In binomial distribution with five Bernoulli’s trials, the probability of one and two success are 0.4096 and 0.2048 respectively. Find the probability of success.

#### Solution

Given: X ~ B(n = 5, p)

The probability of X successes is

P(X = x) = `"^nC_x p^x q^(n - x)`, x = 0, 1, 2,...,n

i.e. P(X = x) = `"^5C_x p^x q^(5 - x)`, x = 0, 1, 2, 3, 4, 5

Probabilities of one and two successes are

P(X = 1) = `"^5C_1 p^1 q^(5 - 1)`

and P(X = 2) = `"^5C_2 p^2 q^(5 - 2)` respectively

Given: P(X = 1) = 0.4096 and P(X = 2) = 0.2048

∴ `("P"("X" = 2))/("P"("X" = 1)) = 0.2048/0.4096`

i.e. `(""^5C_2 p^2 q^(5 - 2))/("^5C_1 p^2 q^(5 - 1)) = 1/2`

i.e. `2 xx ""^5C_2 p^2 q^3 = 1 xx ""^5C_1 "pq"^4`

i.e. `2 xx (5 xx 4)/(1 xx 2) xx "p"^2 "q"^3 = 1 xx 5 xx "pq"^4`

i.e. `20 "p"^2"q"^3 = 5"pq"^4`

i.e. 4p = q

i.e. 4p = 1 - p

i.e. 5p = 1

∴ p = `1/5`

Hence, the probability of success is `1/5`.