Hi All,

We're told that two different cars are selected randomly from a lot of 20 cars (and that all of the cars are either sedans or convertibles). We're asked if the probability that BOTH cars selected will be sedans is greater than 3/4. This is a YES/NO question. We can solve it by TESTing VALUES.

1) At least three-fourths of the cars are sedans.

IF... there are 15 sedans and 5 convertibles, then the probability of selecting 2 sedans is....

(15/20)(14/19) = (3/4)(14/19)

You don't have to actually calculate this value, since we're multiplying 3/4 by a positive fraction, the answer will be LESS then 3/4 and the answer to the question is NO.

IF... there are 19 sedans and 1 convertible, then the probability of selecting 2 sedans is....

(19/20)(18/19) = 18/20 = 90% and the answer to the question is YES.

Fact 1 is INSUFFICIENT

2) The probability that both of the cars selected will be convertibles is less than 1/20.

IF... there are 15 sedans and 5 convertibles, then the probability of selecting 2 convertibles is....

(5/20)(4/19) = 1/19.... This is NOT a match for the information in Fact 2 though, so there MUST BE FEWER than 5 convertibles.

IF... there are 16 sedans and 4 convertibles, then the probability of selecting 2 convertibles is....

(4/20)(3/19) = 3/95

The probability of selecting 2 sedans under these circumstances is...

(16/20)(15/19) = (15/20)(16/19) - since we're multiplying 3/4 by a positive fraction, the answer will be LESS then 3/4 and the answer to the question is NO.

IF... there are 19 sedans and 1 convertible, then the probability of selecting 2 sedans is....

(19/20)(18/19) = 18/20 = 90% and the answer to the question is YES.

Fact 2 is INSUFFICIENT

Combined, we know that there must be FEWER than 5 convertibles. From the work that we did in Fact 2 (above), we have proof that the answer could be NO (if there are 4 convertibles) and YES (if there is just 1 convertible).

Combined, INSUFFICIENT

Final Answer:

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Rich

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