Question

(1) Let X1,.., X4 denote the number of hearts, diamonds, clubs, and spades drawn from 10 draws with replacement from a standard 52 card deck of playing cards (a) Calculate P(X2< 3) (b) Calculate P(X1 3, X3 +X4 2) (c) Calculate E[X2|X1 + X4 = 5] (d) Calculate P(X1 2|X3 = 6)

Answer 1

(a) P(X1 = 11, X2 = 12, X3 = 13, X4 = 14) = 21 Izlzz!(10 – 2; – 12 – 13)*(+) * (1) * (1) *( (10)! /1 1 /1\12 1131110-11-12-13 21!12!13!(10 – 11 – 12 – 13)! (since probability of drawing a heart in a particular trial = probability of drawing a diamond in a particular trial = probability of drawing a club in a particular trial = 13 probability of drawing a spade in a particular trial = = * 110 (10)! 21!12!<3!(10 – 11 – 12 – 13)! 21, 22, 23, 24 > 0, 21 +.22 +33 + 24 = 10 Hence (X1, X2, X3, X4) ~ Multinomial(10, 1/4, 1/4, 1/4, 1/4). Then X2 ~ Binomial(10, 1/4) and so /11 13 10-1 P(X2<3) = P(X2 = 0)+P(X2 = 1)+P(X2 = 2) = I = 0.5256

(6) P(X1 = 21, X2 = 12, X3 = 13, X4 = 14) (10): 11 10 x() 11!C!<3!9(10 – 11 – 12 – 13)! P(X1 = 3, X3 + X4 = 2) = P(X1 = 3, X2 = 5, X3 + X4 = 2) = P(X1 = 3, X2 = 5, X3 = 0, X4 = 2) + P(X1 = 3, X2 = 5, X3 = 2, X4 = 0) + P(X1 = 3, X2 = 5, X3 = 1, X4 = 1) = (0,12 + 2001 + int) 3 x = 0.0096 1/4 1/4 (c) We know, X2, X3 X1 = 11, X4 = 14 - Multinomial (10 – 11 – 14), – 1-(1/4+1/4)'1-(1/4+1/4), Now if 11 + 12 = 5 then X2|X1 + X4 = 5 ~ Binomial(5, 1/2) Hence E(X2 X1 + X4 = 5) = 5/2.

(d) P(X1 > 3, X3 = 6) = P(X1 = 3, X2 = 0, X3 = 6, X4 = 1) + P(X1 = 3, X2 = 1, X3 = 6, X4 = 0) + P(X1 = 4, X2 = 0, X3 = 6, X4 = 0) 10! 10! 10! 1 = ( 3!0!6!1! + 3!1!6!0! + 4!0!6!0!) * 410 = 0.0018 P(X3 = 6) = 1/4)(3/4) (since X3 ~ Binomial(10,1/4)) = 0.0162 P(X1 < 2|X3 = 6) = 1 - P(X1 > 3X3 = 6) = 1- P(X1 > 3, X3 = 6) P(X3 = 6) =1- (0.0018/0.0162) = 0.8889.