A shopkeeper sells TVs. Daily demand X has distribution: P(0)=0.1, P(1)=0.3, P(2)=0.4, P(3)=0.2. The probability he sells at least 2 TVs is:
A shopkeeper sells TVs. Daily demand X has distribution: P(0)=0.1, P(1)=0.3, P(2)=0.4, P(3)=0.2. The probability he sells at least 2 TVs is:
- A. 0.4
- B. 0.6
- C. 0.5
- D. 0.8
Answer: B) 0.6
Explanation: P(X ≥ 2) = P(X=2) + P(X=3) = 0.4 + 0.2 = 0.6.
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