Question

The probability of a man hitting a target is 0.25. He shoots 7 times. What is the probability of his hitting at least twice?

Answer 1

Here n = 7

p = 0.25 = `25/100 = 1/4`

And q = `1 - 1/4 = 3/4`

P(X ≥ 2) = 1 – [P(X = 0) + P(X = 1)]

= `1 - [""^7"C"_0 (1/4)^0 (3/4)^7 + ""^7"C"_1 (1/4)^1 (3/4)^6]`

= `1 - [(3/4)^7 + 7/4(3/4)^6]`

= `1 - (3/4)^6 (3/4 + 7/4)`

= `1 - (3/4)^6 (10/4)`

= `1 - 729/4096 xx 10/4`

= `1 - 7290/16384`

= `(16384 - 7290)/16384`

= `9094/16384`

= `4547/8192`

Hence, the required probability is `4547/8192`.