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

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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.

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Samantha 1 week 2021-10-08T00:38:30+00:00 1 Answer 0

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    2021-10-08T00:39:49+00:00

    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

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