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Sagot :
Applying probability of independent events, it is found that there is a 0.15 = 15% probability that Mine is the first one to win two games in a row.
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If two events, A and B, are independent, the probability of both happening is the multiplication of the probability of each happening, that is:
[tex]P(A \cap B) = P(A)P(B)[/tex]
There are two outcomes in which Mine wins two games in a row:
- Wins the first serving(0.75 probability), then win the second with Kate serving(0.10 probability).
- Wins the first with Kate serving(0.1 probability), then the second with her serving(0.75 probability).
Thus:
[tex]p = 2 \times 0.1 \times 0.75 = 0.15[/tex]
0.15 = 15% probability that Mine is the first one to win two games in a row.
A similar problem is given at https://brainly.com/question/23855473
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