A randomly selected sample population consists of 60% women

Question

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?

A. 95%
B. 90%
C. 80%
D. 75%
E. 60%

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I was able to come up with the double set matrix but could not solve for the probability part. Can you please assist?

Thanks,
Sri

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Bunuel · Accepted Answer

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?

A. 95%
B. 90%
C. 80%
D. 75%
E. 60%

According to the matrix the probability of selecting a colorblind person is 0.6 and the probability of NOT selecting a colorblind person is 0.4.

{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%.

We are asked to find the approximate probability, so the answer is 95%.

Answer: A.

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KarishmaB · Answer

To get to the probability part, all you need is the total % of colorblind people.
Say, there are 100 people - 60 women, 40 men.
Out of 60, 90% women are colorblind i.e. 90% of 60 = 54 women
Out of 40, 15% men are colorblind i.e. 15% of 40 = 6 men

Total 60 people of 100 are colorblind.

P(Selecting colorblind person in 3 tries or less) = 1 - P(Selecting a colorblind person in more than 3 tries) = 1 - (40/100)^3 = 117/125 = approx 95%

ronr34 · Answer

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?

KarishmaB · Answer

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.

rohan567 · Answer

Why did you assume that each selection is without replacement?

KarishmaB · Answer

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.

PrakharGMAT · Answer

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

chetan2u · Answer

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

ShashankDave · Answer

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?

vivsleo · Answer

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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KarishmaB · Answer

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

KarishmaB · Answer

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

JeffTargetTestPrep · Answer

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

EMPOWERgmatRichC · Answer

Hi All,

There are a couple of different ways to solve this problem. Based on the given info, we have….

60% Women 
40% Men

Color-Blind: 
90% of Women = (.9)(60%) = 54% of total group 
15% of Men = (.15)(40%) = 6% of the total group

NOT Color-Blind = everyone else = 40% of the total group

When it comes to dealing with Probabilities, we can either calculate what we WANT or what we DON'T WANT (and then subtract that result from the number 1 to figure out the probability of what we DO want).

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….

Notice the word "approximate"….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 probability of selecting a color-blind person in 3 tries is….

1 - .064 = .936

This is really close to….

Final Answer:  A

GMAT assassins aren't born, they're made, 
Rich

sananoor · Answer

@"VeritasKarishma So it means researcher will try until they dont get colorblind person..It means it is with replacement till they dont get a colorblind?

EMPOWERgmatRichC · Answer

Hi sananoor,

The question asks for the APPROXIMATE 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.').

GMAT assassins aren't born, they're made,
Rich

itisSheldon · Answer

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%

gmexamtaker1 · Answer

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 ?

EMPOWERgmatRichC · Answer

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

A randomly selected sample population consists of 60% women

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