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There are 5 apples in a bag. 4 apples are good but 1 apple is rotten.

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Math Revolution GMAT Instructor
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There are 5 apples in a bag. 4 apples are good but 1 apple is rotten.  [#permalink]

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New post 07 Aug 2017, 01:28
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There are 5 apples in a bag. 4 apples are good but 1 apple is rotten. If you take out 2 apples from the bag, what is the probability that 1 apple selected is rotten?

A. \(\frac{1}{3}\)

B. \(\frac{1}{4}\)

C. \(\frac{2}{5}\)

D. \(\frac{3}{5}\)

E. \(\frac{1}{7}\)

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Re: There are 5 apples in a bag. 4 apples are good but 1 apple is rotten.  [#permalink]

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New post 07 Aug 2017, 05:58
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MathRevolution wrote:
There are 5 apples in a bag. 4 apples are good but 1 apple is rotten. If you take out 2 apples from the bag, what is the probability that 1 apple selected is rotten?

A. \(\frac{1}{3}\)

B. \(\frac{1}{4}\)

C. \(\frac{2}{5}\)

D. \(\frac{3}{5}\)

E. \(\frac{1}{7}\)


Select 2 apples from 5 apples, there are totally 5C2 = 10 selections.

Select 2 apples that is good from 5 apples, there are totally 4C2 = 6 selections.

The probability that select 2 apples from 5 apples, there is one rotten apple, is: 1 - 6/10 = 4/10 = 2/5

The answer is C
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Re: There are 5 apples in a bag. 4 apples are good but 1 apple is rotten.  [#permalink]

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New post 07 Aug 2017, 06:08
Out of 4 good apples 1 good apple can be chosen in 4 ways and 1 rotten apple can be chosen in one way. Therefore total number of favourable ways= 4. Total number of possible ways=10. Thus C.
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Re: There are 5 apples in a bag. 4 apples are good but 1 apple is rotten.  [#permalink]

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New post 07 Aug 2017, 07:57
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Another way can be by making 2 cases

1st case
1st apple to be picked up is rotten, 2nd is not:

1/5*4/4

2nd case

1st apple to be picked up is not rotten, 2nd is rotten:

4/5*1/4


Adding both the cases we get 2/5


Please hit kudos if my answer helped.
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Re: There are 5 apples in a bag. 4 apples are good but 1 apple is rotten.  [#permalink]

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New post 09 Aug 2017, 18:08
==> From probability=want/total, want=selecting one rotten apple and one good apple, and total=selecting 2 of 4 apples, so you get
probability=4C1*1C1/5C2=(4)(1)/(5*4/2!)=4/10=2/5.

The answer is C.
Answer: C
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Re: There are 5 apples in a bag. 4 apples are good but 1 apple is rotten.  [#permalink]

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New post 10 Aug 2017, 10:30
MathRevolution wrote:
There are 5 apples in a bag. 4 apples are good but 1 apple is rotten. If you take out 2 apples from the bag, what is the probability that 1 apple selected is rotten?

A. \(\frac{1}{3}\)

B. \(\frac{1}{4}\)

C. \(\frac{2}{5}\)

D. \(\frac{3}{5}\)

E. \(\frac{1}{7}\)


The number of ways to select the one rotten apple is 1C1 = 1. The number of ways to select 1 good apple is 4C1 = 4, so the total number of ways to select the 2 apples with one of them rotten is 1 x 4 = 4.

The number of ways to select 2 apples from 5 is 5C2 = (5 x 4)/2! = 10.

So, the probability of selecting 2 apples with one of them rotten is 4/10 = 2/5.

Answer: C
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Re: There are 5 apples in a bag. 4 apples are good but 1 apple is rotten.  [#permalink]

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New post 25 Feb 2019, 20:28
Since the total countable items is 5 . As there exists 4 good and 1 rotten apples . simply at once the selection of apples will have 1 good and 1 rotten apple.

As probability is defined as favorable outcomes / total outcomes . Answer is 2/5
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Re: There are 5 apples in a bag. 4 apples are good but 1 apple is rotten.  [#permalink]

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New post 02 Mar 2019, 10:10
MathRevolution wrote:
There are 5 apples in a bag. 4 apples are good but 1 apple is rotten. If you take out 2 apples from the bag, what is the probability that 1 apple selected is rotten?

A. \(\frac{1}{3}\)

B. \(\frac{1}{4}\)

C. \(\frac{2}{5}\)

D. \(\frac{3}{5}\)

E. \(\frac{1}{7}\)


The number of ways to select 2 apples consisting of 1 good apple and 1 rotten apple is selected is 4C1 x 1C1 = 4 x 1 = 4.

The number of ways to select 2 apples from 5 is 5C2 = (5 x 4)/2! = 10.

So the probability is 4/10 = 2/5.

Alternate Solution:

There are two ways to get 1 good and 1 bad apple in this scenario: GB or BG

The probability of GB is 4/5 x 1/4 = 1/5

The probability of BG is 1/5 x 4/4 = 1/5

Since either outcome satisfies the requirement, we add the two probabilities, obtaining 1/5 + 1/5 = 2/5.

Answer: C
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Re: There are 5 apples in a bag. 4 apples are good but 1 apple is rotten.  [#permalink]

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New post 24 Jun 2019, 08:59
Hi, I was wandering if the approach of "1 - counterprobability ( =no rotten apple)" is valid? I.e. 1 - 4/5*3/4= 1 - 12/20= 8/20 = 2/5
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There are 5 apples in a bag. 4 apples are good but 1 apple is rotten.  [#permalink]

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New post 24 Jun 2019, 09:49
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max96hsv wrote:
Hi, I was wandering if the approach of "1 - counterprobability ( =no rotten apple)" is valid? I.e. 1 - 4/5*3/4= 1 - 12/20= 8/20 = 2/5

Yes! Great and totally valid solution. While this method is most commonly used with more complex "at least/more than" problems, any time we can break the scenario down into complementary events (two events where one and only one event must occur), we can solve for one event by finding 1 - the probability of its complement.

In this case, the complement of picking the bad apple is NOT picking the bad apple, so the probability of picking the bad apple = 1 - the probability of NOT picking the bad apple.

On the first pick, we have a 4/5 chance of NOT picking the bad apple (out of the 5 total apples in the bag, 4 are NOT bad). On the second pick, we have a 3/4 chance of NOT picking the bad apple (of the 4 apples left in the bag after the first pick, 3 are NOT bad). So we have a 4/5*3/4 = 12/20 = 3/5 chance of NOT picking the bad apple in either of the two picks. So the chance of picking the bad apple in either of the two picks is 1 - 3/5 = 2/5.
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There are 5 apples in a bag. 4 apples are good but 1 apple is rotten.   [#permalink] 24 Jun 2019, 09:49
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