Question

Find the mean of the number of heads in three tosses of a fair coin.

Answer 1

Let X denote the number of heads.

∴ Possible values of X are 0, 1, 2, 3.

Let P(getting head) = p = `(1)/(2)`

∴ q = 1 – p = `1 - (1)/(2) = (1)/(2)` 

∴ P(X = 0) = P(no head) = qqq = q³ = `(1/2)^3 = (1)/(8)`

P(X = 1) = P(one head) = pqq + qpq + qqp = 3pq²

= `3(1/2)(1/2)^2`

= `(3)/(8)`

P(X = 2) = P(two heads) = ppq + pqp + qpp = 3p²q

= `3(1/2)^2 (1/2)`

= `(3)/(8)`

P(X = 3) = P(three heads) = ppp

= p³

= `(1/2)^3`

= `(1)/(8)`

∴ Mean number of heads

= E(X)

= `sum_(i=1)^4 x_i.p(x_i)`

= `0 xx (1)/(8) + 1 xx (3)/(8) + 2 xx (3)/(8) + 3 xx (1)/(8)`

= `(3)/(2)`

= 1.5