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A randomly selected sample population consists of 60% women

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Re: A randomly selected sample population consists of 60% women [#permalink] New post   15 Feb 2018, 12:48
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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:
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A


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Re: A randomly selected sample population consists of 60% women [#permalink] New post   14 Sep 2018, 01:21
Now correct me if i am wrong :
here the selection should be without replacement ;
pick a number in your hypothesis to deal with :
60 women
40 men

colorblind women = 60x90%=54
colorblind men = 40x15%=6

a total of 60 color blind people out of 100

probability of selecting a colorblind person in no more than three tries = 1-P(selecting NO colorblind person in the three times) = 1-P(NNN) =
1-(40/100 x 39/99 x 38/98) ( and since we are dealing with proximity) = 1-(40/100)^3 = 0.936 approximately 95%
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A randomly selected sample population consists of 60% women [#permalink] New post   05 Oct 2018, 03:55
VeritasKarishma wrote:
vivsleo wrote:
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

Sent from my MotoG3 using GMAT Club Forum mobile app


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


@"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?
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Re: A randomly selected sample population consists of 60% women [#permalink] New post   05 Oct 2018, 14:47
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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.').

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Re: A randomly selected sample population consists of 60% women [#permalink] New post   10 Dec 2018, 11:30
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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Re: A randomly selected sample population consists of 60% women [#permalink] New post   17 Mar 2019, 08:05
chetan2u wrote:
PrakharGMAT wrote:
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



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 1 out of 3 or 2 out of 3 or 3out of three are colorblind- a LONG method
a) 1 out of 3 - .6*.4*.4 *3 = .288
b) 2 out of 3 - .6*.6*.4*3 = .432
c) 3 out of 3 - .6*.6*.6 = .216
add all three .288+.432+.216 = .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



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??
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Re: A randomly selected sample population consists of 60% women [#permalink] New post   23 Feb 2020, 10:13
JeffTargetTestPrep wrote:
gmattesttaker2 wrote:
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%


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


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 ?
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Re: A randomly selected sample population consists of 60% women [#permalink] New post   07 Dec 2020, 14:30
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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Re: A randomly selected sample population consists of 60% women [#permalink] New post   05 Feb 2024, 03:02
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
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Re: A randomly selected sample population consists of 60% women [#permalink]
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