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2 votes
2 votes
Answer is 3

For n=2 p(x)=1/2ie HT TH)

For n=3 p(x)=1/4 ie (HHT and TTH)

 

So for n=k p(x)=1/2^(n-1)

 Let S be E(x)

S=2*1/2+4*1/4+8*1/4......

This is agp series

Solving this we get

S=3
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0 votes

when no of toss n=2 , then probability{HT,TH}={1/2.1/2  ,  1/2 .1/2}

when n=3    , then probability {TTH,HHT}={1/2.1/2.1/2  ,  1/2 .1/2.1/2}

when n=4    , then probability {TTTH,HHHT} ={1/2.1/2.1/2.1/2  ,  1/2 .1/2.1/2.1/2}

So no of tosses = (1/2 .1/2).2 +(1/2 .1/2.1/2).2 +..........................

                          =2(1/4+1/8+1/16............)

                          =1/2+1/4+1/8........................

                           =(1/2)/(1-1/2)=1

So, avg. no of tosses =(total tosses)/2 =1/2

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0 votes
There are two ways that the number of turns needed would be K , first by starting with k-1 heads and last one tail and second by starting with k-1 tails and ending with one head.

$P(X=k) = 0.5^{k-1} \ \ \ \ \ \ E(X) = \sum_{2}^{inf} k*P(X=k) = 3$

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