Class 12 Probability · Very Long Answer5 Marksprobability_distributions

Class 12 Maths · Probability · Very Long Answer #127

Five bad oranges are mixed with20 good ones. Draw4 independently with replacement. Find the distribution and mean of the bad-orange count.

✅ Step-by-Step Solution

Final Answer: P(X=0)=256/625; P(X=1)=256/625; P(X=2)=96/625; P(X=3)=16/625; P(X=4)=1/625. Mean=4/5.

Let X count successes in 4 independent trials. Each succeeds with probability 1/5; hence X~Binomial(4,1/5).
For k=0,…,4, P(X=k)=C(4,k)(1/5)^k(4/5)^(4−k). Evaluate for each k.
The resulting distribution is P(X=0)=256/625; P(X=1)=256/625; P(X=2)=96/625; P(X=3)=16/625; P(X=4)=1/625. Its nonnegative masses sum to 1.
E(X)=ΣxP(X=x)=(0)(256/625)+(1)(256/625)+(2)(96/625)+(3)(16/625)+(4)(1/625)=4/5.
P(X=0)=256/625; P(X=1)=256/625; P(X=2)=96/625; P(X=3)=16/625; P(X=4)=1/625. Mean=4/5.

[1 mark] Let X count successes in 4 independent trials. Each succeeds with probability 1/5; hence X~Binomial(4,1/5).
[1 mark] For k=0,…,4, P(X=k)=C(4,k)(1/5)^k(4/5)^(4−k). Evaluate for each k.
[1 mark] The resulting distribution is P(X=0)=256/625; P(X=1)=256/625; P(X=2)=96/625; P(X=3)=16/625; P(X=4)=1/625. Its nonnegative masses sum to 1.
[1 mark] E(X)=ΣxP(X=x)=(0)(256/625)+(1)(256/625)+(2)(96/625)+(3)(16/625)+(4)(1/625)=4/5.
[1 mark] P(X=0)=256/625; P(X=1)=256/625; P(X=2)=96/625; P(X=3)=16/625; P(X=4)=1/625. Mean=4/5.

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