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The probability of India winning a test ...

The probability of India winning a test match againest England is `(2)/(3)`. Assuming independence from match to match, the probability that in a 7 match series India's third win occurs at the fifth match, is

A

`(8)/(27)`

B

`(16)/(81)`

C

`(8)/(81)`

D

`(32)/(81)`

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem, we need to find the probability that India's third win occurs at the fifth match in a series of seven matches. We can use the concept of the negative binomial distribution for this scenario. ### Step-by-Step Solution: 1. **Identify the probabilities**: - The probability of India winning a match, \( p = \frac{2}{3} \). - The probability of India losing a match, \( q = 1 - p = \frac{1}{3} \). 2. **Understand the scenario**: - We want the third win to occur on the fifth match. This means that in the first four matches, India must have exactly 2 wins and 2 losses. 3. **Calculate the number of ways to arrange the wins and losses**: - The number of ways to choose 2 wins from the first 4 matches can be calculated using the binomial coefficient: \[ \text{Number of ways} = \binom{4}{2} = \frac{4!}{2! \cdot 2!} = 6 \] 4. **Calculate the probability of a specific arrangement**: - The probability of winning 2 matches and losing 2 matches in the first four matches is given by: \[ P(\text{2 wins, 2 losses}) = p^2 \cdot q^2 = \left(\frac{2}{3}\right)^2 \cdot \left(\frac{1}{3}\right)^2 = \frac{4}{9} \cdot \frac{1}{9} = \frac{4}{81} \] 5. **Include the probability of winning the fifth match**: - Since the third win occurs at the fifth match, we multiply the probability of the first four matches by the probability of winning the fifth match: \[ P(\text{3rd win at 5th match}) = \text{Number of ways} \cdot P(\text{2 wins, 2 losses}) \cdot P(\text{win at 5th match}) \] \[ = 6 \cdot \frac{4}{81} \cdot \frac{2}{3} \] 6. **Calculate the final probability**: - Now calculate the final probability: \[ = 6 \cdot \frac{4}{81} \cdot \frac{2}{3} = 6 \cdot \frac{8}{243} = \frac{48}{243} = \frac{16}{81} \] ### Final Answer: The probability that India's third win occurs at the fifth match is \( \frac{16}{81} \).

To solve the problem, we need to find the probability that India's third win occurs at the fifth match in a series of seven matches. We can use the concept of the negative binomial distribution for this scenario. ### Step-by-Step Solution: 1. **Identify the probabilities**: - The probability of India winning a match, \( p = \frac{2}{3} \). - The probability of India losing a match, \( q = 1 - p = \frac{1}{3} \). ...
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MHTCET PREVIOUS YEAR PAPERS AND PRACTICE PAPERS-BINOMIAL DISTRIBUTION-Exercise 2 (MISCELLANEOUS PROBLEM) (Mean and Variance
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  4. If X is a binomial variate with the range {0, 1, 2, 3, 4, 5, 6} and P...

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  5. A dice has four blank faces and two faces marked 3. The change of gett...

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  6. If a dice is thrown twice, the probability of occurrence of 4 atleast ...

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  7. If X be binomial distribution with mean np and variance npq, then find...

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  8. If X is a binomial variate with the range {0, 1, 2, 3, 4, 5, 6} and P...

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  9. A man takes a step forward with probability 0.4 and backward with prob...

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  10. A man takes a forward step with probability (.8) and backward step wit...

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  11. The probability of India winning a test match againest England is (2)/...

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  12. If X and Y are independent binomial vatiates B(5,1/2) and B(7,1/2) and...

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  13. If Aa n dB each toss three coins. The probability that both get the sa...

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  14. A die is thrown 5 times. Find the probability that an odd number will ...

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  15. If X follows a binomial distribution with parameters n=8 and p=1//2, t...

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  16. A die is tossed thrice. If event of getting an even number is a succes...

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  17. The probability that in a family of 5 members,exactly two members have...

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  18. A fair die is thrown 20 times. The probability that on the 10th thr...

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  19. If the mean and the variance of a binomial variable X are 2 and 1 resp...

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  20. From a lot of 15 bulbs which include 5 defectives, a sample of 4 bu...

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