The probability of India winning a test match against West Indies is 1/2. Assuming independence from

Q: Wait, why do we ignore matches 4 and 5? Shouldn't they be part of the calculation?

That is a great question! The condition is that the *second* win occurs at the *third* test. Once that condition is met, the outcome of the 4th and 5th matches does not change the fact that the second win happened at the third test. We are calculating the probability of a specific event sequence, and the subsequent matches are irrelevant to this specific constraint.

Q: Why can't we just use the Binomial Distribution formula $P(X=k) = \binom{n}{k} p^k q^{n-k}$?

The Binomial Distribution tells us the probability of getting exactly $k$ successes in $n$ trials, regardless of the order. Here, the order is strictly fixed: the 3rd match *must* be the 2nd win. Binomial distribution doesn't account for specific positional constraints like 'the 2nd win must be at the 3rd match'.

Q: Why do we add the probabilities of the two cases?

We add them because these two cases—(Win, Lose, Win) and (Lose, Win, Win)—are mutually exclusive. They cannot both happen at the same time. In probability, when you have mutually exclusive scenarios that satisfy your condition, you sum their individual probabilities to find the total probability.

Question

The probability of India winning a test match against West Indies is $1/2$. Assuming independence from match to match the probability that in a $5$ match series India's second win occurs at third test is

Options:

(A) $1/8$

(B) $1/4$

(C) $1/2$

(D) $2/3$

Sigma Percentile · Accepted Answer

### The Art of the Fixed Event: Mastering Probability Constraints

Imagine you are sitting in the stands, the tension of a five-match cricket series hanging in the air. You are not just watching the game; you are analyzing the very fabric of chance.

The problem before us is a classic JEE-style puzzle that tests your ability to look past the noise and focus on the rigid constraints of a scenario. We are told the probability of India winning a match is $P(W) = \frac{1}{2}$, and consequently, the probability of losing is $P(L) = 1 - \frac{1}{2} = \frac{1}{2}$.

The series consists of five matches, but here is the twist: we need the second win to occur exactly at the third test. Let us break this down.

### Phase 1: The Anchor of the Third Match

The most critical step in solving any probability problem with a specific condition is to identify the 'anchor' or the 'fixed event'. The condition states that the second win occurs at the third test. This is not a suggestion; it is a mathematical guarantee.

It means that for our scenario to be valid, the third match **must** be a win. We can lock this in immediately.

If the third match were a loss, the condition would be violated. If the third match were the first win, the condition would be violated. By fixing the third match as a win, we have simplified our universe of possibilities significantly.

### Phase 2: The 'Before' Story

Now, let us look at the first two matches. If the third match is the second win, what must have happened in the first two?

We need exactly one win and one loss in the first two matches. Why? Because if India had won both, the third match would be the third win, not the second.

If they had lost both, the third match would be the first win. Therefore, the logic dictates that we must have exactly one win and one loss in the first two matches. This is the heart of the problem—transforming a complex series into a smaller, manageable puzzle.

### Phase 3: Identifying the Paths

How can we achieve exactly one win in two matches? There are only two ways this can happen:

1. Case 1: India wins the first match and loses the second. The sequence is $(W, L)$.
2. Case 2: India loses the first match and wins the second. The sequence is $(L, W)$.

Now, we append our 'anchor'—the third match, which we know must be a win. So our two valid sequences for the first three matches are $(W, L, W)$ and $(L, W, W)$.

### Phase 4: The Calculation

Because the matches are independent, the probability of a sequence is simply the product of the individual probabilities. For Case 1, the sequence is $(W, L, W)$.

The probability is:

$$P(W, L, W) = P(W) 	imes P(L) 	imes P(W) = \frac{1}{2} 	imes \frac{1}{2} 	imes \frac{1}{2} = \frac{1}{8}$$

For Case 2, the sequence is $(L, W, W)$. The probability is:

$$P(L, W, W) = P(L) 	imes P(W) 	imes P(W) = \frac{1}{2} 	imes \frac{1}{2} 	imes \frac{1}{2} = \frac{1}{8}$$

### Phase 5: The Final Synthesis

We have two mutually exclusive paths that satisfy our condition. To find the total probability, we add them together:

$$Total \ Probability = P(W, L, W) + P(L, W, W) = \frac{1}{8} + \frac{1}{8} = \frac{2}{8} = \frac{1}{4}$$

And there you have it! The beauty of this solution lies in its simplicity.

We ignored the 4th and 5th matches because, once the condition is met at the 3rd match, the rest of the series is irrelevant to the specific question asked. This is the JEE mindset: identify the constraint, isolate the relevant events, and calculate with precision.

You have just mastered the art of the fixed event, and the final answer is **$\frac{1}{4}$**.

Animated Solution for Mathematics - Probability: The probability of India winning a test match against West Indies is $1/2$ . Assuming independence from match to match the probability that in a $5$ match series India's second win occurs at third test is

Select Answer:

Visualized Solution

Visualizing the Series

The "Second Win" Condition

Analyzing the First Two Matches

Identifying Valid Sequences

Probability of Case 1

Probability of Case 2

Final Probability

The Sigma Insight: Random Variables and Probability Distributions

The Art of the Fixed Event

Phase 1

Phase 2

Phase 3

Phase 4

Phase 5

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