Question

A final year student appears for placement interviews in two companies, S and T. Probability of job offer from S = 0.8, from T = 0.6. Let \(p\) = probability of getting offers from both companies. Which option is most appropriate?

• A. \(0 \leq p \leq 0.2\)
• B. \(0.4 \leq p \leq 0.6\)
• C. \(0.2 \leq p \leq 0.4\)
• D. \(0.6 \leq p \leq 1.0\)

Hint:

When calculating intersection bounds, remember: \[ \max(0, P(A)+P(B)-1) \leq P(A \cap B) \leq \min(P(A), P(B)). \] This ensures probabilities remain valid without assuming independence.

Answer 1

The Correct Option is C

Step 1: Recall probability intersection formula.
For two events A and B: \[ P(A \cap B) = P(A) + P(B) - P(A \cup B) \] We know \(P(S) = 0.8\), \(P(T) = 0.6\). So maximum and minimum bounds for intersection can be derived. Step 2: Upper bound of \(p\).
Maximum possible value of \(P(S \cap T)\) is \(\min(P(S), P(T)) = 0.6\). Step 3: Lower bound of \(p\).
Minimum possible value = \(P(S) + P(T) - 1 = 0.8 + 0.6 - 1 = 0.4\). If negative, we take 0, but here it is positive. So range: \[ 0.4 \leq p \leq 0.6 \] Step 4: Correct option check.
This matches option (B), not (C). Correction: The correct range is \(0.4 \leq p \leq 0.6\). Final Answer: \[ \boxed{0.4 \leq p \leq 0.6} \]