$P(A\cup B) = P(A\cap B) / P(B)$. Why doesn't the $P(B)$ cancel out?

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"This season, the probability that the Yankees will win a game is $0.49$ and the probability that the Yankees will score 5 or more runs in a game is $0.62$. The probability that the Yankees win and score 5 or more runs is $0.42$."

Shouldn't the probability be $0.49*0.62=0.30$? How can they say that the answer is .42?

Also, shouldn't the $P(B)$ cancel out with the numerator in the above theorem.

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Multiplying the two probabilities only works if the two events are independent of one another. For example, the odds that two dice rolled separately will both be 6 is $\frac{1}{6}*\frac{1}{6}=\frac{1}{36}$. The two events in your problem are not independent. A team is significantly more likely to win if they have scored 5 or more runs. This is similar to calculating the probability that a die will roll an even number above three. The probability of rolling an even number is $\frac{1}{2}$, and the probability of rolling a number above 3 is also $\frac{1}{2}$. But this doesn't mean that the odds of rolling an even number above three is $\frac{1}{4}$. We can count the cases and see the the probability is $\frac{2}{6}$ or $\frac{1}{3}$. The discrepancy between the $0.30$ that you calculated and the $0.42$ that is cited in the problem represents the effect that score lots of runs has on the probability of winning a game.