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

Answer:

P(exactly two buy yearbooks) = 0.31104

P(at least two buy yearbooks) = 0.76672

Step-by-step 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;

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

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) =  $\binom{6}{2} \times 0.40^{2} \times (1-0.40)^{6-2}$

                             =  $15\times 0.40^{2} \times 0.60^{4}$

                             =  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)

                   =  $1- \binom{6}{0} \times 0.40^{0} \times (1-0.40)^{6-0}-\binom{6}{1} \times 0.40^{1} \times (1-0.40)^{6-1}$

                   =  $1- (1 \times 1 \times 0.60^{6})-(6 \times 0.40^{1} \times 0.60^{5})$

                   =  1 - 0.04666 - 0.18662

                   =  0.76672