S1 P(X1=X2) assistance

Can anybody assist please?
Q. A random number generator in a computer game produces values that can be modelled by the descrete random variable X with probability distribution given by:
P(X = r) = kr! for r = 0,1,2,3,4
P(X = r) = 0 Otherwise
Two independent values of X are generated. Let these values be X1 and X2.
(iii) Show that P(X1=X2) is a little greater than 0.5.
(iv) Given that X1=X2, find the probability that X1 and X2 are each equal to 4.

Reply 1

m:)ckel

(iii)

P(X=0) = k
P(X=1) = k
P(X=2) = 2k
P(X=3) = 6k
P(X=4) = 24k

These probabilities must add up to 1,

=> 34k = 1
k = 1/34

P(X1=X2) = P(X=0)*P(X=0) + ..... + P(X=4)*P(X=4)

P(X1=X2) = (1/34²)[1 + 1 + 4 + 36 + 576] = 0.535

(iv)

P(X1,X2 =4 | X1=X2) = P(X1,X2 = 4 AND X1=X2) / P(X1=X2)

= (1/5) / (0.535)

[The]

= 0.374

I'm certain of the first answer. Not as sure with the second.

Reply 2

Jonny W

mockel

The numerator is 1/5 since there are 5 ways in which X1=X2, but we are only choosing one of those ways.

The five ways in which X1 can equal X2 have different probabilities.

P(X1 = 4 and X2 = 4)
= P(X1 = 4)*P(X2 = 4) . . . . . since X1 and X2 are independent
= (24/34)^2

Answer
= (24/34)^2 / (309/578)
= 96/103

Reply 3

m:)ckel

Oh yeah, of course.