Question

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

Answer 1

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