In a horse race, there are 18 horses numbered from 1 to 18. The probability that horse 1 would win is $ \dfrac{1}{6} $ , horse 2 is $ \dfrac{1}{10} $ and horse 3 is $ \dfrac{1}{8} $ . Assuming that a tie is impossible, the chance that one of the three horses wins the race is
(a) $ \dfrac{143}{420} $
(b) $ \dfrac{119}{120} $
(c) $ \dfrac{47}{420} $
(d) $ \dfrac{1}{5} $

Question

In a horse race, there are 18 horses numbered from 1 to 18. The probability that horse 1 would win is $ \dfrac{1}{6} $ , horse 2 is $ \dfrac{1}{10} $ and horse 3 is $ \dfrac{1}{8} $ . Assuming that a tie is impossible, the chance that one of the three horses wins the race is
(a) $ \dfrac{143}{420} $
(b) $ \dfrac{119}{120} $
(c) $ \dfrac{47}{420} $
(d) $ \dfrac{1}{5} $

Vedantu Content Team · Accepted Answer

Hint:In this question, we are given the probability of winning each horse. Also, we should note that as a tie is impossible, only one horse can win the race. Therefore, the winning of one horse is a mutually exclusive event of winning another horse. Therefore, the probability of winning one of the three horses should be equal to the sum of the probabilities of winning of the individual horses. Therefore, we can add the given probabilities of winning of each horse to obtain the required answer.

Complete step-by-step answer:
We are given that probability that horse 1 would win is $ \dfrac{1}{6} $ , horse 2 is $ \dfrac{1}{10} $ and horse 3 is $ \dfrac{1}{8} $ …………..(1.1)
Also, it is given that a tie is impossible. Therefore, two horses cannot win at the same time. However, we should note that in a race, one of the horses should win. This means that the events are mutually exclusive, i.e. if one horse wins, the other horses lose.
As we know that the probability of occurring of either of a set of mutually exclusive events is equal to the sum of their individual probabilities, from (1.1), we can write that
The probability that one horse wins= Sum of the individual probabilities of winning of the horses
$ =\dfrac{1}{6}+\dfrac{1}{10}+\dfrac{1}{8}=\dfrac{80+48+60}{480}=\dfrac{188}{480}=\dfrac{47\times 4}{120\times 4}=\dfrac{47}{120} $
Which matches option (c). Therefore, option (c) is the correct answer to this question.

Note:We should note that we could add the individual probabilities to get the answer considering that two horses cannot win at the same time. Otherwise we have to consider the probabilities of all events in which at least one of three horses wins i.e. all three horses win, horse 1 wins but horse 2 and 3 loose and so on to get the answer to this question.