Q.A randomly selected sample population consists of 60% women and 40% men. 90% of the women and 15% of the men are colorblind. For a certain experiment, scientists will select one person at a time until they have a colorblind subject. What is the approximate probability of selecting a colorblind person in no more than three tries?
95%
90%
80%
75%
60%
OA is A.
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P(selecting a colorblind person in 3 or fewer tries) = 1 - P(NOT selecting a colorblind person in 3 or fewer tries).ash4gmat wrote:Q.A randomly selected sample population consists of 60% women and 40% men. 90% of the women and 15% of the men are colorblind. For a certain experiment, scientists will select one person at a time until they have a colorblind subject. What is the approximate probability of selecting a colorblind person in no more than three tries?
95%
90%
80%
75%
60%
Let the total number of people = 100, implying that there are 60 women and 40 men.
Since 90% of the 60 women are colorblind, 10% of the 60 women are NOT colorblind.
Thus, the number of non-colorblind women = (10/100)(60) = 6.
Since 15% of the 40 men are colorblind, 85% of the 40 men are NOT colorblind.
Thus, the number of non-colorblind men = (85/100)(40) = 34.
Total number of non-colorblind people = 6+34 = 40.
P(not selecting a colorblind person in 3 or fewer tries):
P(not selecting a colorblind person on the first try) = 40/100. (Of the 100 people, 40 are not colorblind.)
P(not selecting a colorblind person on the second try) = 39/99. (Of the 99 remaining people, 39 are not colorblind.)
P(not selecting a colorblind person on the third try) = 38/98. (Of the 98 remaining people, 38 are not colorblind.)
Thus:
P(not selecting a colorblind person in 3 or fewer tries) ≈ 40/100 * 39/99 * 38/98 = 4/10 * (a bit less than 4/10) * (a bit less than 4/10) = a bit less than 64/1000 = a bit less than 6.4%.
P(selecting a colorblind person in 3 or fewer tries):
1 - (a bit less than 6.4) = a bit more than 93.6%.
The correct answer is A.
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Jeff@TargetTestPrep
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If the sample population has 60% women and 40% men, and 90% of the women and 15% of the men are colorblind, then the probability that a randomly selected person is colorblind is (0.6)(0.9) + (0.4)(0.15) = 0.54 + 0.06 = 0.6. This also means that the probability that a randomly selected person is not colorblind is 0.4.ash4gmat wrote:Q.A randomly selected sample population consists of 60% women and 40% men. 90% of the women and 15% of the men are colorblind. For a certain experiment, scientists will select one person at a time until they have a colorblind subject. What is the approximate probability of selecting a colorblind person in no more than three tries?
95%
90%
80%
75%
60%
We need to determine the probability of selecting a colorblind person in no more than 3 tries.
Let's calculate the probability for each possible scenario:
Scenario 1: A colorblind person is chosen on the first try.
P(colorblind person is chosen on the first try) = 0.6
Scenario 2: A colorblind person is chosen on the second try. (That is, a non-colorblind person is chosen on the first try.)
P(colorblind person is chosen on the second try) = 0.4 x 0.6 = 0.24
Scenario 3: A color blind person is chosen on the third try. (That is, a non-colorblind person is chosen on each of the first two tries.)
P(colorblind person is chosen on the third try) = 0.4 x 0.4 x 0.6 = 0.096
Thus, the probability of selecting a colorblind person on no more than three tries is 0.6 + 0.24 + 0.096 = 0.936 = 93.6%, which is approximately 95%.
Answer: A
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