is transferred from Bag I to Bag II and then a ball is drawn from Bag II. The ball so drawn is found to be red in colour. Find the probability that the transferred ball is black.

CBSE | Class 12 | Maths | NCERT | Probability

Bag I contains 3 red and 4 black balls and Bag II contains 4 red and 5 black balls. One ball is transferred from Bag I to Bag II and then a ball is drawn from Bag II. The ball so drawn is found to be red in colour. Find the probability that the transferred ball is black.

Solution:

Let us first suppose that the events denoted by $A_1$ are the occurrences in which a red ball is transferred from bag I to bag II. In addition, $A_2$ denotes the occurrence in which a black ball is transported from bag I to bag II.

$\therefore \mathrm{P}\left(\mathrm{A}_{1}\right)=3 / 7$

And, $P\left(A_{2}\right)=4 / 7$

Let $\mathrm{X}$ be the event that the drawn ball is red $\therefore$ when red ball is transferred from bag I to II,

$P\left(X \mid A_{1}\right)=\frac{5}{10}$

$=\frac{1}{2}$

And, when black ball is transferred from bag I to II,

$P\left(X \mid A_{2}\right)=\frac{4}{10}$

$=\frac{2}{5}$

Hence, $^{P\left(A_{2} \mid X\right)=\frac{P\left(A_{2}\right) P\left(X \mid A_{2}\right)}{P\left(A_{1}\right) P\left(X \mid A_{1}\right)+P\left(A_{2}\right) P\left(X \mid A_{2}\right)}}$