A) \[\frac{1}{2}\]
B) \[\frac{1}{3}\]
C) \[\frac{1}{4}\]
D) None of these
Correct Answer: B
Solution :
Let \[X\] denotes the number of tosses required. Then \[P(X=r)={{(1-p)}^{r-1}}.\] \[p,\] for \[r=1,\,2,\,3\,......\] Let \[E\] denote the event that the number of tosses required is even. Then \[P(E)=P[(X=2)\cup (X=4)\cup (X=6)\cup ........]\] \[P(E)=P(X=2)+P(X=4)+P(X=6)+......\] \[P(E)=(1-p)p+{{(1-p)}^{3}}p+{{(1-p)}^{5}}p+.......=\frac{1-p}{2-p}\] But we are given that \[P(E)=\frac{2}{5},\] then we get \[p=\frac{1}{3}.\]
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