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......

• A. \(0.3\)
• B. \(0.4\)
• C. \(0.5\)
• D. \(0.6\)

Hint:

Use inclusion-exclusion for exactly one of three events.

Answer 1

The Correct Option is B

Step 1: Find each "only" probability:
Only $A$: $0.6 - 0.3 - 0.2 + 0.1 = 0.2$. Only $B$: $0.5 - 0.3 - 0.2 + 0.1 = 0.1$. Only $C$: $0.4 - 0.2 - 0.2 + 0.1 = 0.1$.

Step 2: Add:
$0.2 + 0.1 + 0.1 = 0.4$.

Final Answer:
The probability is $0.4$, option (B). \[ \boxed{0.4} \]