Question

For a Binomial distribution n = 6 and p = 0.3, find the probability of getting exactly 3 successes.

Answer 1

Given x ∼ B (n, p)

i.e; Given n = 6, p = 0.3

Let x be the number of success in n = 6 trials

∴ x = 0, 1, 2, 3, 4, 5, 6

∴ x ∼ B (6, 0.3)

∴  p.m.f. is p(X = x) = nC_x . p^x . q^n-x

Since p + q = 1 ⇒ q = 1 − p

⇒ q = 1 − 0.3 = 0.7

∴ q ∝ 0.7

∴ p (X = x) = 6C_x . (0.3)^x . (0.7)^6-x

∴ Required probability = p (X = 3)

= 6C₃ . (0.3)³ . (0.7)^{(6 − x)}

∴ p(X = 3) = `[6 xx 5 xx 4]/[1 xx 2 xx 3] xx (0.3)^3 xx (0.7)^3`

∴ p(X = 3) = 0.1852