Suppose you just received a shipment of eight televisions. Three of the televisions are defective. If two televisions are randomly​ selected

Question

Suppose you just received a shipment of eight televisions. Three of the televisions are defective. If two televisions are randomly​ selected, compute the probability that both televisions work. What is the probability at least one of the two televisions does not​ work?

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1 week 2021-09-15T20:14:49+00:00 1 Answer 0

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    2021-09-15T20:15:49+00:00

    Answer: The probability that both televisions work is \dfrac{5}{14}.

    The probability at least one of the two televisions does not​ work is \dfrac{9}{14}.

    Step-by-step explanation:

    Given : Number of televisions received = 8

    Number of defective televisions=3

    Number of good television = 8-3=  5

    Then , the number of ways to select any two television out of 8 = ^8C_2

    The number of ways to select two working televisions out of three = ^5C_2

    Now ,  If two televisions are randomly​ selected, the the probability that both televisions work

    =\dfrac{^5C_2}{^8C_2}=\dfrac{\dfrac{5!}{2!3!}}{\dfrac{8!}{2!6!}}=\dfrac{\dfrac{5\times4\times3!}{2\times3!}}{\dfrac{8\times7\times6!}{2\times6!}}=\dfrac{5}{14}

    Hence, the probability that both televisions work is \dfrac{5}{14}.

    Also , the probability at least one of the two televisions does not​ work = 1- P( both televisions work)

    =1-\dfrac{5}{14}=\dfrac{14-5}{14}\\\\=\dfrac{9}{14}

    Hence, the probability at least one of the two televisions does not​ work is \dfrac{9}{14}

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