Question

at high school 40% of the students buy yearbooks. you select 6 students at random. Find P(exactly two buy yearbooks) and P(at least two buy yearbooks). Explain.

Answer 1

P(exactly two buy yearbooks) = 0.31104

P(at least two buy yearbooks) = 0.76672

Explanation:

We are given that at high school 40% of the students buy yearbooks.

You select 6 students at random.

The above situation can be represented through binomial distribution;

`P(X = r) = \binom{n}{r} * p^(r) * (1-p)^(n-r);x=0,1,2,3,.......`

where, n = number trials (samples) taken = 6 students

r = number of success

p = probability of success which in our question is probability that

students buy yearbooks, i.e; p = 0.40

Let X = Number of students who buy yearbooks

So, X ~ Binom(n = 6, p = 0.40)

(a) Now, Probability that exactly two buy yearbooks is given by = P(X = 2)

P(X = 2) =
`\binom{6}{2} * 0.40^(2) * (1-0.40)^(6-2)`

=
`15* 0.40^(2) * 0.60^(4)`

= 0.31104

(b) Probability that at least two buy yearbooks is given by = P(X
`\geq` 2)

P(X
`\geq` 2) = 1 - P(X = 0) - P(X = 1)

=
`1- \binom{6}{0} * 0.40^(0) * (1-0.40)^(6-0)-\binom{6}{1} * 0.40^(1) * (1-0.40)^(6-1)`

=
`1- (1 * 1 * 0.60^(6))-(6 * 0.40^(1) * 0.60^(5))`

= 1 - 0.04666 - 0.18662

= 0.76672