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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?

Sagot :

Answer:

[tex]\frac{10}{56} = 0.1786[/tex] 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 [tex]1 - \frac{3}{8} = \frac{8}{8} - \frac{3}{8} = \frac{5}{8}[/tex] probability that Shaun loses.

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

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

What is the probability that Shaun loses both games?

This is:

[tex]p = \frac{5}{8} \times \frac{2}{7} = \frac{5*2}{8*7} = \frac{10}{56} = 0.1786[/tex]

[tex]\frac{10}{56} = 0.1786[/tex] probability that Shaun loses both games

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