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%.
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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.
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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 = 0.4for 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 methoda) 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 = .936we 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
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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