Class 12 Maths · Probability · Very Long Answer #128

Question

An urn has5 red,2 black balls. Draw2 uniformly without replacement. X is the number of black balls. State its possible values, explain whether it is a random variable, and find E(X).

Accepted Answer

Final Answer: P(X=0)=10/21; P(X=1)=10/21; P(X=2)=1/21. Mean=4/7. X takes0,1,2 and is a random variable because it assigns a real count to every sample outcome. Step 1: Let X count the designated type in a uniform sample of 2 drawn without replacement from 7 items, 2 of the designated type. Step 2: There are C(7,2)=21 equally likely samples. For admissible k, P(X=k)=C(2,k)C(5,2−k)/C(7,2). Step 3: The resulting distribution is P(X=0)=10/21; P(X=1)=10/21; P(X=2)=1/21. Its nonnegative masses sum to 1. Step 4: E(X)=ΣxP(X=x)=(0)(10/21)+(1)(10/21)+(2)(1/21)=4/7. Step 5: P(X=0)=10/21; P(X=1)=10/21; P(X=2)=1/21. Mean=4/7. X takes0,1,2 and is a random variable because it assigns a real count to every sample outcome.

Class 12 Maths · Probability · Very Long Answer #128

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