Two teams A and B play a series of four matches. If the probability that team A wins a match is 2/3,then the probability that team A wins threematches, loses one and the third win occurs inthe fourth match isa)8/27b)16/27c)8/81d)32/81Correct answer is option 'A'. Can you explain this answer?

Question

Answer

To find the probability that team A wins three matches, loses one, and the third win occurs in the fourth match, we can break it down into smaller probabilities and then multiply them together.

Let's analyze each condition step by step:

1. Probability of team A winning a match:
Given that the probability of team A winning a match is 2/3, we can say that the probability of team A losing a match is 1 - 2/3 = 1/3.

2. Probability of team A winning three matches and losing one:
We need to calculate the probability of team A winning three matches and losing one match. Since the order of the wins and loss matters in this case, we can calculate it using the binomial probability formula.

The probability of winning three matches out of four is calculated as:
P(Winning 3 matches) = (4C3) * (2/3)^3 * (1/3)^1
Here, (4C3) represents the number of ways to choose 3 matches out of 4, which is 4.

So, P(Winning 3 matches) = 4 * (2/3)^3 * (1/3)^1 = 32/81.

3. Probability of the third win occurring in the fourth match:
Since we know that team A wins three matches and loses one, the third win can only occur in the fourth match. So, the probability of the third win occurring in the fourth match is 1.

4. Multiplying the probabilities:
To find the probability that team A wins three matches, loses one, and the third win occurs in the fourth match, we multiply the probabilities calculated in steps 2 and 3.

P(Winning 3 matches, losing 1, and third win in fourth match) = P(Winning 3 matches) * P(Third win occurs in fourth match)
= (32/81) * 1 = 32/81.

Hence, the correct answer is option A, 32/81.

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