A basket contains 5 apples of which one is rotten. 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 the rotten apple?

A. 1/5
B. 3/10
C. 2/5
D. 1/2
E. 3/5


Direct combinatorial approach:

\(Probability=\frac{# \ of \ favorable \ outcomes}{total \ # \ of \ outcomes}\):

Favorable outcomes: \(C^1_1*C^1_4\) --> \(C^1_1\) choosing one rotten out of 1 and \(C^1_4\) choosing any for the second apple;

Total # of outcomes: \(C^2_5\), choosing 2 apples out of 5;

\(P=\frac{C^1_1*C^1_4}{C^2_5}=\frac{2}{5}\).


Reverse combinatorial approach:

Let's count the probability of the opposite event and subtract this value from 1. The opposit event will be if out of 2 apples both are good, so \(P=1-\frac{C^2_4}{C^2_5}=\frac{2}{5}\).


Direct probability approach:

\(P=\frac{1}{5}*\frac{4}{4}+\frac{4}{5}*\frac{1}{4}=\frac{2}{5}\): probability of {first apple rotten, second apple good} plus the probability of {first apple good, second apple rotten}.


Reverse probability approach:

Again 1- the probability of the opposite event --> \(P=1-\frac{4}{5}*\frac{3}{4}=\frac{2}{5}\).

Answer: C.


Check Probability chapter of Math Book for more on this issues: http://gmatclub.com/forum/math-probability-87244.html

Hope it helps.
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p (both apples are not spoiled) = 4C2/5C2 = 4*3/5*4 = 3/5

p (one of the apple spoiled) = 1-3/5 = 2/5

This is the best explanation on probability I have ever got = using all the approaches. Thanks! I now know how Combinations/Direct Probability etc. all tie together!

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Hope it helps.
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+1 C

Total number of combinations: 5!/3! = 20
Number of combinations that include the spoiled apple:
When the spoiled apple is the first apple picked: 1*4= 4
When the spoiled apple is the second apple picked: 4*1 = 4
Total: 8

Probability: 8/20 = 2/5

C.
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No. of apples that are not spoiled = 4
No. of apples that are spoiled = 1

The question is asking us to find out the probability of finding one spoiled apple among the 2 randomly selection ...

We can have 2 scenarios , Scenario A) where the first apple selected is not spoiled , and the second apple selected is spoiled and Scenario B) where the first apple selected is spoiled and the second selected is not spoiled ..

Scenario A) We have four apples out of five that are not spoiled , therefore probability is 4/5 , the second apple is a spoiled one out of the remaining 4 , therefore we have overall probability of this outcome as (4/5) x (1/4) = 4/20

Scenario B ) We have only one apple out of 5 that is spoiled so probability is 1/5 , the second apple can be one from any 4 remaining thefore 4/4... Probability of this outcome is (1/5) x (4/4) = 1/5 ....

We must add these two scenarios therefore we get 4/20 + 1/5 = 8/20 = 2/5 (C)
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We need to determine the probability of selecting a spoiled apple and a non-spoiled apple when selecting two apples.

The number of ways to select the spoiled apple is 1C1 = 1. The number of ways to select a good apple is 4C1 = 4. Thus, the spoiled apple and a good apple can be selected in 1 x 4 = 4 ways.

The number of ways to select 2 apples from 5 is 5C2 = (5 x 4)/2! = 20/2 = 10.

Thus, the probability of selecting the spoiled apple and a good apple is 4/10 = 2/5.

Alternate Solution:

There are two outcomes that satisfy the requirement that the spoiled apple (S) is chosen along with a good apple (G), either S-G or G-S. The probability of S-G is (1/5)(4/4) = 1/5. The probability of G-S is (4/5)(1/4) = 1/5. Since either outcome satisfies our requirement, we add these two probabilities: 1/5 + 1/5 = 2/5.

Answer: C
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Hi All

We're told that we have 4 regular apples and 1 spoiled apple. We're asked - if you grab 2 apples, then what's the probability of getting the spoiled apple?

The question can be solved in a couple of ways. Here's how to track each of the different outcomes that matches what we're looking for:

1st regular, 2nd spoiled = (4/5)(1/4) = 4/20
1st spoiled, 2nd regular = (1/5)(4/4) = 4/20

Total ways to get 1 spoiled and 1 regular = 4/20 + 4/20 = 8/20 = 2/5

Final Answer:

GMAT assassins aren't born, they're made,
Rich
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\(5\,\,{\text{apples}}\,\,\left\{ \begin{gathered}
\,1\,\,{\text{spoiled}} \hfill \\
\,4\,\,{\text{good}} \hfill \\
\end{gathered} \right.\)

\(? = P\left( {{\text{extract 2,}}\,\,{\text{1}}\,\,{\text{spoiled}}} \right)\)

\({\text{total}}:\,\,C\left( {5,2} \right) = 10\,\,\,{\text{equiprobables}}\)

\({\text{favorable:}}\,\,{\text{4}}\,\,\,\,\left( {{\text{spoiled}} + {\text{any}}\,\,{\text{good}}} \right)\)

\(? = \frac{4}{{10}} = \frac{2}{5}\)


This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.
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