Question

In a horse race there are 18 horses numbered from 1 to 18. The probability that horse 1 would win is $\frac{1}{6}$, horse 2 is $\frac{1}{10}$ and 3 is $\frac{1}{8}$. Assuming a tie is impossible, the chance that one of the three horses wins the race, is

• A. $\frac{143}{420}$
• B. $\frac{119}{120}$
• C. $\frac{47}{120}$
• D. $\frac{1}{5}$

Answer 1

The correct option is C $\frac{47}{120}$

Give,18 horses are numbered as 1,2,3,...,18 respectively.
The probability that horse1 would win the race,$P\left(H1\right)=\frac{1}{6}$
The probability that horse2 would win the race,$P\left(H2\right)=\frac{1}{10}$
The probability that horse3 would win the race,$P\left(H3\right)=\frac{1}{8}$
The probability that one of the horse1 or horse2 or horse3 win the race$=P\left(H1\right)+P\left(H2\right)+P\left(H3\right)$
$\therefore Therequiredprobability=\frac{1}{6}+\frac{1}{10}+\frac{1}{8}=\frac{47}{120}$