Question

Your favorite team is in the final playoffs. You have assigned a probability of 60% that it will win the championship. Past records indicate that when teams win the championship, they win the first game of the series 70% of the time. When they lose the series, they win the first game 25% of the time. The first game is over; your team has lost. What is the probability that it will win the series?

Answer 1

Answer on Question #61185 – Math – Statistics and Probability

Question

Your favorite team is in the final playoffs. You have assigned a probability of $60\%$ that it will win the championship. Past records indicate that when teams win the championship, they win the first game of the series $70\%$ of the time. When they lose the series, they win the first game $25\%$ of the time. The first game is over; your team has lost. What is the probability that it will win the series?

Solution

First we will define events. Let $W$ be the event that your team wins the championship and $F$ be the event that they win the first game (so $W^c$ is the event they lose the championship and $F^c$ is the event that they lose the first game). Given

$$P (W) = 0. 6, \quad P (F | W) = 0. 7, \quad P (F | W ^ {c}) = 0. 2 5,$$

and using the formulae $P(A) + P(A^c) = 1$ and $P(B^{c}|A) + P(B|A) = 1$ , we find

$$P \left(W ^ {c}\right) = 1 - P (W) = 1 - 0. 6 = 0. 4, P \left(F ^ {c} | W\right) = 1 - P (F | W) = 1 - 0. 7 = 0. 3,$$ $$P \left(F ^ {c} \mid W ^ {c}\right) = 1 - P \left(F \mid W ^ {c}\right) = 1 - 0. 2 5 = 0. 7 5,$$

and complete the following probability tree:

Now, since we want to compute $P(W|F^c) = \frac{P(W \cap F^c)}{P(F^c)}$ , we only need to find $P(W \cap F^c)$ and $P(F^c)$ . Next,

$$P (W \cap F ^ {c}) = P (F ^ {c} | W) P (W) = 0. 3 \cdot 0. 6 = 0. 1 8.$$

We note that $P(F^c) = P(W \cap F^c) + P(W^c \cap F^c)$ (the team either loses the first game and wins the series or loses the first game and loses the series, and these are mutually exclusive events).

We already know $P(W \cap F^c)$ , so we only need to find $P(W^c \cap F^c)$ :

$$P \left(W ^ {c} \cap F ^ {c}\right) = P \left(F ^ {c} \mid W ^ {c}\right) P \left(W ^ {c}\right) = 0. 7 5 \cdot 0. 4 = 0. 3.$$

So $P(F^c) = P(W \cap F^c) + P(W^c \cap F^c) = 0.18 + 0.3 = 0.48$ . Thus,

$$P (W | F ^ {c}) = \frac {P (W \cap F ^ {c})}{P (F ^ {c})} = \frac {0 . 1 8}{0 . 4 8} = 0. 3 7 5.$$

Answer: 0.375.

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