Question

Urn A contains 2 red and 4 white balls, and urn B contains 1 red and 1 white ball. A ball is randomly chosen from urn A and put into urn B, and a ball is then chosen from urn B. What is the conditional probability that the transferred ball was white given that a white ball is selected from urn B?

Answer 1

Answer: 0.7991

Step-by-step explanation:

Given : Urn A contains 2 red and 4 white balls, and urn B contains 1 red and 1 white ball.

Let $E_1$ be the event that the transferred ball was white.

and $E_2$ be the event that the ball drawn from urn B was white.

Then by Bayes theorem, the conditional probability that the transferred ball was white given that a white ball is selected from urn B will be :-

$P(E_1|E_2)=\dfrac{P(E_1\cap E_2)}{P(E_2)}$

Where,

$P(E_1\cap E_2)=P(E_2|E_1) P(E_1)=\dfrac{2}{3}\times\dfrac{4}{6}=0.4444$ [ by conditional probability formula]

$P(E_2)=P(E_2|E_1)\times P(E_1)+P(E_2|E_1^c)\times P(E_1^c)$ [By law of total probability]

i.e. $P(E_2)=\dfrac{2}{3}\times \dfrac{4}{6}+\dfrac{1}{3}\times \dfrac{2}{6}=0.5556$

Now, $P(E_1|E_2)=\dfrac{0.444}{0.5556}=0.79913606911\approx0.7991$

Hence, the required probability =0.7991