A randomly selected sample population consists of 60% women

As I mentioned above, it is implied in the question. We don't know the size of the sample population so it is not possible to change the probability. Also the word population is used so the assumption is that we select one person, find he is not colorblind, then let him be. We do not remove him from the population. Then we pick another person and check.

Hi PrakharGMAT,

the Q asks us - Prob of choosing a colorblind in not more than 3 choices... OR a colorblind is choosen in any of the three chances....

now 60% are W, 90% are colorblind - so .54 ..
40% are M. 15% are colrblind - .06...
total .54+.06 = 0.6..
so NOT a colorblind = 1-0.6 = 0.4

for this you can do it in two ways,,..

1) take ways in which the colourblind is the first, second or the third : a slightly longer method
a) first: 0.6
b) Second : 0.4*0.6=0.24
c) Third : 0.4*0.4*0.6=0.096
add all three =0.096+0.24+0.6= .936
we have multiplied a and b by 3 because in three ways the colorblind or non-colorblind can be choosen - CNN, NCN and NNC

2) the easier way is to find the way wherein NONE of the 3 choosen are colorblind and then subtract that from 1..

so prob that no colorblind is choosen in 3 chances = .4*.4*.4
and prob that atleast one in three is colorblind = 1-none are colorblind = 1-.4*.4*.4 = .936

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.

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

Yes, that is correct. But instead of calculating all that - get a subject on first try, second try or third try, it is easier to calculate "not get a subject on each of the first three tries."
See the solutions given above.

Note that the answer would be the same.
First try = 60/100
Second try = 40/100 * 60/100
Third try = 40/100 * 40/100 * 60/100

Total = 3/5 + 6/25 + 12/125 = 117/125

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,
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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

Doesn't the probability of selection change after the first, second and third?
We are reducing the number of non-color blind people with each try, are we not?

We don't know the size of the sample population so it is not possible to change the probability. Also the word population is used so the assumption is that we select one person, find he is not colorblind, then let him be. We do not remove him from the population. Then we pick another person and check.

Why did you assume that each selection is without replacement?

It isnt' mentioned any where in the question that they are replaced. How could the probability remain same i.e 0.4 throughout ?

Hi VeritasPrepKarishma / chetan2u,

I am not able to understand this method-

{The probability of selecting a colorblind person in three tries} = 1 - {the probability of NOT selecting a colorblind person in three tries} = 1 - 0.4*0.4*0.4 = 1 - 0.064 = 0.936 = 93.6%.

_________________________________________

I would have used this method if the question have asked about to find the probability of ATLEAST 3 colorblind person.
Then I first find the probability of of person who NOT colorblind and then subtract it from 1.
_________________________________________

Can you please assist..?

Thanks and Regards,
Prakhar

A question about calculating the {the probability of NOT selecting a colorblind person in three tries}. Isn't this (4/10)*(3/9)*(2/8)=(24/720)=1/30?
1-(1/30)=29/30=96.67%

Hi VeritasPrepKarishma

Because you have said that we don't know the sample size, wouldn't

1 - P(probability of choosing no colorblind in 3 or less tries)

give us

Probability of choosing a colorblind in more than 3 tries + Probability of never choosing a colorblind?

I have one doubt .. question says in the experiment scientist will look for a color blind subject until they find one ... It should mean that if he finds on the first turn he would not go for another check .
Please clarify

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We know that "90% of the women and 15% of the men are colorblind".
Eventually, they are bound to pick a colorblind person.

Hi Karishma,

As Rohan567 pointed out, we can't replace the selected persons back in to the lot since we know the status of the person once we select them.

So, even if we do without the replacement we will receive the correct answer (but it takes a few more seconds to calculate)

This is how I did it.

From the question we understand that out of total population 60% are colour blind

Hence, by removing the percentage and assuming real numbers we get 60 people out of the 100 are colour blind.

Hence the probability of selecting a colour blind person in the first 3 tries is

= 1- (Probability of not selecting a colour blind person in the first 3 attempts)

= 1- ( 40/100 * 39/99 * 38/98)

=1- (494/8085)

=1-0.06..

=0.94 = 95% (approx)

Ans A