Re: Probability doubt

Posted by Amit Goyal on
URL: http://discussion-forum.276.s1.nabble.com/Probability-doubt-tp7586579p7586585.html

Pr(X = 0) = (1/2)
Pr(X = 1) = (1/2)(1/3) = (1/6)
Pr(X = 2) = (1/2)(2/3)(1/4) = (1/12)
Pr(X = 3) = (1/2)(2/3)(3/4)(1/5) = (1/20)
Pr(X = 4) = (1/2)(2/3)(3/4)(4/5) = (1/5)