Class 12 Probability · Long Answer3 Marksprobability_distributions

Class 12 Maths · Probability · Long Answer #70

Four bad oranges are mixed with 16 good ones. Draw two uniformly without replacement. Find the distribution and mean of the number of bad oranges.

✅ Step-by-Step Solution

Final Answer: P(X=0)=12/19; P(X=1)=32/95; P(X=2)=3/95. Mean=2/5.

Let X count the designated type in a uniform sample of 2 drawn without replacement from 20 items, 4 of the designated type.
There are C(20,2)=190 equally likely samples. For admissible k, P(X=k)=C(4,k)C(16,2−k)/C(20,2). The probabilities are P(X=0)=12/19; P(X=1)=32/95; P(X=2)=3/95. Their sum is 1.
E(X)=ΣxP(X=x)=(0)(12/19)+(1)(32/95)+(2)(3/95)=2/5.
P(X=0)=12/19; P(X=1)=32/95; P(X=2)=3/95. Mean=2/5.

[0.5 marks] Let X count the designated type in a uniform sample of 2 drawn without replacement from 20 items, 4 of the designated type.
[1 mark] There are C(20,2)=190 equally likely samples. For admissible k, P(X=k)=C(4,k)C(16,2−k)/C(20,2). The probabilities are P(X=0)=12/19; P(X=1)=32/95; P(X=2)=3/95. Their sum is 1.
[1 mark] E(X)=ΣxP(X=x)=(0)(12/19)+(1)(32/95)+(2)(3/95)=2/5.
[0.5 marks] P(X=0)=12/19; P(X=1)=32/95; P(X=2)=3/95. Mean=2/5.

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