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

Question

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

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

Bunuel · Accepted Answer

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: https://gmatclub.com/forum/math-probability-87244.html

Hope it helps.

AndersonBound · Answer

Total Apples = 5
Spoiled Apples = 1
Not Spoiled Apples = 4

Possible outcomes: 
(Apple 1) Not Spoiled, (Apple 2) Spoiled; = (4/5)*(1/4) = 4/20 or 2/10.
 OR 
(Apple 1) Spoiled, (Apple 2) Not Spoiled; = (1/5)*(4/4) = 4/20 or 2/10.

Total Probability = 2/10 + 2/10 = 4/10 or  2/5 .

x2suresh · Answer

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

shash · Answer

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

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

shash · Answer

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!

rg1 · Answer

@ Bunuel, you're awesome. Do you focus on anything else apart form solving Quant questions on GC?:)
Just kidding. Kudos +1..

Bunuel · Answer

21. Combinatorics/Counting Methods 
     Theory   
 Math: Combinatorics  
 Combinatorics Made Easy! 
 1 hour Combinatorics Video Lesson 
 Combinations while dealing with similar and dissimilar things 
 When Permutations & Combinations and Data Sufficiency Come Together 
     Questions   
 Advanced Topic: Probability and Combinations Questions With Solutions 
 Letter Arrangements in Words 
 Seating Arrangements in a Row and around a Table 
 Constructing Numbers, Codes and Passwords 
 DS Questions 
 PS Questions

22. Probability 
     Theory   
 Math: Probability 
 Probability Made Easy! 
 GMAT Tip of the Week: 99 Problems But Probability Ain't One 
     Questions   
 Probability and Geometry Problems 
 Advanced Topic: Probability and Combinations Questions With Solutions 
 DS Questions 
 PS Questions

For more:
 ALL YOU NEED FOR QUANT ! ! ! 
 Ultimate GMAT Quantitative Megathread

Hope it helps.

mymbadreamz · Answer

Bunuel, you are awesome!

metallicafan · Answer

+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.

EvaJager · Answer

Computing directly probabilities and using the complementary event:

P(spoiled apple selected) = 1 - P(no spoiled apple selected) = 1 - P(first apple not spoiled)*P(second apple also not spoiled) = 1 - (4/5)*(3/4) = 1 - 3/5 = 2/5.

Answer C

vomhorizon · Answer

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)

mbaiseasy · Answer

Anagram helps.

ScottTargetTestPrep · Answer

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

mtk10 · Answer

Good Apple= A
Spoiled Apple = S

Total ways of selecting two appls = ( A * S) + ( S*S)

What i want to ask is why are we not consideringt the order of selection here. For example ( A*S) could also be taken as ( S*A). Why is the order not improtant here?

Is it because the word   has been mentioned and that makes this a combination question?

EMPOWERgmatRichC · Answer

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

GMAT assassins aren't born, they're made, 
Rich

BrentGMATPrepNow · Answer

The two approaches shown above are the best (faster) approaches, but here's one more.
Since there are so few objects involved (5 apples), we should be able to quickly answer the question by simply  listing and counting

Let A, B, C, D, and  E  represent the 5 apples, and let  E  represent the  SPOILED APPLE

We want to select 2 apples at random. So, let's list all of the possible outcomes: 
1) AB
2) AC
3) AD
4) A E 
5) BC
6) BD
7) B E 
8) CD
9) C E 
10) D E

So, there are  10  possible outcomes
Of those  10  possible outcomes,   4   outcomes include the  SPOILED APPLE

So, P(selection includes the spoiled apple) =   4  / 10  = 2/5

Answer: C

Cheers,
Brent

fskilnik · Answer

5\,\,{	ext{apples}}\,\,\left\{ \begin{gathered}
 \,1\,\,{	ext{spoiled}} \hfill \
 \,4\,\,{	ext{good}} \hfill \ 
\end{gathered} ight.

? = P\left( {{	ext{extract 2,}}\,\,{	ext{1}}\,\,{	ext{spoiled}}} ight)

{	ext{total}}:\,\,C\left( {5,2} ight) = 10\,\,\,{	ext{equiprobables}}

{	ext{favorable:}}\,\,{	ext{4}}\,\,\,\,\left( {{	ext{spoiled}} + {	ext{any}}\,\,{	ext{good}}} ight)

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

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

Regards,
Fabio.

GK002 · Answer

What am I missing here?

There is 1 Bad apple and 4 Good apples - so BGGGG are possible options.

We are selecting two.

Total Favorable outcomes / Total outcomes

Total favorable outcomes are 2 because you can select BG or GB

Total outcomes is 5C2 = 10 possibilities:

Therefore, 2/10 = 1/5

Please help me understand where I went wrong? Thanks

GK002 · Answer

NVM - I think I got it. Favorable outcomes are actually 4:

1. B, G1
2. B, G2
3. B, G3
4. B, G4

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