Class 12 Probability · Very Long Answer5 Marksprobability_distributions

Class 12 Maths · Probability · Very Long Answer #124

Roll a pair of independent fair dice4 times independently. X counts doublets (equal faces). Find its distribution and mean.

✅ Step-by-Step Solution

Final Answer: P(X=0)=625/1296; P(X=1)=125/324; P(X=2)=25/216; P(X=3)=5/324; P(X=4)=1/1296. Mean=2/3.

Let X count successes in 4 independent trials. Each succeeds with probability 1/6; hence X~Binomial(4,1/6).
For k=0,…,4, P(X=k)=C(4,k)(1/6)^k(5/6)^(4−k). Evaluate for each k.
The resulting distribution is P(X=0)=625/1296; P(X=1)=125/324; P(X=2)=25/216; P(X=3)=5/324; P(X=4)=1/1296. Its nonnegative masses sum to 1.
E(X)=ΣxP(X=x)=(0)(625/1296)+(1)(125/324)+(2)(25/216)+(3)(5/324)+(4)(1/1296)=2/3.
P(X=0)=625/1296; P(X=1)=125/324; P(X=2)=25/216; P(X=3)=5/324; P(X=4)=1/1296. Mean=2/3.

[1 mark] Let X count successes in 4 independent trials. Each succeeds with probability 1/6; hence X~Binomial(4,1/6).
[1 mark] For k=0,…,4, P(X=k)=C(4,k)(1/6)^k(5/6)^(4−k). Evaluate for each k.
[1 mark] The resulting distribution is P(X=0)=625/1296; P(X=1)=125/324; P(X=2)=25/216; P(X=3)=5/324; P(X=4)=1/1296. Its nonnegative masses sum to 1.
[1 mark] E(X)=ΣxP(X=x)=(0)(625/1296)+(1)(125/324)+(2)(25/216)+(3)(5/324)+(4)(1/1296)=2/3.
[1 mark] P(X=0)=625/1296; P(X=1)=125/324; P(X=2)=25/216; P(X=3)=5/324; P(X=4)=1/1296. Mean=2/3.

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