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%
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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).
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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