A box has 6 red hats and 5 green hats. what is the probability of drawing atleast one green hat in 2 consecutive drawings if the hat is not replaced?
a. 10/11
b. 8/11
c. 7/12
d. 5/13
e. 2/7
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Probability
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ranjani2002
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rohit_gmat
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This can be solved by "brute force"ranjani2002 wrote:A box has 6 red hats and 5 green hats. what is the probability of drawing atleast one green hat in 2 consecutive drawings if the hat is not replaced?
a. 10/11
b. 8/11
c. 7/12
d. 5/13
e. 2/7
Possibilities are : GG, RG & GR
GG = 5/11 x 4/10 = 20/110
RG = 6/11 x 5/10 = 30/110
GR = 5/11 x 6/10 = 30/110
P(at least one green) = P (GG or RG or GR) = 20/110 + 30/110 + 30/110 = 80/110 = 8/11
Answer is B
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frank1
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i have not seen much easier question in probablities...
any way i think easier way to do this would be
both 1-Both r
ie 1-(6/11 x 5/10)=8/11
any way i think easier way to do this would be
both 1-Both r
ie 1-(6/11 x 5/10)=8/11
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