Five horses are in a race. Mr. A selects two of the horses at random and bets on them. The probabili

Q: Wait, why do we use combinations instead of permutations? Does the order in which Mr. A picks the horses matter?

That is a fantastic question! In this scenario, Mr. A is betting on a pair of horses. Whether he picks (Horse 1, Horse 2) or (Horse 2, Horse 1), he is holding the exact same ticket. Since the order of selection does not change the outcome of his bet, we use combinations. If the order mattered—like if he were betting on who comes in 1st and who comes in 2nd—then we would use permutations.

Q: Why can't we just say the probability is $1/5 + 1/5 = 2/5$?

I love that you are looking for intuitive shortcuts! While your intuition leads to the correct answer here, it is important to understand why. You are essentially saying there is a $1/5$ chance for the first horse to be the winner and a $1/5$ chance for the second horse to be the winner. This works because the events are mutually exclusive (only one horse can win). However, for more complex problems, always rely on the fundamental definition of probability: $\frac{\text{Favorable Outcomes}}{\text{Total Outcomes}}$. It is the safest way to avoid traps!

Q: Is there a faster way to solve this without calculating combinations?

Yes! There is a beautiful shortcut often used in competitive exams. If you are selecting $r$ items from a total of $n$ items, the probability that any *specific* item (like our winning horse) is included in your selection is always $\frac{r}{n}$. Here, $r=2$ and $n=5$, so the probability is simply $\frac{2}{5}$. It is a powerful tool to keep in your arsenal for quick verification!

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

Options:

(A) 2/5

(B) 4/5

(C) 3/5

(D) 1/5

Sigma Percentile · Accepted Answer

### The Thrill of the Track: Visualizing the Probability

Imagine you are standing at the edge of a race track. The air is electric, the crowd is roaring, and five magnificent horses—let us call them $H_1, H_2, H_3, H_4,$ and $H_5$—are lined up at the starting gate.

Mr. A, a seasoned bettor, steps up to the window. He does not know which horse will win, but he has a strategy: he selects two horses at random to bet on.

Our goal is to calculate the probability that his ticket contains the ultimate winner. This is not just a math problem; it is a story of chance, selection, and the elegance of combinatorics.

### Phase 1: Defining the Sample Space

Before we can determine the probability of winning, we must understand the universe of possibilities. Mr. A is choosing two horses out of five.

In the language of mathematics, we are looking for the number of ways to choose $r=2$ items from a set of $n=5$. Since the order of the horses on his ticket does not matter, we use the combination formula, denoted as ${}^nC_r$.

The total number of possible pairs is given by ${}^5C_2$. Let us break this down:

$$
{}^5C_2 = \frac{5!}{2!(5-2)!} = \frac{5 	imes 4}{2 	imes 1} = 10
$$

There are exactly $10$ unique pairs of horses Mr. A could walk away with. This is our sample space, the denominator of our probability fraction. It represents every possible scenario that could unfold at the betting window.

### Phase 2: The Winning Condition

Now, let us turn our attention to the race itself. In any standard race, there is exactly one winner. For the sake of our calculation, let us assume $H_1$ is the champion.

For Mr. A to win his bet, his ticket must include $H_1$. Think of this as a game of 'fixing' the winner.

If we force $H_1$ to be on Mr. A's ticket, we have already used one of his two slots. Now, he only needs to choose one more horse to complete his pair.

He can choose from the remaining four horses: $H_2, H_3, H_4,$ or $H_5$. This is represented as ${}^4C_1$, which is simply $4$.

So, the favorable outcomes are the pairs: $(H_1, H_2), (H_1, H_3), (H_1, H_4),$ and $(H_1, H_5)$. There are exactly $4$ ways for Mr. A to hold a winning ticket.

### Phase 3: The Final Calculation and the Shortcut

We have our total outcomes ($10$) and our favorable outcomes ($4$). The probability $P(E)$ is the ratio of the two:

$$
P(E) = \frac{n(E)}{n(S)} = \frac{4}{10}
$$

Simplifying this fraction, we get **$\frac{2}{5}$**.

But wait—there is a secret weapon. In the world of JEE, time is your most precious resource.

There is a powerful shortcut for this exact type of problem: whenever you select $r$ objects from a total of $n$ objects, the probability that any specific object is included in your selection is always $\frac{r}{n}$.

In our case, $r=2$ and $n=5$, so the probability is $\frac{2}{5}$. It is elegant, it is fast, and it is a concept that will serve you well throughout your journey in physics and mathematics.

Animated Solution for Mathematics - Probability: 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

Select Answer:

Visualized Solution

Visualizing the Race

Defining the Sample Space

Total Possible Selections

Calculating $^{5} C_{2}$

Identifying the Winner

Condition for Winning

Favorable Outcomes Setup

Calculating Favorable Outcomes

Probability Formula

Substituting the Values

Final Probability

Generalizing the Concept

The Sigma Insight: Classical Definition of Probability

The Thrill of the Track

Phase 1

Phase 2

Phase 3

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