Question

Four womens college basketball teams are participating in a single illumination holiday basketball tournament if one team is favored in its semi-final match by odds of 1.40 to 1.60 and another squad is favored in its contest by odds of 2.30 to 1.70 what is the probability that both teams win their games, neither favored team wins its game or at least one of the favored teams wins its games

Answer 1

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:

$P(A \cap B) = P(A)P(B)$

One team is favored in its semi-final match by odds of 1.40 to 1.60:

Thus it's probability of winning is:

$p_1 = \frac{1.6}{1.4 + 1.6} = \frac{1.6}{3.0} = 0.5333$

Another squad is favored in its contest by odds of 2.30 to 1.70:

Thus it's probability of winning is:

$p_2 = \frac{2.3}{4} = 0.575$

The probability of both winning is:

$p = 0.5333(0.575) = 0.3066$

0.3066 = 30.66% probability that both teams win their games.

The probability of neither winning is:

$p = (1 - 0.5333)(1 - 0.575) = 0.1983$

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:

$p = 1 - 0.1983 = 0.8017$

0.8017 = 80.17% probability that at least one of the favored teams wins its games.

A similar problem is given at brainly.com/question/24935451