Question:
If the probabilities of a student succeeding in the entrance tests for institutes A, B and C are \(0.6\), \(0.5\) and \(0.4\) respectively, while the probability of succeeding in both A and B is \(0.3\), in both B and C is \(0.2\), in both A and C is \(0.2\), and in all three is \(0.1\), then the probability that the student succeeds in exactly one of these tests is......
If the probabilities of a student succeeding in the entrance tests for institutes A, B and C are \(0.6\), \(0.5\) and \(0.4\) respectively, while the probability of succeeding in both A and B is \(0.3\), in both B and C is \(0.2\), in both A and C is \(0.2\), and in all three is \(0.1\), then the probability that the student succeeds in exactly one of these tests is......
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Use inclusion-exclusion for exactly one of three events.
Updated On: Oct 1, 2026
- \(0.3\)
- \(0.4\)
- \(0.5\)
- \(0.6\)
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The Correct Option is B
Solution and Explanation
Step 1: Understanding the Concept:
Let \(A\), \(B\), \(C\) be the events of success in each test. Exactly one success means only one of the three happens.
Step 2: Key Formula or Approach:
\(P(\text{exactly one}) = \sum P(A) - 2\sum P(A\cap B) + 3P(A\cap B\cap C)\).
Step 3: Detailed Explanation:
\(\sum P(A) = 0.6 + 0.5 + 0.4 = 1.5\).
\(\sum P(A\cap B) = 0.3 + 0.2 + 0.2 = 0.7\).
\(P(A\cap B\cap C) = 0.1\).
\[ P = 1.5 - 2(0.7) + 3(0.1) = 1.5 - 1.4 + 0.3 = 0.4 \]
Final Answer:
The probability is \(0.4\), option (B). \[ \boxed{0.4} \]
Let \(A\), \(B\), \(C\) be the events of success in each test. Exactly one success means only one of the three happens.
Step 2: Key Formula or Approach:
\(P(\text{exactly one}) = \sum P(A) - 2\sum P(A\cap B) + 3P(A\cap B\cap C)\).
Step 3: Detailed Explanation:
\(\sum P(A) = 0.6 + 0.5 + 0.4 = 1.5\).
\(\sum P(A\cap B) = 0.3 + 0.2 + 0.2 = 0.7\).
\(P(A\cap B\cap C) = 0.1\).
\[ P = 1.5 - 2(0.7) + 3(0.1) = 1.5 - 1.4 + 0.3 = 0.4 \]
Final Answer:
The probability is \(0.4\), option (B). \[ \boxed{0.4} \]
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