BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

A basket has 5 apples

This topic has expert replies
Post new topic Post Reply
Previous Topic Next Topic
Senior | Next Rank: 100 Posts
Posts: 31
Joined: Sat Aug 23, 2008 8:23 pm
Location: Toronto
Thanked: 1 times
GMAT Score:610

A basket has 5 apples

by ebehboud » Sat May 02, 2009 11:46 am
A basket has 5 apples, one is spoiled. If Henry picked 2 apples simultaneously and at random, what is probability that the 2 selected apples will include the spoiled one.
a. 1⁄5
b. 2⁄10
c. 2⁄5
d. 1⁄2
e. 3⁄5
Top
Source: — Problem Solving |
Junior | Next Rank: 30 Posts
Posts: 28
Joined: Wed Apr 15, 2009 11:35 pm
Thanked: 1 times

by mixpanda » Sat May 02, 2009 2:58 pm
IMO the answer is C.) 2/5

SOLUTION A:

The formula for combinations (since order is not an issue) is:

n!/(k!(n-k)!)

This is what we can use to figure out how many different ways there are to arrange the 5 apples so that there are groups of 2 chosen.

5!/(2!(5-2)!)
5!/(2!3!)

5x4x3x2x1 / 2x1 x 3x2x1

Cancel out

5x4x3x2x1 / 2x1 x 3x2x1

So 20/2 = 10. There are 10 different groups of 2 that can be created by choosing 2 apples

Since one of them has to be the rotten apple, it has to be paired with one of the four good apples. So there are 4 favorable outcomes out of 10 possible outcomes. 4/10 = 2/5.

SOLUTION B:

Draw out 5 circles to represent the 5 apples.

O O @ O O

The third one is rotten.

Then draw lines from each apple to each other apple, representing the different possible pairs.

If you draw the lines you can see that there are 10 different pairs possible. Since order is not an issue here, it doesn't matter whether the rotten apple is picked first or second.

EDIT: Ignore the lines - the spacing is off in the post.

/ - - - - - - \
// - - - - -\ \
/// - - -\ \ \
////- \ \ \ \
O O @ O O

That's 4 pairs.

/ - - - - - \
// - - - \ \
///- -\ \ \
O O @ O O

That's 3 more pairs.

/ - - - \
//- -\ \
O O @ O O

That's 2 more pairs.


/- -\
O O @ O O

That's 1 more pair.

So 10 pairs, of which four of them have a rotten apple. 2/5.
Top
Senior | Next Rank: 100 Posts
Posts: 31
Joined: Sat Aug 23, 2008 8:23 pm
Location: Toronto
Thanked: 1 times
GMAT Score:610

by ebehboud » Sat May 02, 2009 4:41 pm
Very well done. Thanks

basically the answer is

(1x 4C1) / (5C2) = 2/5
Top
Master | Next Rank: 500 Posts
Posts: 101
Joined: Tue Apr 07, 2009 12:30 am
Thanked: 2 times
GMAT Score:750

by Svedankae » Wed May 13, 2009 1:00 pm
would it be wrong to simply say that there is 1/5 chance that the rotten apple is the first one that you chose and again a 1/5 chance that it is the second one that you chose (simoultaneous choice)?

So what it comes down to 1/5 + 1/5 = 2/5

or is that bogus?
Top
Newbie | Next Rank: 10 Posts
Posts: 4
Joined: Sun Mar 30, 2008 2:50 am
Thanked: 2 times

by gmahendru » Sun May 31, 2009 11:07 am
or... may be like this..

Case1: First apple rotten i.e. 1/5 and second good i.e. 4/4=1
Therefore, 1/5*1 = 1/5

Case2: First apple good i.e. 4/5 and second rotten i.e. 1/4
Therefore, 4/5*1/4 = 1/5

Adding both cases: 1/5 + 1/5 = 2/5
Top
Junior | Next Rank: 30 Posts
Posts: 12
Joined: Sat May 16, 2009 9:02 pm

by Adevas » Sun May 31, 2009 11:52 am
i agree

(Probability of bad apple) 1/5 * 4/4 (P of good apple) + 4/5 (P Good Apple) * 1/4 (P Bad Apple) = 2/5
Top

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 3380
Joined: Mon Mar 03, 2008 1:20 am
Thanked: 2256 times
Followed by:1535 members
GMAT Score:800

by lunarpower » Fri Jun 05, 2009 2:46 am
there's also a clever angle from which you can look at this problem.

the deal is that the five apples are separated into two groups - the 2 that are selected, and the 3 that are not selected - at random. also, a random apple is spoiled.
therefore, since both events are random (neither is predicated upon the other), it makes no difference which is presumed to have occurred first (the selection or the spoilage).

in the problem statement, the 1 apple is presumed to be spoiled already, before selection occurs.
if you consider the problem the other way, though, it's much easier to solve.
i.e., assume that the apples are already separated into the 2 "chosen" ones (#1 or #2) and the 3 "non-chosen" ones (#3, #4, #5). then one of them spoils, at random. the question is:
what's the probability that the apple that spoils is #1 or #2?
this probability is of course 2 out of 5, and we're done.

note that not all problems can be re-cast in this way, but it's golden when it works.
Ron has been teaching various standardized tests for 20 years.

--

Pueden hacerle preguntas a Ron en castellano
Potete chiedere domande a Ron in italiano
On peut poser des questions à Ron en français
Voit esittää kysymyksiä Ron:lle myös suomeksi

--

Quand on se sent bien dans un vêtement, tout peut arriver. Un bon vêtement, c'est un passeport pour le bonheur.

Yves Saint-Laurent

--

Learn more about ron
Top
User avatar
Master | Next Rank: 500 Posts
Posts: 170
Joined: Tue May 26, 2009 12:00 pm
Thanked: 5 times

by yogami » Tue Jun 09, 2009 1:55 am
I will take an alternative approach. Probability of picking up only good apples is 4C2/5C2 = 3/5. Whats the opposite of this? At least one of them is kaput. So take the complement 1- 3/5 = 2/5 is the answer


This is applicable to any general probability question like this. 10 apples in the basket, 4 are spoilt and 2 are picked at random so whats the probability that at least one of them is bad. 1 - 6C2/10C2.

Wht is the probability of at least one being good? 1 - 4C2/10C2
Wats the probability of exactly one good and one bad? 6C1*4C1/10C2

Correct me if I am wrong plz..
200 or 800. It don't matter no more.
Top

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 3380
Joined: Mon Mar 03, 2008 1:20 am
Thanked: 2256 times
Followed by:1535 members
GMAT Score:800

by lunarpower » Tue Jun 09, 2009 2:22 am
yogami wrote:I will take an alternative approach. Probability of picking up only good apples is 4C2/5C2 = 3/5. Whats the opposite of this? At least one of them is kaput. So take the complement 1- 3/5 = 2/5 is the answer


This is applicable to any general probability question like this. 10 apples in the basket, 4 are spoilt and 2 are picked at random so whats the probability that at least one of them is bad. 1 - 6C2/10C2.

Wht is the probability of at least one being good? 1 - 4C2/10C2
Wats the probability of exactly one good and one bad? 6C1*4C1/10C2

Correct me if I am wrong plz..
this is all excellent.

a good takeaway for anyone who happens to be reading this post:
if you don't immediately see how to calculate a probability, you should consider whether it might be easier to calculate the probability of the OPPOSITE EVENT.

if the opposite event's probability is easier to calculate, then you should just calculate that probability and subtract it from 1.
Ron has been teaching various standardized tests for 20 years.

--

Pueden hacerle preguntas a Ron en castellano
Potete chiedere domande a Ron in italiano
On peut poser des questions à Ron en français
Voit esittää kysymyksiä Ron:lle myös suomeksi

--

Quand on se sent bien dans un vêtement, tout peut arriver. Un bon vêtement, c'est un passeport pour le bonheur.

Yves Saint-Laurent

--

Learn more about ron
Top
Master | Next Rank: 500 Posts
Posts: 171
Joined: Mon Jun 01, 2009 8:59 pm
Thanked: 8 times

yes

by brick2009 » Thu Jun 11, 2009 8:01 pm
Sve...that is the actual way of solving it...
its simple..other solns are detailed...but lengthy...
Svedankae wrote:would it be wrong to simply say that there is 1/5 chance that the rotten apple is the first one that you chose and again a 1/5 chance that it is the second one that you chose (simoultaneous choice)?

So what it comes down to 1/5 + 1/5 = 2/5

or is that bogus?
Top

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 3380
Joined: Mon Mar 03, 2008 1:20 am
Thanked: 2256 times
Followed by:1535 members
GMAT Score:800

Re: yes

by lunarpower » Wed Jun 17, 2009 12:09 am
brick2009 wrote:would it be wrong to simply say that there is 1/5 chance that the rotten apple is the first one that you chose and again a 1/5 chance that it is the second one that you chose (simoultaneous choice)?

So what it comes down to 1/5 + 1/5 = 2/5

or is that bogus?
nope, that's legitimate.

just remember that you can't add probabilities unless the events in question are mutually exclusive.
in this case, that criterion is fulfilled, since the rotten apple clearly can't be both the first and second one. so you have the green light to add the probabilities together.
Ron has been teaching various standardized tests for 20 years.

--

Pueden hacerle preguntas a Ron en castellano
Potete chiedere domande a Ron in italiano
On peut poser des questions à Ron en français
Voit esittää kysymyksiä Ron:lle myös suomeksi

--

Quand on se sent bien dans un vêtement, tout peut arriver. Un bon vêtement, c'est un passeport pour le bonheur.

Yves Saint-Laurent

--

Learn more about ron
Top
Senior | Next Rank: 100 Posts
Posts: 60
Joined: Thu Sep 24, 2009 2:38 pm
Location: NYC
Thanked: 33 times
Followed by:9 members
GMAT Score:710

by benjiboo » Fri Oct 16, 2009 5:11 pm
yogami wrote:I will take an alternative approach. Probability of picking up only good apples is 4C2/5C2 = 3/5. Whats the opposite of this? At least one of them is kaput. So take the complement 1- 3/5 = 2/5 is the answer


This is applicable to any general probability question like this. 10 apples in the basket, 4 are spoilt and 2 are picked at random so whats the probability that at least one of them is bad. 1 - 6C2/10C2.

Wht is the probability of at least one being good? 1 - 4C2/10C2
Wats the probability of exactly one good and one bad? 6C1*4C1/10C2

Correct me if I am wrong plz..
I think there is an easier way to explain this. Lets work with "10 apples in the basket, 4 are spoiled and 2 are picked at random together"

There are 10 apples, and we are choosing 2, so the total number of ways to choose is 10c2=45

1)whats the probability that at least one of them is bad?

We want the (# of ways to choose at least 1 bad) / total number of ways to choose.


The # of ways to choose at least one bad is: We can choose two apples at once and have both be bad, or we can choose two apples at once and have one be bad and one be good. Lets write this mathematically:

Choose one bad and one good: 4c1 * 6c1 (there are 4 bad apples, we must choose 1, and 6 good apples, we must choose 1)

Choose both bad: 4c2 (we have 4 bad apples, and must choose two).

Notice how I bolded the word OR. That is because we add these two, not multiply. It is sort of like probability rules, where: probability of 1 and of 2 occurring = multiply together, while the probability of 1 or 2 occurring is added together.

Now put everything together:

(4c1 * 6c1) + (4c2) / 45 (the 45 comes from the 10c2).

---------

Im going to try and explain it a little differently just in case someone did not understand.

Pretend you have two spots: "__ and __", and each spot represents a picking of an apple that u are grabbing out of the bag.

If you must grab at least one bad apple, then fill the first spot with the choice of 4 bad apples, and ur choosing one: 4c1. Now we have "4c1 and __"

For the second spot, we can have either a bad apple, or a good apple. If we choose a good apple, you are doing 6c1, so we have "4c1 and 6c1" in our hand. They are happening together, so multiply them!

Now we start over, put all the apples back in the bag. The second way to choose is if we took two bad apples together. This is common sense, its the ONLY OTHER WAY TO HAVE ATLEAST 1 BAD APPLE! There are 4 bad apples, and we want to choose 2 of them. Thats 4c2.

Now add these together, and put them over 10c2.

Hope this helps!
Top

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 3380
Joined: Mon Mar 03, 2008 1:20 am
Thanked: 2256 times
Followed by:1535 members
GMAT Score:800

by lunarpower » Sat Oct 17, 2009 2:07 am
ok people, i just want to step in here with a little perspective.

remember the following:
the gmat is never going to write a problem that REQUIRES you to combine multiple "c" and "p" formulas.
it's NEVER going to happen.

this doesn't mean that you won't be able to solve problems with such a method - in many cases, you will - but, if you do, you'll be doing WAY too much work.

--

here's a simple solution to this problem.

note that the desired event - "the spoiled apple is one of the two apples selected" is a complex event.

when you're given a complex event, you should consider the OPPOSITE event. often, that's a simple event (i.e., only one possibility).

that's what happens here:
the opposite event is "NEITHER of the two selected apples is the spoiled apple". i.e., both of them are good.
since there are 4 good apples, the probability of that event is (4/5)(3/4), or 3/5.

since that's the opposite event, our desired event has a probability of 1 - 3/5 = 2/5.
Ron has been teaching various standardized tests for 20 years.

--

Pueden hacerle preguntas a Ron en castellano
Potete chiedere domande a Ron in italiano
On peut poser des questions à Ron en français
Voit esittää kysymyksiä Ron:lle myös suomeksi

--

Quand on se sent bien dans un vêtement, tout peut arriver. Un bon vêtement, c'est un passeport pour le bonheur.

Yves Saint-Laurent

--

Learn more about ron
Top
Senior | Next Rank: 100 Posts
Posts: 60
Joined: Thu Sep 24, 2009 2:38 pm
Location: NYC
Thanked: 33 times
Followed by:9 members
GMAT Score:710

by benjiboo » Fri Oct 30, 2009 7:31 pm
well said Ron. I always enjoy reading your post.
Top
Newbie | Next Rank: 10 Posts
Posts: 2
Joined: Wed Nov 04, 2009 11:31 am

by koolkartin » Wed Nov 04, 2009 12:25 pm
2/5

I always used to get confused with Npr, Ncr and the Probability of occurrence.

Desired / Total is the most relevant in this scenario;
Last edited by koolkartin on Sun Jan 24, 2010 8:36 pm, edited 1 time in total.
Top
Post new topic Post Reply