Question

Five horses are in a race. Mr. A selects two of the horses at random and bets on them. The probability that Mr. A selected the winning horse is:
A. $\dfrac{3}{5}$
B. $\dfrac{1}{5}$
C. $\dfrac{2}{5}$
D. $\dfrac{4}{5}$

Answer 1

Hint: Here we can find the probability that none of the horses that Mr. A selects wins. After that we can subtract this from $1$ and get the probability that one of the selected horses win because we know that the sum of the probabilities is $1$.

Complete Step by Step Solution:
Here we are given that five horses are in a race and Mr. A selects two of the horses at random and bets on them.
Let the five horses that are there in the race be ${H_1},{H_2},{H_3},{H_4},{H_5}$
Now let us suppose that probability of the event where one of the horse that is selected by Mr. A wins be $P\left( A \right)$ and let us say that probability that none of the horses that is selected by Mr. A wins be $P\left( {\overline A } \right)$.
Now we can say that $P\left( A \right) + P\left( {\overline A } \right) = 1$
Hence we can find the probability that none of the horses that are selected wins.
This can be found by multiplying the probabilities of the events that the first horse selected does not win and then the probability that out of the rest $4$ even the second selected horse also does not win.
$P\left( {\overline A } \right) = P\left( {{\text{first selected horse does not win}}} \right) \times P\left( {{\text{second selected horse out of left 4 does not win}}} \right)$$ - - (1)$
$P\left( {{\text{first selected horse does not win}}} \right) = \dfrac{{{\text{number of left horse}}}}{{{\text{total number of students}}}} = \dfrac{4}{5}$.
$P\left( {{\text{second selected horse also out of rest 4 does not win}}} \right) = \dfrac{{{\text{total horses}} - {\text{two horse that Mr A selected}}}}{{{\text{total horses}} - {\text{first selected horse}}}}$$P\left( {{\text{second selected horse also out of rest 4 does not win}}} \right) = \dfrac{{5 - 2}}{{5 - 1}} = \dfrac{3}{4}$.
Let us substitute these values in the equation (1) and we will get:
$P\left( {\overline A } \right) = \dfrac{4}{5} \times \dfrac{3}{4} = \dfrac{3}{5}$
Now we have got the value of $P\left( {\overline A } \right)$
We also know that $P\left( A \right) + P\left( {\overline A } \right) = 1$
We need to find the value of $P\left( A \right)$ so let us substitute the value we have found in the above equation we will get:
$
  P\left( A \right) + \dfrac{3}{5} = 1 \\
  P\left( A \right) = 1 - \dfrac{3}{5} = \dfrac{2}{5} \\
 $
Hence we get that probability that he selects the winning horse$ = \dfrac{2}{5}$

Hence option C) is the correct option.

Note:
Here we can also find the total cases by selecting $2{\text{ out of 5}}$ by ${}^5{C_2} = 10$ now we have got the total cases as $10$ and as one out of those $2$ has to win then one out of the other four can be selected in $4$ ways.
So required probability$ = \dfrac{4}{{10}} = \dfrac{2}{5}$.