Two urns contain respectively 2 red, 3 white, and 3 red, 5 white balls. One ball is drawn at random from the first urn and transferred into the second one. A ball is then drawn from the second urn and it turns out that the ball is red. What will be the probability that the transferred ball was white by using bayes theorem.

Let A₁ and A₂ denote the events that the transferred ball from the first urn to the second is white and red respectively.

probability of white ball

P(A₁) = \$\frac(3)(3)\$ + 2 = \$\frac(3)(2)\$

probability of red ball

P(A₂) = \$\frac(2)(3)\$ + 2 = \$\frac(2)(5)\$

let X denote the event of drawing a red ball from the second urn after the occurrence of A₁ or A₂

probability of the second urn after occurance of A1

P(A₁/X) = \$\frac(3)(9)\$ = ⅓

probability of the second urn after occurance of A2

P(X/A₂) = 3 + ⅓ + 5+ 1 = \$\frac(4)(9)\$

Bayes theorem \$(P(E_1/A) = P(E_1) * P(A/E_1)) / (P(E_1) * P(A/E_1) + P(E_2) * P(A/E_2))\$

Simplification

\$\frac(3)(5) \times \frac(1)(3) / \frac(3)(5) \times \frac(1)(3) + \frac(2)(5) \times (4)(9)\$

After simplification we get

\$\frac(9)(11)\$

Answer

B