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.