ExplanationQuantity A asks for the probability “without replacement,” so that means you have to take into account that there will be one marble less in the total after each draw.

The probability of first choosing a red marble is \(\frac{5}{12}\), a second red marble is \(\frac{4}{11}\), and then a green marble is \(\frac{3}{10}\).

This is an “and” probability problem, so you have to multiply the probability of each event together: \(\frac{5}{12}\times\frac{4}{11}\times\frac{3}{10}=\frac{60}{1320}=\frac{1}{22}\).

For Quantity B, you do the same thing, but the total stays the same for each draw: \(\frac{5}{12}\times\frac{4}{12}\times\frac{4}{12}=\frac{80}{1728}=\frac{5}{108}\).

Quantity B is greater.
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Sandy

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