Question

A coin with probably of heads = 1/2 is tossed over and over. X is defined as the number of tosses needed to obtain the pattern: heads, tails (a toss that results in heads is followed by tails on the next toss).

a. Find the probability mass function of X

b. find the expected value of X.

Answer 1

a)

here as to get the above pattern last 2 flips shoups Head and then tail ; therefore we need to arrange first (x-2) flips in such a way that No tail precedes H for which there are (x-1) number of ways

hence Probability mass function of X =P(X=x)=(x-1)*(1/2)^x

b) expected value of X E(X)= xP(x) = x*(x-1)*(1/2)^x =(1/4)* x*(x-1)*(1/2)^x-2 =(1/4)*2/(1-1/2)³ =4

(Note:as x*(x-1)*(a)^x-2 =2a/(1-x)³ ; therefore x*(x-1)*(1/2)^x-2 =2/(1-1/2)³ )