In a large school, 80% of the pupil like Mathematics. A visitor to the school asks each of 4 pupils, chosen at random, whether they like Mathematics.

Find the probability that the visitor obtains answer yes from at least 2 pupils:

a. when the number of pupils questioned remains at 4.

b. when the number of pupils questioned is increased to 8.

#### Solution

Let X = number of pupils like Mathematics.

p = probability that pupils like Mathematics

∴ p = 80% = `80/100 = 4/5`

and q = 1 - p = `1 - 4/5 = 1/5`

Given: n = 4

∴ X ~ B `(4, 4/5)`

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

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

i.e. p(x) = `"^4C_x (4/5)^x (1/5)^(4 - x)` x = 0, 1, 2, 3, 4

**(a)**. P(visitor obtains the answer yes from at least 2 pupils when the number of pupils questioned remains at 4) = P(X ≥ 2)

= P(X = 2) + P(X = 3) + P(X = 4)

`= ""^4C_2 (4/5)^2 (1/5)^(4 - 2) + ""^4C_3 (4/5)^3 (1/5)^(4 - 3) + "^4C_4 (4/5)^4 (1/5)^(4 - 4)`

`= (4 xx 3)/(1 xx 2) xx 16/5^2 xx 1/5^2 + 4 xx 64/5^3 xx 1/5 + 1 xx 256/5^4`

`= 96/5^4 + 256/5^4 + 256/5^4`

`= (96 + 256 + 256)1/5^4`

`= 608/5^4 = 608/625`

**(b). **P(the visitor obtains the answer yes from at least 2 pupils when number of pupils questioned is increased to 8)

= P(X ≥ 2)

= 1 - P(X < 2)

= 1 - [P(X = 0) + P(X = 1)]

`= 1 - [""^8C_0 (4/5)^0 (1/5)^(8 - 0) + ""^8C_1 (4/5)^1 (1/5)^(8 - 1)]`

`= 1 - [1 (1) (1/5)^8 + (8)(4/5)(1/5)^7]`

`= 1 - [1/5^8 + 32/5^8]`

`= 1 - 33/5^8`.