In a chess club the probability that Shaun will beat Mike is 3/8 . The probability that Shaun will beat Tim is 5/7 . (Assume all

Question

In a chess club the probability that Shaun will beat Mike is 3/8 .
The probability that Shaun will beat Tim is 5/7 .
(Assume all the games are independent of one another)
If Shaun plays 1 game with Mike and then 1 game with Tim, what is the probability that Shaun loses both games?

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Thái Dương 6 months 2021-07-20T02:59:19+00:00 1 Answers 40 views 0

Answers ( )

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    2021-07-20T03:00:24+00:00

    Answer:

    \frac{10}{56} = 0.1786 probability that Shaun loses both games

    Step-by-step explanation:

    Games are independent, so we find each separate probability, and multiply them.

    In a chess club the probability that Shaun will beat Mike is 3/8.

    So 1 - \frac{3}{8} = \frac{8}{8} - \frac{3}{8} = \frac{5}{8} probability that Shaun loses.

    The probability that Shaun will beat Tim is 5/7 .

    So 1 - \frac{5}{7} = \frac{2}{7} probability that Shaun loses.

    What is the probability that Shaun loses both games?

    This is:

    p = \frac{5}{8} \times \frac{2}{7} = \frac{5*2}{8*7} = \frac{10}{56} = 0.1786

    \frac{10}{56} = 0.1786 probability that Shaun loses both games

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