Class 12 Probability · Long Answer3 Marksprobability_distributions

Class 12 Maths · Probability · Long Answer #64

Draw twice independently with replacement from an urn with equal positive numbers of red and green balls. Find the distribution and mean of the number of red balls.

✅ Step-by-Step Solution

Final Answer: P(X=0)=1/4; P(X=1)=1/2; P(X=2)=1/4. Mean=1.

Let X count successes in 2 independent trials. Each succeeds with probability 1/2; hence X~Binomial(2,1/2).
For k=0,…,2, P(X=k)=C(2,k)(1/2)^k(1/2)^(2−k). Evaluate for each k. The probabilities are P(X=0)=1/4; P(X=1)=1/2; P(X=2)=1/4. Their sum is 1.
E(X)=ΣxP(X=x)=(0)(1/4)+(1)(1/2)+(2)(1/4)=1.
P(X=0)=1/4; P(X=1)=1/2; P(X=2)=1/4. Mean=1.

📋 CBSE Stepwise Marking Scheme

[0.5 marks] Let X count successes in 2 independent trials. Each succeeds with probability 1/2; hence X~Binomial(2,1/2).
[1 mark] For k=0,…,2, P(X=k)=C(2,k)(1/2)^k(1/2)^(2−k). Evaluate for each k. The probabilities are P(X=0)=1/4; P(X=1)=1/2; P(X=2)=1/4. Their sum is 1.
[1 mark] E(X)=ΣxP(X=x)=(0)(1/4)+(1)(1/2)+(2)(1/4)=1.
[0.5 marks] P(X=0)=1/4; P(X=1)=1/2; P(X=2)=1/4. Mean=1.

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Class 12 Maths · Probability · Long Answer #64

✅ Step-by-Step Solution

📋 CBSE Stepwise Marking Scheme

Master Chapter 13: Probability

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