expectation

Question

A fair coin is tossed repeatedly till both head and tail appear atleast once. Average no of tosses required is ?

Comments

3 ?

Answer 1

answer 3

let the random variable X be number of tosses ..here we need at least two tosses to get a success

X                    2             3            4               5                    ..........

Prob(X=i)        2/4        2/8          2/16           2/32                 ........

E(x)=             2*2/4 + 3 *2/8 + 4*2/16 +......  (i)

E(x) /2=                     2*2/8 + 3*2/16 +.........(ii)

(i)-(ii) is 2*2/4 + (3-2)2/8 + (4-3)2/16 +.....

E(x)/2 => 1+2/8+2/16+2/32+...

E(x)/2=>1+ 2(1/8 + 1/16 + 1/32 +...)                      used formula of sum of infinte gp series

E(x)/2=>1 +2* (1/8 / (1 -  1/2))

E(x)/2=>1+1/2

E(x)/2=>3/2

=>3

now let me tell you why P(x=3 ) is 2/8

lets enumerate the possibilities (0 for head ,1 for tall )

000

001
010
011
100
101
110
111

the red ones are the only favourable cases because we came to the third toss only when we get first two tosses equal either it can be 00 or 11 now if next we get a 1 after 00 or 0 after 11 thats a success .so similarly try the other ones