Question 379323: A newspaper advertises six different concerts, four different plays, and two different basketball games. If a couple selects three activities at random, find the probability that they will attened:
a. Three Concerts
B. Two plays and one basketball game
C One movie, one play, and one basketball game
Answer by ewatrrr(24785)   (Show Source):
You can put this solution on YOUR website!
Hi,
12 different choices of which 6 are different concerts P(a concert) = 6/12 = 1/2
P(attending three concerts) = 6/12 * 5/11 * 4/10
12 different choices of which 4 are different plays and 2 are different BB games
P(Two plays and one basketball game) = 4/12 * 3/11 * 2/10
P(One movie, one play, and one basketball game) = 0 (movie not a option)
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