Question

question
this season, the probability that the yankees will win a game is 0.6 and the probability that the yankees will score 5 or more runs in a game is 0.44. the probability that the yankees lose and score fewer than 5 runs is 0.32. what is the probability that the yankees would score fewer than 5 runs when they lose the game?
round your answer to the nearest thousandth.
answer attempt 1 out of 2
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Explanation:

Step1: Find probability of losing

The probability of winning is \( 0.6 \), so the probability of losing is \( 1 - 0.6 = 0.4 \).

Step2: Apply conditional probability formula

Let \( A \) be the event of scoring fewer than 5 runs, and \( B \) be the event of losing. We need \( P(A|B)=\frac{P(A\cap B)}{P(B)} \). We know \( P(A\cap B) = 0.32 \) and \( P(B)=0.4 \). So \( P(A|B)=\frac{0.32}{0.4}=0.8 \). Wait, no, wait—wait, the problem says "the probability that the Yankees lose and score fewer than 5 runs is 0.32". Wait, no, let's re - read. Wait, the problem: "the probability that the Yankees will win a game is 0.6 and the probability that the Yankees will score 5 or more runs in a game is 0.44. The probability that the Yankees lose and score fewer than 5 runs is 0.32. What is the probability that the Yankees would score fewer than 5 runs when they lose the game?"

So, let's define:

Let \( L \) be the event that the Yankees lose, \( W \) be the event that they win. \( S\geq5 \) be scoring 5 or more, \( S < 5 \) be scoring fewer than 5.

We know \( P(W)=0.6 \), so \( P(L)=1 - P(W)=1 - 0.6 = 0.4 \).

We know \( P(L\cap S < 5)=0.32 \).

We need to find \( P(S < 5|L)=\frac{P(L\cap S < 5)}{P(L)} \) (by the definition of conditional probability \( P(A|B)=\frac{P(A\cap B)}{P(B)} \) where \( A = S < 5 \), \( B = L \))

So \( P(S < 5|L)=\frac{0.32}{0.4}=0.8 \)? Wait, no, wait, 0.32 divided by 0.4: \( 0.32\div0.4 = 0.8 \)? Wait, 0.4 times 0.8 is 0.32. But let's check again. Wait, maybe I misread the numbers. Wait, the problem says "the probability that the Yankees lose and score fewer than 5 runs is 0.32". And \( P(L)=0.4 \). So \( P(S < 5|L)=\frac{0.32}{0.4}=0.8 \). But let's confirm the formula. Conditional probability: \( P(A|B)=\frac{P(A\cap B)}{P(B)} \), where \( A \) is scoring fewer than 5 runs, \( B \) is losing. So yes, \( P(S < 5|L)=\frac{P(S < 5\cap L)}{P(L)}=\frac{0.32}{0.4}=0.8 \). Wait, but the problem says "round your answer to the nearest thousandth". 0.8 is 0.800 when rounded to the nearest thousandth? Wait, no, 0.8 is exactly 0.800 when rounded to three decimal places. Wait, but maybe I made a mistake. Wait, let's re - check the problem statement again.

Wait, the problem: "This season, the probability that the Yankees will win a game is 0.6 and the probability that the Yankees will score 5 or more runs in a game is 0.44. The probability that the Yankees lose and score fewer than 5 runs is 0.32. What is the probability that the Yankees would score fewer than 5 runs when they lose the game? Round your answer to the nearest thousandth."

So, \( P(L)=1 - 0.6 = 0.4 \)

\( P(S < 5\cap L)=0.32 \)

\( P(S < 5|L)=\frac{P(S < 5\cap L)}{P(L)}=\frac{0.32}{0.4}=0.8 \). When rounded to the nearest thousandth, 0.8 is 0.800? Wait, no, 0.8 is already to one decimal place. But 0.8 as a decimal to the thousandth place is 0.800. Wait, but maybe I messed up the numbers. Wait, is the probability of losing and scoring fewer than 5 runs 0.32, and the probability of losing is 0.4. So 0.32 divided by 0.4 is 0.8. So the answer is 0.800 when rounded to the nearest thousandth? Wait, no, 0.8 is equal to 0.800 when considering three decimal places.

Wait, maybe I made a mistake in the problem interpretation. Let's re - read:

"The probability that the Yankees will win a game is 0.6 and the probability that the Yankees will score 5 or more runs in a game is 0.44. The probability that the Yankees lose and score fewer than 5 runs is 0.32. What is the probability that the Yankees would score fewer than 5 runs when they lose the game? Round your answer to the nearest thousandth."

So, the formula for conditional probability is correct. \( P(A|B)=\frac{P(A\cap B)}{P(B)} \). Here, \( A \): score fewer than 5 runs, \( B \): lose the game. \( P(B)=1 - P(win)=1 - 0.6 = 0.4 \). \( P(A\cap B)=0.32 \). So \( P(A|B)=\frac{0.32}{0.4}=0.8 \). So when rounded to the nearest thousandth, it's 0.800? Wait, no, 0.8 is 0.800 in three decimal places. Wait, but maybe the problem has a different approach. Wait, maybe the "probability that the Yankees will score 5 or more runs in a game is 0.44" is a red herring? Because we are asked about scoring fewer than 5 runs when they lose, and we are given the probability of losing and scoring fewer than 5 runs, and the probability of losing. So yes, the conditional probability formula applies here. So \( \frac{0.32}{0.4}=0.8 \), which is 0.800 when rounded to the nearest thousandth.

Wait, but let's check the division again: 0.32 divided by 0.4. 0.4 times 0.8 is 0.32. So yes, that's correct.

Answer:

\( 0.800 \)