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13 At the end of the 2002 Major League Baseball season eight of the thirty teams

ID: 3404955 • Letter: 1

Question

13 At the end of the 2002 Major League Baseball season eight of the thirty teams had lost 90 or more games. Twelve of the thirty teams had a total home attendance of under two million fans. Four of the 30 teams (Tampa Bay, Kansas City, Detroit and Milwaukee) had a home attendance under two million and lost 90 or more games. a) If a team is chosen at random from the thirty teams, what is the chance that it either lost 90 games or more during 2002 or had a home atten- dance under 2 million? b) If a team is chosen at random from the thirty teams, what is the chance that it either had a home attendance over 2 million or lost less than 90 games? c) If a team is chosen at random from the thirty teams, what is the chance that it had a home attendance over 2 million and lost less than 90 games?

Explanation / Answer

Let A be the event that the team lost 90 or more games

B be the event that there was a home attendance under two million fans.

Given P(A) = 8/30 = 4/15

P(B) = 12/30 = 2/5

P(A and B) = 4/30 = 2/15

a) We need to find the chance that either the team lost 90 percent r more games or had a home attendance under two million

We need P(A U B) = P(A) + P(B) - P(A and B) = 4/15 + 2/5 - 2/15 = 8/15

Required probability = 8/15

b) P(~ A or ~ B) = 1-P(A and B)= 1 - 2/15 = 13/15

c) P(~A and ~B) = 1- P(AUB) = 1- 8/15= 7/15

We use De Morgan's laws of set theory in the last two parts.

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