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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]

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05 Oct 2018, 14:47
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]

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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]

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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.
_________________________________________

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??
Re: A randomly selected sample population consists of 60% women   [#permalink] 17 Mar 2019, 08:05

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