Answer 1

Let X = number of successes, i.e. number of odd numbers.

p = probability of getting an odd number in a single throw of a die

∴ p = `3/6 = 1/2` and q = `1 - "p" = 1 - 1/2 = 1/2`

Given: n = 6

∴ X ∼ B`(6, 1/2)`

The p.m.f. of X is given by

`p("X = x") = "^nC_x  p^x  q^(n - x)`

i.e. p(x) = `"^6C_x (1/2)^x (1/2)^(6 - x)`

` = "^6C_x (1/2)^6,` x = 0, 1, 2, ...,6

P(5 successes) = P[X = 5]

`= p(5) = "^6C_5 (1/2)^6`

`= "^6C_1 xx 1/64`       .....`[because ""^nC_x = "^nC_(n - x)]`

`= 6/64 = 3/32`

Hence, the probability of 5 successes is `3/32`