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100 students appeared in two examinations. 60 passed the first, 50 passed the other and 30 passed both. Find the probability that a student selected at random has failed in exactly one.
- a)1/6
- b)1/2
- c)1/3
- d)1/4
- e)1/5
Correct answer is option 'B'. Can you explain this answer?
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100 students appeared in two examinations. 60 passed the first, 50 pas...
To solve this problem, we can use the principle of inclusion-exclusion.
Let's denote the event of passing the first examination as A and the event of passing the second examination as B. We are asked to find the probability that a student has failed in exactly one of the exams, which is the probability of the event (A ∩ ¬B) ∪ (¬A ∩ B), where ¬A represents the complement of event A (failing the first exam) and ¬B represents the complement of event B (failing the second exam).
We are given that 60 students passed the first exam (A), 50 students passed the second exam (B), and 30 students passed both exams (A ∩ B).
To find the probability of (A ∩ ¬B) ∪ (¬A ∩ B), we need to subtract the probability of (A ∩ B) from the sum of the probabilities of A ∩ ¬B and ¬A ∩ B.
Let's calculate each of these probabilities:
- Probability of A ∩ ¬B: This represents the event of passing the first exam (A) but failing the second exam (¬B). Since 60 students passed the first exam and 30 passed both exams, there are 60 - 30 = 30 students who passed the first exam but failed the second exam. So, the probability of A ∩ ¬B is 30/100.
- Probability of ¬A ∩ B: This represents the event of failing the first exam (¬A) but passing the second exam (B). Similarly, since 50 students passed the second exam and 30 passed both exams, there are 50 - 30 = 20 students who failed the first exam but passed the second exam. So, the probability of ¬A ∩ B is 20/100.
- Probability of A ∩ B: This represents the event of passing both exams (A ∩ B), which we are given as 30/100.
Now, let's calculate the probability of (A ∩ ¬B) ∪ (¬A ∩ B):
(A ∩ ¬B) ∪ (¬A ∩ B) = (A ∩ ¬B) + (¬A ∩ B) - (A ∩ B)
= 30/100 + 20/100 - 30/100
= 20/100
= 1/5
Therefore, the probability that a student selected at random has failed in exactly one of the exams is 1/5, which corresponds to option E.
Let's denote the event of passing the first examination as A and the event of passing the second examination as B. We are asked to find the probability that a student has failed in exactly one of the exams, which is the probability of the event (A ∩ ¬B) ∪ (¬A ∩ B), where ¬A represents the complement of event A (failing the first exam) and ¬B represents the complement of event B (failing the second exam).
We are given that 60 students passed the first exam (A), 50 students passed the second exam (B), and 30 students passed both exams (A ∩ B).
To find the probability of (A ∩ ¬B) ∪ (¬A ∩ B), we need to subtract the probability of (A ∩ B) from the sum of the probabilities of A ∩ ¬B and ¬A ∩ B.
Let's calculate each of these probabilities:
- Probability of A ∩ ¬B: This represents the event of passing the first exam (A) but failing the second exam (¬B). Since 60 students passed the first exam and 30 passed both exams, there are 60 - 30 = 30 students who passed the first exam but failed the second exam. So, the probability of A ∩ ¬B is 30/100.
- Probability of ¬A ∩ B: This represents the event of failing the first exam (¬A) but passing the second exam (B). Similarly, since 50 students passed the second exam and 30 passed both exams, there are 50 - 30 = 20 students who failed the first exam but passed the second exam. So, the probability of ¬A ∩ B is 20/100.
- Probability of A ∩ B: This represents the event of passing both exams (A ∩ B), which we are given as 30/100.
Now, let's calculate the probability of (A ∩ ¬B) ∪ (¬A ∩ B):
(A ∩ ¬B) ∪ (¬A ∩ B) = (A ∩ ¬B) + (¬A ∩ B) - (A ∩ B)
= 30/100 + 20/100 - 30/100
= 20/100
= 1/5
Therefore, the probability that a student selected at random has failed in exactly one of the exams is 1/5, which corresponds to option E.
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