The probability that GATE CS/IT question has an error is 1/20 and 65 q...
Using Binomial Distribution

Probability that x questions have an error

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The probability that GATE CS/IT question has an error is 1/20 and 65 q...
Understanding the Problem
The problem is about calculating the probability of errors in questions on a GATE CS/IT exam. Given that the probability of a question having an error is 1/20 (or 0.05), we can analyze the situation using the complementary probability.
Calculating the Complementary Probability
- The probability that a question does not have an error is 1 - (1/20) = 19/20.
- For 65 questions, the probability that none of them have errors is (19/20)^65.
- The probability that at least 2 questions have errors can be calculated as:
P(at least 2 errors) = 1 - P(0 errors) - P(1 error)
Calculating the Individual Probabilities
1. No Errors:
- P(0 errors) = (19/20)^65
2. Exactly One Error:
- For exactly one error, we choose 1 question to have an error and the rest not:
- P(1 error) = C(65, 1) * (1/20)^1 * (19/20)^64
- C(65, 1) = 65 (the number of ways to choose 1 question from 65)
Therefore, P(1 error) = 65 * (1/20) * (19/20)^64.
Final Calculation
Combining these probabilities:
P(at least 2 errors) = 1 - [(19/20)^65 + 65 * (1/20) * (19/20)^64].
This can be factored as:
P(at least 2 errors) = 1 - [a * (19/20)^64], where a = (19/20) + 65 * (1/20).
Calculating a:
- a = (19/20) + 65 * (1/20) = (19 + 65) / 20 = 84 / 20 = 4.2.
Conclusion
Thus, the value of "a" is 4.2, which matches the given correct answer.