Answer :
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