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Class 12 Probability · Long Answer3 Marksprobability_distributions

Class 12 Maths · Probability · Long Answer #68

Draw two distinct cards uniformly without replacement from cards numbered 1,2,3,4. X is their sum. Find E(X).

✅ Step-by-Step Solution

Final Answer: P(X=3)=1/6; P(X=4)=1/6; P(X=5)=1/3; P(X=6)=1/6; P(X=7)=1/6. Mean=5.
  1. The six equally likely unordered pairs are {1,2},{1,3},{1,4},{2,3},{2,4},{3,4}.
  2. Their sums are 3,4,5,5,6,7. Thus the probabilities are 1/6,1/6,2/6,1/6,1/6. The probabilities are P(X=3)=1/6; P(X=4)=1/6; P(X=5)=1/3; P(X=6)=1/6; P(X=7)=1/6. Their sum is 1.
  3. E(X)=ΣxP(X=x)=(3)(1/6)+(4)(1/6)+(5)(1/3)+(6)(1/6)+(7)(1/6)=5.
  4. P(X=3)=1/6; P(X=4)=1/6; P(X=5)=1/3; P(X=6)=1/6; P(X=7)=1/6. Mean=5.

📋 CBSE Stepwise Marking Scheme

  • [0.5 marks] The six equally likely unordered pairs are {1,2},{1,3},{1,4},{2,3},{2,4},{3,4}.
  • [1 mark] Their sums are 3,4,5,5,6,7. Thus the probabilities are 1/6,1/6,2/6,1/6,1/6. The probabilities are P(X=3)=1/6; P(X=4)=1/6; P(X=5)=1/3; P(X=6)=1/6; P(X=7)=1/6. Their sum is 1.
  • [1 mark] E(X)=ΣxP(X=x)=(3)(1/6)+(4)(1/6)+(5)(1/3)+(6)(1/6)+(7)(1/6)=5.
  • [0.5 marks] P(X=3)=1/6; P(X=4)=1/6; P(X=5)=1/3; P(X=6)=1/6; P(X=7)=1/6. Mean=5.

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