1. Key Concepts and Formulas

• Probability of an Event ($P(E)$): For any event $E$, its probability is defined as the ratio of the number of favorable outcomes for $E$ to the total number of possible outcomes in the sample space. $P(E) = {{\text{Number of favorable outcomes for } E} \over {\text{Total number of possible outcomes}}}$
• Complementary Probability: If $E$ is an event, its complement, denoted as $\overline{E}$ (or $E'$), is the event that $E$ does not occur. The sum of the probabilities of an event and its complement is always 1: $P(E) + P(\overline{E}) = 1$ This implies $P(E) = 1 - P(\overline{E})$, which is a very useful formula when it's simpler to calculate the probability of an event not happening.
• Combinations ($\binom{n}{k}$): When selecting $k$ items from a set of $n$ distinct items, and the order of selection does not matter, the number of ways to do this is given by the combination formula: $\binom{n}{k} = {{n!} \over {k!(n-k)!}}$

2. Step-by-Step Solution

Understanding the Problem and Defining Events: We have 5 horses in a race. Mr. A selects 2 horses at random. We need to find the probability that one of the horses Mr. A selected is the winning horse. Let's define our events:

• Let $E$ be the event that "Mr. A selected the winning horse."
• Let $\overline{E}$ be the event that "Mr. A did not select the winning horse." This means both horses he selected are non-winning horses.

We will use complementary probability because calculating $P(\overline{E})$ is simpler. It involves only one scenario: Mr. A picks two non-winning horses. Calculating $P(E)$ directly would involve Mr. A picking exactly one winning horse and one non-winning horse.

Step 1: Calculate the total number of ways Mr. A can select 2 horses from 5.

• This represents the total number of possible outcomes (our sample space). Since the order in which Mr. A picks the horses does not matter for his bet, we use combinations.
• Total number of horses ($n$) = 5
• Number of horses Mr. A selects ($k$) = 2
• Total possible outcomes = $\binom{5}{2}$. $\binom{5}{2} = {{5!} \over {2!(5-2)!}} = {{5!} \over {2!3!}} = {{5 \times 4 \times 3!} \over {(2 \times 1) \times 3!}} = {{5 \times 4} \over 2} = 10$ So, there are 10 distinct pairs of horses Mr. A could have selected.

Step 2: Calculate the number of ways Mr. A can select 2 horses such that NEITHER is the winning horse (Favorable outcomes for event $\overline{E}$).

• For Mr. A not to select the winning horse, both of his chosen horses must come from the pool of non-winning horses.
• Number of winning horses = 1
• Number of non-winning horses = Total horses - Winning horses = $5 - 1 = 4$.
• Mr. A needs to select 2 horses from these 4 non-winning horses.
• Number of favorable outcomes for $\overline{E}$ = $\binom{4}{2}$. $\binom{4}{2} = {{4!} \over {2!(4-2)!}} = {{4!} \over {2!2!}} = {{4 \times 3 \times 2!} \over {(2 \times 1) \times 2!}} = {{4 \times 3} \over 2} = 6$ Thus, there are 6 ways Mr. A could have selected two horses, neither of which is the winner.

Step 3: Calculate the probability of event $\overline{E}$ (Mr. A did not select the winning horse).

• Using the probability formula: $P(\overline{E}) = {{\text{Number of favorable outcomes for } \overline{E}} \over {\text{Total number of possible outcomes}}} = {6 \over {10}} = {3 \over 5}$ This means there is a $\frac{3}{5}$ probability that Mr. A fails to select the winning horse.

Step 4: Calculate the probability of event $E$ (Mr. A selected the winning horse).

• Using the complementary probability rule: $P(E) = 1 - P(\overline{E})$ $P(E) = 1 - {3 \over 5}$ $P(E) = {5 \over 5} - {3 \over 5}$ $P(E) = {2 \over 5}$ Therefore, the probability that Mr. A selected the winning horse is $\frac{2}{5}$.

3. Common Mistakes & Tips

• Embrace Complementary Probability: Always consider if calculating $P(\overline{E})$ is simpler than calculating $P(E)$ directly. This is a common and powerful strategy in JEE problems, especially when $P(E)$ involves multiple cases.
• Combinations vs. Permutations: This is a crucial distinction. Use combinations ($\binom{n}{k}$) when the order of selection does not matter (e.g., selecting a group of horses for a bet). Use permutations ($P(n,k)$) when the order does matter (e.g., arranging horses in specific finishing positions). In this problem, Mr. A just "selects" two horses, implying order doesn't matter for his bet.
• Direct Calculation as a Check: For verification, you could also calculate $P(E)$ directly: Mr. A must select 1 winning horse from the 1 available, AND 1 non-winning horse from the 4 available. $P(E) = {{\binom{1}{1} \times \binom{4}{1}} \over {\binom{5}{2}}} = {{1 \times 4} \over {10}} = {4 \over {10}} = {2 \over 5}$ Both methods consistently yield the same result, confirming the answer.

4. Summary

This problem is a straightforward application of probability principles, particularly the concept of complementary probability combined with combinations. By first calculating the probability that Mr. A does not select the winning horse (i.e., he selects two non-winning horses), we could easily determine the probability of him selecting the winning horse. The use of combinations ensured accurate counting of the total possible selections and the specific unfavorable selections.

5. Final Answer

The final answer is $\boxed{{2 \over 5}}$, which corresponds to option (A).