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Sagot :
Using probability of independent events, it is found that there is a:
- 0.3066 = 30.66% probability that both teams win their games.
- 0.1983 = 19.83% probability that neither favored team wins its game.
- 0.8017 = 80.17% probability that at least one of the favored teams wins its games.
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- If two events, A and B, are independent, the probability of both happening is the multiplication of the probabilities of each happening, that is:
[tex]P(A \cap B) = P(A)P(B)[/tex]
One team is favored in its semi-final match by odds of 1.40 to 1.60:
Thus it's probability of winning is:
[tex]p_1 = \frac{1.6}{1.4 + 1.6} = \frac{1.6}{3.0} = 0.5333[/tex]
Another squad is favored in its contest by odds of 2.30 to 1.70:
Thus it's probability of winning is:
[tex]p_2 = \frac{2.3}{4} = 0.575[/tex]
The probability of both winning is:
[tex]p = 0.5333(0.575) = 0.3066[/tex]
0.3066 = 30.66% probability that both teams win their games.
The probability of neither winning is:
[tex]p = (1 - 0.5333)(1 - 0.575) = 0.1983[/tex]
0.1983 = 19.83% probability that neither favored team wins its game.
At least one is complementary to neither, thus, the probability of at least one is:
[tex]p = 1 - 0.1983 = 0.8017[/tex]
0.8017 = 80.17% probability that at least one of the favored teams wins its games.
A similar problem is given at https://brainly.com/question/24935451
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