Q2 >>> its

**2**Let "x" indicate the number of tosses of the coin

The number of tosses of the coin would be

* 1 if a head appears on the 1st throw

* 2 if a tail appears on the 1st throw and a head appears on the 2nd throw

* 3 if a tail appears on the 1st 2nd throws and a head appears on the 3rd throw

* ... ... ... and so on..

We know, In a single throw with a coin, Probability of:

P(H) = 1/2 = P(T)

Now, probablity distribution would be

x P(X=x) x*P

1 1/2 1/2

2 ( 1/2 ) ^2 2 * ( 1/2 ) ^2

3 ( 1/2 ) ^3 3 * ( 1/2 ) ^3

. . .

. . .

. . .

. . .

Therefore , expected number of tosses = summation x*P = 1/2 + 2 * ( 1/2 ) ^2 + 3 * ( 1/2 ) ^3 +...

solving the above >> we get expected no = 2

:)