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Bunuel
A bowl contains 10 apples, 2 of which are bad. If someone randomly selected four apples from the bowl, what is the probability that at least one of the apples is bad?

A. 1/5
B. 1/3
C. 1/2
D. 2/3
E. 4/5

Probability (Atleast one apple is bad) = 1 - (No apple is bad) = 1 - (8C4)/(10C4) = 1- 70/210 = 1- (1/3) = 2/3

Answer: Option D

ALTERNATIVE

Required Probability = Probability (Exactly one apple is bad + exactly two apples are bad ) = (2C1 *8C3 + 2C2*8C2) / 10C4 = (2*56+1*28)/210 = 140/210 = 2/3

Answer: Option D
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GMATinsight
Bunuel
A bowl contains 10 apples, 2 of which are bad. If someone randomly selected four apples from the bowl, what is the probability that at least one of the apples is bad?

A. 1/5
B. 1/3
C. 1/2
D. 2/3
E. 4/5

Probability (Atleast one apple is bad) = 1 - (No apple is bad) = 1 - (8C4)/(10C4) = 1- 70/210 = 1- (1/3) = 2/3

Answer: Option D

ALTERNATIVE

Dear

Required Probability = Probability (Exactly one apple is bad + exactly two apples are bad ) = (2C1 *8C3 + 2C2*8C2) / 10C4 = (2*56+1*28)/210 = 140/210 = 2/3

Answer: Option D

Hi GMATinsight

I understood the ways above but I tried to make other way but it yielded different result

Required Probability = Probability (Exactly one apple is bad + exactly two apples are bad ) = (1/2) (3/8) + (2/2) (2/8) = 7/16

Can you please where i went wrong?
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Bunuel
A bowl contains 10 apples, 2 of which are bad. If someone randomly selected four apples from the bowl, what is the probability that at least one of the apples is bad?

A. 1/5
B. 1/3
C. 1/2
D. 2/3
E. 4/5

Probability (Atleast one apple is bad) = 1 - (No apple is bad) = 1 - (8C4)/(10C4) = 1- 70/210 = 1- (1/3) = 2/3

Answer: Option D

ALTERNATIVE

Dear

Required Probability = Probability (Exactly one apple is bad + exactly two apples are bad ) = (2C1 *8C3 + 2C2*8C2) / 10C4 = (2*56+1*28)/210 = 140/210 = 2/3

Answer: Option D

Hi GMATinsight

I understood the ways above but I tried to make other way but it yielded different result

Required Probability = Probability (Exactly one apple is bad + exactly two apples are bad ) =(1/2) (3/8) + (2/2) (2/8) = 7/16

Can you please where i went wrong?


The highlighted part is the wrong way of calculating probability. Please check my ALTERNATIVE solution carefully... That's the right way.

Mo2men
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Prob. of atleast one is picking one or two;

Ways of picking one = 8C3*2=8*7*6*2/(3*2)=16*7
Ways of picking two=8C2=8*7/2=4*7
So Ways of picking atleast one=20*7=140
Total ways=10C4=210. So Prob.=140/210=2/3 IMO D
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