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apple probability

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by B166418 » Sun Nov 25, 2012 8:19 am
A basket contains 5 apples ,of which 1 is spoiled and the rest are good ,If Henry is to select 2 apples from the basket simultaneously and at random,what is the probability that the 2 apples selected will include spoiled apple

1/5
3/10
2/5
1/2
3/5
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by Anindya Madhudor » Sun Nov 25, 2012 9:12 am
First, find the number of ways 2 apples can be selected simultaneously from 5 apples. This is given by 5C2=10.

Then, find how many ways two apples can be selected such that you have the spoiled apple and any other apple. There are 4 ways to select the spoiled apple and any other apple.

So, the answer is 4/10=2/5.
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by Brent@GMATPrepNow » Sun Nov 25, 2012 9:14 am
B166418 wrote:A basket contains 5 apples ,of which 1 is spoiled and the rest are good ,If Henry is to select 2 apples from the basket simultaneously and at random,what is the probability that the 2 apples selected will include spoiled apple

A) 1/5
B) 3/10
C) 2/5
D) 1/2
E) 3/5
P(selection includes spoiled apple) = 1 - P(selection does not include spoiled apple)

Let's take a closer look at P(selection does not include spoiled apple)
= P(1st apple good AND 2nd apple good)
= P(1st apple good) x P(2nd apple good)
= (4/5) x (3/4)
= 3/5

Aside
P(1st apple good) = 4/5 because there are 5 apples and 4 of them are good
P(2nd apple good) = 3/4 because, once a good apple is selected 1st, there are 4 apples remaining, and 3 of them are good

So...P(selection includes spoiled apple) = 1 - (selection does not include spoiled apple)
= 1 - 3/5
= [spoiler]2/5 = C[/spoiler]

Cheers,
Brent
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by B166418 » Sun Nov 25, 2012 10:10 am
perfect Brent
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by Titanp321 » Sun Nov 25, 2012 7:57 pm
Another approach can be taken. It is very similar to Brent's idea, but without the 1-P(x) trick.

First, we want to think about all the situation in which the person does pick the spoiled apple. If we go beyond the "simultaneous" comment and think about this problem as removal without replacement, then we can identify two situations (call them case I and case II) in which the spoiled apple is picked. So, either case I OR case II can produce the desired event. Meaning, the individual probabilities for case I and case II can be added.

Moreover, the probability of pulling apples from the basket one after the other can be thought of as independent events.

(1) The person picks the spoiled apple first and a non-spoiled apple second.
P(picking a spoiled apple)*P(picking a non-spoiled apple)= 1/5*4/4 = 1/5


(2) The person picks the non-spoiled apple first and the spoiled apple second.
P(picking a non-spoiled apple)*P(picking a spoiled apple)= 4/5*1/4 = 1/5

P(picking a spoiled apple) = P(case 1 OR case 2) = P(case 1) + P(case 2) = 1/5+1/5 = 2/5


(there is no subtraction of P(case 1 AND case 2) because the event are mutually exclusive)[/b]
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by GMATGuruNY » Mon Nov 26, 2012 4:26 am
B166418 wrote:A basket contains 5 apples ,of which 1 is spoiled and the rest are good ,If Henry is to select 2 apples from the basket simultaneously and at random,what is the probability that the 2 apples selected will include spoiled apple

1/5
3/10
2/5
1/2
3/5
An alternate approach:

P = (good outcomes)/(total possible outcomes).

Total possible outcomes:
Number of ways to select 2 apples from 5 options = 5C2 = (5*4)/(2*1) = 10.

Good outcomes:
In a good outcome, the spoiled apple is combined with one of the 4 unspoiled apples.
Number of unspoiled apples that could be combined with the spoiled apple = 4.

P(good outcome):
(combinations with a spoiled apple)/(total possible combinations) = 4/10 = 2/5.

The correct answer is C.
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