If the probability of success in a single trial is 0.01. How many trials are required in order to have a probability greater than 0.5 of getting at least one success?

#### Solution

Let X = number of successes.

p = probability of success in a single trial

∴ p = 0.01

and q = 1 - p = 1 - 0.01 = 0.99

∴ X ~ B(n, 0.01)

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

P(X = x) = `"^nC_x p^x q^(n - x)`

i.e. p(x) = `"^nC_x (0.01)^x (0.99)^(n - x)` x = 1, 2,....,n

P(at least one success)

= P(X ≥ 1) = 1 - P(X < 1)

= 1 - P(X = 0) = 1 - p(0)

`= 1 - "^nC_0 (0.01)^0 (0.99)^(n - 0)`

`= 1 - 1(1)(0.99)^n`

`= 1 - (0.99)^n`

Given: P(X ≥ 1) > 0.5

i.e. `1 - (0.99)^n > 0.5`

i.e. 1 - 0.5 > `(0.99)^"n"`

i.e. 0.5 > `(0.99)^"n"`

i.e. `(0.99)^"n" < 0.5`

i.e. log `(0.99)^"n"` < log (0.5)

i.e. n log (0.99) < log 0.5

i.e. n < `("log" 0.5)/("log" 0.99)`

i.e. n < 68.96

∴ n = 68

Hence, the number of trials required in order to have probability greater than 0.5 of getting at least one success is `("log" 0.5)/("log" 0.99)` OR 68