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An industrial chemical that will retard the spread of fire has been developed. The local sales representatives have determined from past experience 48% of call will result in an order.

- What is the probability of that the first order will cause on 4th sales call of the day?

- If 8 sales calls are made on a day, what is the probability of getting exactly 6 orders?

- If 4 sales calls are made before lunch, what is the probability that one or less will result in an order.

1 Answer

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  • In first question we have to find the probability of that first order will cause on 4th sales call of the day.

    Here there is geometric distribution.

    p=48%=0.48

    q = 1-p = 1-0.48 = 0.52

    we have to determine first order will cause on 4th sales call hence x=4

    According to geometric distribution

    p(X=x) = p*q(x-1)

    p(X=4) = (0.48)*(0.52)(4-1)

    =0.067

  • In the second question we have to find probability that exactly 6 order cause from 8 sales calls.

    Here x=6, n=8.

    As n<30 we use Binomial distribution.

    According to binomial distribution.

    p(X=x) =($^nC_x$) * ($p^x$) * ($q^{n-x}$)

    p(X=6) = ($^8C_6$) * ($0.48^6$) * ($0.52^{8-6}$)

    =0.092

  • In last question we have to find the probability of that one or less result in order from 4.

    So n=4, x=1 or x=0

    p(X=0)+p(X=1)

    = ($^4C0$)($0.48^0$)($0.52^{(4-0)})+(^4C1$)($0.48^1$)($0.52^{(4-1)})$

    =0.343

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