Veritas Prep OFFICIAL EXPLANATION

First, find the percentage of colorblind people in the population. For the women, 90% of 60% is 54%, and for the men 15% of 40% is 6%, so 60% of the entire population is colorblind. The probability of selecting a colorblind person on the first, second, or third try is then:

first try= 60%

not on the first, but then on the second = 40% * 60% = 24%

not on the first, not on the second,. but then on the third = 40% * 40% * 60% = 9.6% (round to 10 since the answers are approximate)

total = 60% + 24% + 10% = 94%, which is closest to 95%
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i have an issue with your solution

we have multiplied a and b by 3 because in three ways the colorblind or non-colorblind can be choosen - CNN, NCN and NNC.....


The question clearly mentions" until they find a colourblind subject " it can only by C, NC, NNC ............... do you agree??

Hey Jeff,
Could you please explain why do we take for granted that there is going to be replacement, what I mean is shouldn't scenario 2 be 40/100x60/99 and senario 3 40/100x39/99x60/98 ?

Hi UNSTOPPABLE12,

The question asks for the APPROXIMATE probability (not "the probability"), so it does not matter whether you 'replace' any of the people selected or not. From a calculation-standpoint, it's considerably easier to assume that each selection is put back into the 'pool' for the next selection (otherwise, the calculation could get REALLY 'math heavy.').

In probability questions, you can typically use the following equation to your advantage:

Want + Don't Want = 1

When a probability question gives us multiple tries to accomplish 1 task, it's usually fastest to calculate the probability that we DO NOT accomplish the task, then subtract that from the number 1. Instead of calculating the probability of picking a color-blind person in 3 tries, we'll calculate the probability that we DO NOT select a color-blind person in 3 tries….

Again, the word "approximate" means that we can keep things simple….

(NOT)(NOT)(NOT) = (.4)(.4)(.4) = .064 This is the probability of NOT selecting a color-blind person in 3 tries.

The approximate probability of selecting a color-blind person in 3 tries is….

1 - .064 = .936 = approximately 93.6%

GMAT assassins aren't born, they're made,
Rich
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It turns out that 60% of the sample population is colorblind.

What is the approximate probability of selecting a colorblind person in no more than three tries? Instead of calculating the probability of selecting a colorblind person after one try, calculating the probability of selecting a colorblind person after two tries, etc.. it's easier to calculate the probability of NOT selecting a colorblind person in three tries and then subtracting that probability from 1.

2/5 * 2/5 * 2/5 = 8/125

1 - P(not selecting a colorblind person in 3 tries) = P(selecting a colorblind person in no more than 3 tries)

1 - 8/125 = 117/125

Closest answer is A. 95%.
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chetan2u
I believe there's some room of debatable issue here. The question says scientist are picking till they get a colorblind subject. With this logic, Option A , B, And C make a mistake. In option A The probability you calculated shows A case where scientists picked a Non Color blind even after getting a color blind. same issue with B and C.
The same thing can be repeated for people doing 1-.4*.4*.4

Yes, you are correct. May be when I posted the solution, I misread the question.
The ways will be
NNC-0.4*0.4*0.6=0.096
NC-0.4*0.6=0.24
C-0.6
Total 0.096+0.24+0.6=0.936