Darkphyx
08.05.2021 •
Mathematics
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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Ответ:
11%
Step-by-step explanation:
Shaun's percentage of winning against Mike is 8/11 or 73%
Shaun's percentage of winning against Tim is 7/12 or 58%
Shaun's percentage of losing against Mike is 100-73 = 27%
Shaun's percentage of losing against Tim is 100-58 = 42%
Shaun's percentage of losing against Mike and Tim is .27 x .42 which is approximately 11%
Ответ:
the first one the third one and the fourth one
Step-by-step explanation: