A randomly selected sample population consists of 60% women

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

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%.

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

correct option should be A

We can break it sown in cases to make it simple. We know there are 60 color-blind people in totla out of 100. So, in a single try selecting the subject has a probability of 0.4 and not selecting has 0.4.

We shall have 3 tries. Let's denote each try as a letter (i.e, A, B, C). If we are able to get the subject in a try we write the letter itself but if we don't we put a complement sign on it (i.e, A' or B' or C').

1st try:
case 1 : We get the subject . So A (*)
case 2 : we can't select a subject. A'

2nd try: (only if we have been able to get a subject in the 1st try)
case 1: we get our subject. So A'B (*)
cae 2: we can'r get our subject. A'B'

3rd try: (if we haven't got the color-blind subject in both of 1st and 2nd try)
only possible case : we get the 1 color-bling person.. A'B'C (*)

Now we should sum all the cases in which there has been a positive outcome. So P = A+A'B+A'B'C
=> P = 0.6 + 0.4*0.6 + 0.4*0.4*0.6 = 0.6*(1+0.4+0.16) = 0.6*1.56 = 0.936
=> \(P\approx{0.95}\)

So 95% should be the approximate answer.

Hence, correct option should be A

Many seem confused by the replacement vs not replacement issue.

If the size of the population had been specified then it would make sense NOT to put that person back into the population for the next selections.

So if there were 60 and we're selecting 1 then that would leave 59 and the probability would be adjusted for that revised base.

Likewise if there were 600 then the base for the next probability would be revised to 599.

But the two probabilities would then not be the same and so the population size makes a difference when not replacing.

So knowing the population size is required when not replacing.

Therefore the only way to calculate the probabilities is to assume replacement because then our not knowing the population size doesn't matter.