Question

Let X be the number of successes in 'n' independent Bernoulli trials with probability of 3 success p = `3/4`. The least value of 'n' so that P(X ≥ 1) ≥ 0.9375 is ______.

पर्याय

• 2

• 1

• 4

• 3

Answer 1

Let X be the number of successes in 'n' independent Bernoulli trials with probability of 3 success p = `3/4`. The least value of 'n' so that P(X ≥ 1) ≥ 0.9375 is 2.

Explanation:

We have, p = `3/4`, q = `1 - "p" = 1/4`

It is given that P(X ≥ 1) ≥ 0.9375

= 1 - P(X = 0) ≥ 0.9375

= 1 - ⁿC₀ (p⁰) (g)^n-0 ≥ 0.9375

`= 1 - (1/4)^"n" ≥ 0.9375`

`= 1 - 0.9375 ≥ (1/4)^"n"`

`= 0.0625 ≥ (1/4)^"n"`

`= 625/10000 ≥ (1/4)^"n"`

`= 1/16 ≥ (1/4)^"n"`

= 16 ≥ 4ⁿ

⇒ n = 2