you can use binomial probability for this question
let x denote getting an odd no
n= No of times die is tossed = 3
p=proabability of getting an odd number in one throw = 1/2
using
P(X=r)=nCr prqn-r
now P(X>=1)= prob of getting an odd no atleast once= 1-P(X=0)
P(X>=1)=1-3C0(1/2)0(1/2)3
=1-(1/8)=7/8
P(X<=1)=prob of getting an odd no atmost once
= P(X=0) + P(X=1)
= 3C0(1/2)0(1/2)3 + 3C1(1/2)1(1/2)2
= (1/8) + 3*(1/8)
=1/2
Moderation Team
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![[tex] \\ \mbox{The first question ..P(atleast once) = 1-P(never odd)} \\ \\\mbox{That is all times we shud get even} \\ \\ P = 1-(\frac{1}{2})(\frac{1}{2})(\frac{1}{2}) = 1-\frac{1}{8} = \frac{7}{8} \\ \\ \mbox{Atmost once means either no odd at all or only once}\\ \\ P=(\frac{1}{2})(\frac{1}{2})(\frac{1}{2})+3\*(\frac{1}{2})(\frac{1}{2})(\frac{1}{2}) = \frac{1}{2}[\tex]](http://alt1.artofproblemsolving.com/Forum/latexrender/pictures/5/2/a/52a00393a1873ceed45eb8ce2548c788482aaadc.gif)
