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A basket contains 5 apples, of which 1 is spoiled and the

by BTGmoderatorLU » Fri Sep 14, 2018 6:49 pm

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Source: GMAT Prep

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 possibility that the 2 apples selected will include the spoiled apple?

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

The OA is C.
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by Jay@ManhattanReview » Fri Sep 14, 2018 7:36 pm
BTGmoderatorLU wrote:Source: GMAT Prep

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 possibility that the 2 apples selected will include the spoiled apple?

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

The OA is C.
There are 4 good and 1 bad apple.

Probability of picking 2 apples such that 1 is good and the other is bad = (4C1 * 1C1) / (5C2) = 4/(5.4/1.2) = 2/5

The correct answer: C

Hope this helps!

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by GMATGuruNY » Sat Sep 15, 2018 1:37 am
BTGmoderatorLU wrote:Source: GMAT Prep

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 possibility that the 2 apples selected will include the spoiled apple?

A. 1/5
B. 3/10
C. 2/5
D. 1/2
E. 3/5
P(spoiled apple is selected) = 1 - P(spoiled apple is NOT selected).

P(1st apple is not spoiled) = 4/5. (Of the 5 apples, 4 are not spoiled).
P(2nd apple is not spoiled) = 3/4. (Of the 4 remaining apples, 3 are not spoiled).
Since we want both events to happen, we multiply the fractions:
4/5 * 3/4 = 3/5.

Thus, P(spoiled apple is selected) = 1 - 3/5 = 2/5.

The correct answer is C.
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by fskilnik@GMATH » Sun Sep 16, 2018 7:06 am
BTGmoderatorLU wrote:Source: GMAT Prep

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 possibility that the 2 apples selected will include the spoiled apple?

A. 1/5
B. 3/10
C. 2/5
D. 1/2
E. 3/5
\[5\,\,{\text{apples}}\,\,\left\{ \begin{gathered}
\,1\,\,{\text{spoiled}} \hfill \\
\,4\,\,{\text{good}} \hfill \\
\end{gathered} \right.\]

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

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

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

\[? = \frac{4}{{10}} = \frac{2}{5}\]


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

Regards,
Fabio.
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by Scott@TargetTestPrep » Tue Sep 18, 2018 5:55 pm
BTGmoderatorLU 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 possibility that the 2 apples selected will include the spoiled apple?

A. 1/5
B. 3/10
C. 2/5
D. 1/2
E. 3/5
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 one 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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by [email protected] » Tue Sep 18, 2018 7:46 pm
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: C

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