In a horse racing competition there were 18 numbered 1 to 18.the organisers assigned a probability of winning the race to each horse such that horse1 would win is 1/7,for horse 2 it is 1/8 and for horse 3 it is 1/8.Assuming that tie is not possible.find chance that one of these three will win the race?
Solution(By Examveda Team)
Probability of Horse 1, 2 and 3 winning is 1/7, 1/8, 1/8 respectively.
Probability of Horse 1, 2 and 3 not winning is (1-1/7), (1-1/8), (1-1/8) i.e., 6/7, 7/8,7/8 respectively.
There are three possible situations which will satisfy the given conditions:
a) Horse 1 win and horses 2 and 3 lose. Probability of this event happening = (1/7*7/8*7/8) = 7/64
b) Horse 2 win and horses 1 and 3 lose. Probability of this event happening = (6/7*1/8*7/8) = 6/64
c) Horse 3 win and horses 1 and 2 lose. Probability of this event happening = (6/7*7/8*1/8) = 6/64
Hence, the required probability is = (7/64 + 6/64 + 6/64) = 19/64.
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