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A litter holds four tabby and five black kittens, and two are chosen without replacement. What is the probability that both are tabby?
AIt is 4/9 times 4/9
BIt is 4/9 times 5/8
CIt is 1/2 times 1/2
DIt is 4/9 times 3/8
Answer & Solution
Correct answer: D. It is 4/9 times 3/8
1. The litter holds nine kittens in all, four tabby and five black.
2. The first pick is tabby with probability 4/9.
3. That kitten is not replaced, so three tabby kittens remain among eight.
4. The second pick is tabby with probability 3/8.
5. The joint probability is therefore 4/9 times 3/8.
6. Writing 4/9 times 4/9 would assume replacement, which the adoption plainly does not allow.
_Source: OpenStax Introductory Business Statistics (CC BY 4.0), Ch 3 "Probability Topics", section 3.4 Contingency Tables and Probability Trees_
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