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16 Jul 2024, 05:32
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We can split the science experiment into three outcomes
1. Mouse finds no treats
2. Mouse finds one of the treats
3. Most finds both treats
(1) In the first statement we are provided with the probability that the mouse finds one of the treats is 1/5. That is great but with only one probability for one outcome we do not know how the probability for the other two outcomes will be split.
(2) In this statement, we are again given the probability of one outcome, that the mouse finds both treats, but with the two other outcomes being unkown, similar to the first statement, it is not sufficient to answer the question.
When we combine the statements we know that the probability the mouse finds one of the treats is 1/5 and that the probability the mouse finds both treats is 3/10. If we go to our 3 possible outcomes and fill in their probabilities, it would look like this:
1. Probability Mouse finds no treats - ?
2. Probability Mouse finds one of the treats - 1/5 or 2/10
3. Probability Mouse finds both treats - 3/10
Since these are the only three possible outcomes, we know that the sum of the probabilities of 1,2 and 3 must add up to 1. Therefore, we can calculate that the probability that the Mouse finds no treats is 1 - (2/10+3/10). Therefore the probability that the mouse finds no treats is 1/2.
Therefore, the answer is C.
1. Mouse finds no treats
2. Mouse finds one of the treats
3. Most finds both treats
(1) In the first statement we are provided with the probability that the mouse finds one of the treats is 1/5. That is great but with only one probability for one outcome we do not know how the probability for the other two outcomes will be split.
(2) In this statement, we are again given the probability of one outcome, that the mouse finds both treats, but with the two other outcomes being unkown, similar to the first statement, it is not sufficient to answer the question.
When we combine the statements we know that the probability the mouse finds one of the treats is 1/5 and that the probability the mouse finds both treats is 3/10. If we go to our 3 possible outcomes and fill in their probabilities, it would look like this:
1. Probability Mouse finds no treats - ?
2. Probability Mouse finds one of the treats - 1/5 or 2/10
3. Probability Mouse finds both treats - 3/10
Since these are the only three possible outcomes, we know that the sum of the probabilities of 1,2 and 3 must add up to 1. Therefore, we can calculate that the probability that the Mouse finds no treats is 1 - (2/10+3/10). Therefore the probability that the mouse finds no treats is 1/2.
Therefore, the answer is C.
Bunuel
16 Jul 2024, 05:34
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Total = {Probability of only 1+ Probability of Both + Probability of none}
Probability of none = Total - {Probability of only 1 + Probability of Both}
(1) The probability that the mouse will find only one of the two treats is 1/5.
(Probability of Both) is not given. Insufficient.
(2) The probability that the mouse will find both treats is 3/10.
(Probability of Only 1) is not given. Insufficient.
Combining 1 and 2 we can solve for Probability of none = \(1- ({\frac{3}{10}-\frac{1}{5}})\)
Probability of none= \(1- \frac{5}{10} = \frac{1}{2}\)
IMO C.
Probability of none = Total - {Probability of only 1 + Probability of Both}
(1) The probability that the mouse will find only one of the two treats is 1/5.
(Probability of Both) is not given. Insufficient.
(2) The probability that the mouse will find both treats is 3/10.
(Probability of Only 1) is not given. Insufficient.
Combining 1 and 2 we can solve for Probability of none = \(1- ({\frac{3}{10}-\frac{1}{5}})\)
Probability of none= \(1- \frac{5}{10} = \frac{1}{2}\)
IMO C.
RenB
Joined: 13 Jul 2022
Last visit: 02 Mar 2026
Posts: 388
Given Kudos: 304
Location: India
Concentration: Finance, Nonprofit
Schools: Stanford '26 Wharton '26 (D) Booth '26 (D) Yale '26 (D) CBS '26 (A$$$) Darden '26 (WL) HBS '26 (D) Ross '26 (A$$$$) LBS '26 (D) Tuck '26 (A$$) Kellogg '26 (A)
GMAT Focus 1: 715 Q90 V84 DI82 
GPA: 3.74
WE:Corporate Finance (Consulting)
Schools: Stanford '26 Wharton '26 (D) Booth '26 (D) Yale '26 (D) CBS '26 (A$$$) Darden '26 (WL) HBS '26 (D) Ross '26 (A$$$$) LBS '26 (D) Tuck '26 (A$$) Kellogg '26 (A)
GMAT Focus 1: 715 Q90 V84 DI82
Posts: 388
16 Jul 2024, 05:34
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Bunuel
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