What is the probability that an ordinary year has 53 sundays

Explanation

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

(i) `.^3C_2(1/2)^2(1-1/2)^1`

`= 3 xx (1/2)^3= 3/8`

(ii) 0 heads `.^3C_0(1/2)^0(1/2)^3 = 3 xx (1/2)^3 = 3/8`

1 head `.^3C_1(1/2)^1 (1/2)^2= 3 xx (1/2)^3 = 3/8`

2 heads `3/8`

`:. 1/8 + 3/8 + 3/8 = 7/8`

(iii) 1 head, 2 tail `.^3C_1(1/2)(1/2)^2= 3/8`

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