Probabilities of three students A, B and C to passan examination arc respectively 1/3,1/4 and 1/5.The probability that exactly one student will pass is :a)5/12b)7/30c)13/30d)3/5Correct answer is option 'C'. Can you explain this answer?

Question

Answer

Required probability

Answer

Understanding the Problem
We need to calculate the probability that exactly one of the three students (A, B, C) passes the examination. The probabilities of passing for each student are:
- A: 1/3
- B: 1/4
- C: 1/5
Calculating Individual Probabilities
First, we determine the probabilities that each student fails:
- Probability that A fails: 1 - 1/3 = 2/3
- Probability that B fails: 1 - 1/4 = 3/4
- Probability that C fails: 1 - 1/5 = 4/5
Scenarios for Exactly One Pass
There are three scenarios where exactly one student passes the exam:
1. A passes, B and C fail
- Probability = (1/3) * (3/4) * (4/5) = 1/3 * 3/4 * 4/5 = 1/5
2. B passes, A and C fail
- Probability = (1/4) * (2/3) * (4/5) = 1/4 * 2/3 * 4/5 = 2/15
3. C passes, A and B fail
- Probability = (1/5) * (2/3) * (3/4) = 1/5 * 2/3 * 3/4 = 1/10
Summing the Probabilities
Now, we add the probabilities of these three scenarios:
- Probability (exactly one passes) = (1/5) + (2/15) + (1/10)
To add these fractions, we can find a common denominator:
- Common denominator = 30
Converting each fraction:
- 1/5 = 6/30
- 2/15 = 4/30
- 1/10 = 3/30
Now, summing these:
- Total = 6/30 + 4/30 + 3/30 = 13/30
Conclusion
Thus, the probability that exactly one student will pass the examination is 13/30, aligning with option 'C'.

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