Worked solution · Exam P
This Exam P sample reference tests Hypergeometric Distribution. This is a hypergeometric composition probability for the first five positions. Choosing three of six items from one type and two of four from the other gives C(6,3)C(4,2)/C(10,5)=10/21, so choice E.
Independent solution
Setup
The unordered set occupying the first five positions is uniformly selected from all five-item subsets of the ten objects.
Model
A capacity-filling set requires the specified three-two composition across the two types.
Compute
Divide the favorable subset count by the total subset count.
Answer
The probability is 10/21.
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Error diagnosis
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Original practice · fully worked
A box contains 4 blue, 5 amber, and 3 white tokens. Tokens are drawn one at a time without replacement until a blue token appears. Calculate the probability that the first blue token appears on the third draw.
The first two draws must come from the eight nonblue tokens and the third from the four blue tokens. The probability is (8/12)(7/11)(4/10)=28/165=0.169697, so choice C.
Setup
Group amber and white together as eight nonblue tokens. The target sequence is nonblue, nonblue, then blue.
Model
Update the available counts after each draw because sampling is without replacement.
Compute
Multiply the three sequential conditional probabilities.
Answer
The first blue token appears on draw three with probability 28/165.
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