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Question

# A biased coin with probability p,0<p<1, of heads is tossed until a head appears for the first time. If the probability that the number of tosses required is even is 2/5, then p=........

A
2/5
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B
2/3
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C
1/3
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D
3/5
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Solution

## The correct option is C 1/3Let x denote the number of tosses required.Then P(x=r)=(1−P)r−1P;r=1,2,3....Let E denote the event that the number of tosses required is even. Then,P(E)=P{(x=2)∪(x=4)∪(x=6)∪...}=P(x=2)+P(x=4)+P(x=6)+....=p(1−p)+p(1−p)3+p(1−p)5+..=p(1−p)1−(1−p)2=p(1−p)2p−p2=1−p2−pAs we are given that P(E)=25, we get5(1−p)=2(2−p)⇒1−3p=0⇒p=13

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