Class 12 Probability · Very Long Answer5 Marksprobability_distributions

Class 12 Maths · Probability · Very Long Answer #126

An urn contains3 white,6 red balls. Draw4 independently with replacement. Find the distribution and mean of the red count.

✅ Step-by-Step Solution

Final Answer: P(X=0)=1/81; P(X=1)=8/81; P(X=2)=8/27; P(X=3)=32/81; P(X=4)=16/81. Mean=8/3.

Let X count successes in 4 independent trials. Each succeeds with probability 2/3; hence X~Binomial(4,2/3).
For k=0,…,4, P(X=k)=C(4,k)(2/3)^k(1/3)^(4−k). Evaluate for each k.
The resulting distribution is P(X=0)=1/81; P(X=1)=8/81; P(X=2)=8/27; P(X=3)=32/81; P(X=4)=16/81. Its nonnegative masses sum to 1.
E(X)=ΣxP(X=x)=(0)(1/81)+(1)(8/81)+(2)(8/27)+(3)(32/81)+(4)(16/81)=8/3.
P(X=0)=1/81; P(X=1)=8/81; P(X=2)=8/27; P(X=3)=32/81; P(X=4)=16/81. Mean=8/3.

[1 mark] Let X count successes in 4 independent trials. Each succeeds with probability 2/3; hence X~Binomial(4,2/3).
[1 mark] For k=0,…,4, P(X=k)=C(4,k)(2/3)^k(1/3)^(4−k). Evaluate for each k.
[1 mark] The resulting distribution is P(X=0)=1/81; P(X=1)=8/81; P(X=2)=8/27; P(X=3)=32/81; P(X=4)=16/81. Its nonnegative masses sum to 1.
[1 mark] E(X)=ΣxP(X=x)=(0)(1/81)+(1)(8/81)+(2)(8/27)+(3)(32/81)+(4)(16/81)=8/3.
[1 mark] P(X=0)=1/81; P(X=1)=8/81; P(X=2)=8/27; P(X=3)=32/81; P(X=4)=16/81. Mean=8/3.

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