Class 12 Probability · Very Long Answer5 Marksprobability_distributions

Class 12 Maths · Probability · Very Long Answer #82

An urn contains 4 white and 3 red balls. Draw three uniformly without replacement. Let X be the red count. Find E(X).

✅ Step-by-Step Solution

Final Answer: P(X=0)=4/35; P(X=1)=18/35; P(X=2)=12/35; P(X=3)=1/35. Mean=9/7.

Let X count the designated type in a uniform sample of 3 drawn without replacement from 7 items, 3 of the designated type.
There are C(7,3)=35 equally likely samples. For admissible k, P(X=k)=C(3,k)C(4,3−k)/C(7,3).
The resulting distribution is P(X=0)=4/35; P(X=1)=18/35; P(X=2)=12/35; P(X=3)=1/35. Its nonnegative masses sum to 1.
E(X)=ΣxP(X=x)=(0)(4/35)+(1)(18/35)+(2)(12/35)+(3)(1/35)=9/7.
P(X=0)=4/35; P(X=1)=18/35; P(X=2)=12/35; P(X=3)=1/35. Mean=9/7.

[1 mark] Let X count the designated type in a uniform sample of 3 drawn without replacement from 7 items, 3 of the designated type.
[1 mark] There are C(7,3)=35 equally likely samples. For admissible k, P(X=k)=C(3,k)C(4,3−k)/C(7,3).
[1 mark] The resulting distribution is P(X=0)=4/35; P(X=1)=18/35; P(X=2)=12/35; P(X=3)=1/35. Its nonnegative masses sum to 1.
[1 mark] E(X)=ΣxP(X=x)=(0)(4/35)+(1)(18/35)+(2)(12/35)+(3)(1/35)=9/7.
[1 mark] P(X=0)=4/35; P(X=1)=18/35; P(X=2)=12/35; P(X=3)=1/35. Mean=9/7.

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