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