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A bowl contains 10 apples, 2 of which are bad. If someone randomly sel

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Bunuel
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A bowl contains 10 apples, 2 of which are bad. If someone randomly sel [#permalink] New post   14 Jun 2019, 00:39
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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
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D
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Re: A bowl contains 10 apples, 2 of which are bad. If someone randomly sel [#permalink] New post   14 Jun 2019, 00:47
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Bunuel wrote:
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 of at least 1 bad apple = 1 - prob of no bad apple
= 1 - 8c4/10c4
= 1 - (8*7*6*5)/(10*9*8*7)
= 1 - 1/3
= 2/3

IMO Option D

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Re: A bowl contains 10 apples, 2 of which are bad. If someone randomly sel [#permalink] New post   14 Jun 2019, 02:01
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Bunuel wrote:
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


P of not bad = 8/10 * 7/9*6/8*5/7 = 1/3
1-1/3 ; 2/3 P of atleast one is bad
IMO D
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Re: A bowl contains 10 apples, 2 of which are bad. If someone randomly sel [#permalink] New post   14 Jun 2019, 04:07
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Bunuel wrote:
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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Re: A bowl contains 10 apples, 2 of which are bad. If someone randomly sel [#permalink] New post   14 Jun 2019, 04:29
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GMATinsight wrote:
Bunuel wrote:
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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Re: A bowl contains 10 apples, 2 of which are bad. If someone randomly sel [#permalink] New post   14 Jun 2019, 04:34
Expert Reply
Mo2men wrote:
GMATinsight wrote:
Bunuel wrote:
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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A bowl contains 10 apples, 2 of which are bad. If someone randomly sel [#permalink] New post   14 Jun 2019, 06:14
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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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Re: A bowl contains 10 apples, 2 of which are bad. If someone randomly sel [#permalink] New post   16 Feb 2024, 09:26
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Re: A bowl contains 10 apples, 2 of which are bad. If someone randomly sel [#permalink]
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