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What is the formula if you have 20 people on a basketball team 45 people on a football team 10 are on both--how many ways can a person be selected from either team?
Let P(A) = 1/10; P(A) = probability of selecting one people on a Basketball team
P(B) = 1/35; P(B) = probability of selecting one people on a football team
P(C) = 1/10 = probability of selecting one people who plays in both teams
P(D) = probability of selecting from either team.
P(D) = P(A) + P(B) - P(C)
P(D) = 1/10 + 1/35 - 1/10
P(D) = 1/35 or 0.0286
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