The overall percentage of failure in a certain examination is 0.30. What is theprobability that out of a group of 6 candidates at least 4 passed the examination?(a) 0.74(b) 0.71(c) 0.59(d) 0.67
Question
The overall percentage of failure in a certain examination is 0.30. What is theprobability that out of a group of 6 candidates at least 4 passed the examination?(a) 0.74(b) 0.71(c) 0.59(d) 0.67
Solution
To solve this problem, we need to use the binomial probability formula:
P(X=k) = C(n, k) * (p^k) * ((1-p)^(n-k))
where:
- P(X=k) is the probability of k successes in n trials
- C(n, k) is the combination of n items taken k at a time
- p is the probability of success
- (1-p) is the probability of failure
- n is the number of trials
- k is the number of successes
In this case, the probability of passing (success) is 1 - 0.30 = 0.70 (since the probability of failure is 0.30). We want to find the probability that at least 4 out of 6 candidates pass, which means 4, 5, or 6 candidates could pass.
So, we need to calculate P(X=4) + P(X=5) + P(X=6).
P(X=4) = C(6, 4) * (0.70^4) * ((1-0.70)^(6-4)) P(X=5) = C(6, 5) * (0.70^5) * ((1-0.70)^(6-5)) P(X=6) = C(6, 6) * (0.70^6) * ((1-0.70)^(6-6))
Adding these three probabilities together will give us the total probability that at least 4 out of 6 candidates pass the exam.
After calculating, you will find that the answer is approximately 0.7443, which is closest to (a) 0.74.
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