Question

An urn contains balls numbered 1 through 6. Balls are repeatedly selected one at a time and with replacement. Let Xz be the number of the selection on which the first 3 appears, and let X4 be the number of the selection on which the first 4 appears. Let Px. y. (x3|x4) be the conditional distribution of X3, given that X4 = x4. (a) Find Px, x,(5|3) (b) Find Px, x,(315)

Answer 1

a) The conditional probability here is computed using Bayes theorem as:

$P(X_3 = 5 | X_4 = 3) = \frac{P(X_3 = 5, X_4 = 3)}{P(X_4 = 3)}$

$P(X_3 = 5 | X_4 = 3) = \frac{(4/6)^2*(1/6)*(5/6)*(1/6)}{(5/6)^2*(1/6)}$

$P(X_3 = 5 | X_4 = 3) = \frac{4^2}{5*6^2} = 0.0889$

Therefore 0.0889 is the required probability here

b) The conditional probability here is again computed using Bayes theorem as:

$P(X_3 = 3 | X_4 = 5) = \frac{P(X_3 = 3, X_4 = 5)}{P(X_4 = 5)}$

$P(X_3 = 3 | X_4 = 5) = \frac{(4/6)^2*(1/6)*(5/6)*(1/6)}{(5/6)^4*(1/6)}$

$P(X_3 = 3 | X_4 = 5) = \frac{4^2}{5^3} = 0.128$

Therefore 0.128 is the required probability here.