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An urn holds three red and eight blue marbles, and two are drawn without replacement. What is the probability that both are red?
AIt is 9/121
BIt is 6/110
CIt is 6/121
DIt is 5/110
Answer & Solution
Correct answer: B. It is 6/110
1. The first draw gives red with probability 3/11.
2. The first marble is not replaced, so only two red marbles remain among ten.
3. The second draw gives red with probability 2/10.
4. Multiplying gives 3/11 times 2/10, which is 6/110, while 9/121 would be the answer with replacement.
_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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