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

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Intern
Joined: 17 Apr 2010
Posts: 38
A bowl contains 10 apples, 2 of which are bad. If someone randomly se  [#permalink]

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23 Apr 2010, 12:10
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Question Stats:

83% (00:34) correct 17% (13:06) wrong based on 7 sessions

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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?

2/3

I found this approach:

First selecting the probability of all good apples, then subtracting that from one in order to find the probability of at least selecting one bad apple.
P(all good)=
(8/10)*(7/9)*(6/8)*(5/7) = 1/3

P(all good)= 1/3

P(at least one bad)= 1-P(all good)

Can anybody suggest an alternative approach?

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Manager
Joined: 13 Dec 2009
Posts: 116
Re: A bowl contains 10 apples, 2 of which are bad. If someone randomly se  [#permalink]

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23 Apr 2010, 12:23
1
gmatt14 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?

I found this approach:

First selecting the probability of all good apples, then subtracting that from one in order to find the probability of at least selecting one bad apple.
P(all good)=
(8/10)*(7/9)*(6/8)*(5/7) = 1/3

P(all good)= 1/3

P(at least one bad)= 1-P(all good)

Can anybody suggest an alternative approach?

required probability will be = probability of(one bad apple selected and 3 good apple selected ) + probability of( 2 bad apple selected and 2 good apple selected )

$$= \frac{ 2C1*8C3}{10C4}+\frac{ 2C2*8C2}{10C4} =\frac{ 2}{3}$$
Intern
Joined: 17 Apr 2010
Posts: 38
Re: A bowl contains 10 apples, 2 of which are bad. If someone randomly se  [#permalink]

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23 Apr 2010, 12:30
Sorry, but what do you mean by the variable "C?"
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Status: Nothing comes easy: neither do I want.
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Re: A bowl contains 10 apples, 2 of which are bad. If someone randomly se  [#permalink]

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23 Apr 2010, 12:31
gmatt14 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?

I found this approach:

First selecting the probability of all good apples, then subtracting that from one in order to find the probability of at least selecting one bad apple.
P(all good)=
(8/10)*(7/9)*(6/8)*(5/7) = 1/3

P(all good)= 1/3

P(at least one bad)= 1-P(all good)

Can anybody suggest an alternative approach?

the one stated above by u is the best approach.
Alternative approach::

atleast one means case1+case 2

case1-- one bad 9 good total ways = 2c1*8c3
case2 2 bad 8 go0d total ways = 2c2*8c2

total ways = 10c4

probability = $$\frac{{2c1*8c3 + 2c2*8c2}}{10c4}$$ = 2/3
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Manager
Joined: 13 Dec 2009
Posts: 116
Re: A bowl contains 10 apples, 2 of which are bad. If someone randomly se  [#permalink]

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23 Apr 2010, 12:35
gmatt14 wrote:
Sorry, but what do you mean by the variable "C?"

http://gmatclub.com/forum/math-combinatorics-87345.html
Intern
Joined: 17 Apr 2010
Posts: 38
Re: A bowl contains 10 apples, 2 of which are bad. If someone randomly se  [#permalink]

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23 Apr 2010, 12:36
I lost track.......Can you please break it down...........
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Re: A bowl contains 10 apples, 2 of which are bad. If someone randomly se  [#permalink]

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20 Sep 2017, 06:29
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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