Question

A fair coin is tossed 12 times. Find the probability of getting  at least 2 heads .

Answer 1

Let X denote number of heads obtained in 12 tosses.

∴ X = 0, l , 2, 3, 4, 5 , 6, 7, 8, 9, 10, 11, 12

n = 12 

p : Probability of getting head in a single toss. 

∴ p = `1/2`

q = `1 - 1/2 = 1/2`

The binomial distribution is 

x ~ B `(12 , 1/2)`

∴ p.m.f. is  P(X = X) = `"^n C _x  p^x  q^(n-x)`

P(X ≥ 2) = 1 - P(X < 2) 

`1 - ["^12 C_0 p^0 q^12` + `"^12 C _1  p^1 q^11]`

`= 1 [(1/2)^12 + 12 (1/2)^12]`

`= 1 - [13/2^12]`

`= 4083/4096`

= 0.9968

∴ Probability of getting at least 2 heads is 0.9968 .