Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

Two pieces of fruit are selected out of a group of 8 pieces [#permalink]
29 Sep 2010, 14:39

1

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

95% (hard)

Question Stats:

40% (02:51) correct
60% (02:31) wrong based on 75 sessions

Two pieces of fruit are selected out of a group of 8 pieces of fruit consisting only of apples and bananas. What is the probability of selecting exactly 2 bananas?

(1) The probability of selecting exactly 2 apples is greater than ½.

(2) The probability of selecting 1 apple and 1 banana in either order is greater than 1/3.

To know the probability of selecting 2 bananas, we need to know exactly how many bananas are in the original group of 8. Thus, we can rephrase the question as “How many of the 8 pieces of fruit are bananas?”

Statement (1): INSUFFICIENT. We should work out the scenarios in which this condition is true.

Obviously, if all 8 pieces of fruit are apples, the probability is 1. So let’s start from this end.

If there are 7 apples and 1 banana, then the probability of 2 apples is (7/8)(6/7) = 42/56, which is greater than ½.

If there are 6 apples and 2 bananas, then the probability of 2 apples is (6/8)(5/7) = 30/56, which is still greater than ½. (28/56 would be exactly ½.)

If there are 5 apples and 3 bananas, then the probability of 2 apples is (5/8)(4/7) = 20/56, which is NOT greater than ½.

So we know from this statement that the number of apples could be 6, 7, or 8, and the number of bananas could be 0, 1, or 2. This narrows down the possibilities, but we do not know exactly how many bananas there are.

Statement (2): INSUFFICIENT. Again, we should work out the scenarios in which this condition is true. Let’s start with the scenario in which it would be most likely to pick one of each type of fruit: 4 apples and 4 bananas.

If there are 4 apples and 4 bananas, then the probability of picking 1 of each can be found this way:

1) Pick an apple, then a banana: (4/8)(4/7) = 16/56.

2) Pick a banana, then an apple: (4/8)(4/7) = 16/56.

Add: 16/56 + 16/56 = 32/56. This is obviously greater than 1/3.

Alternatively and more easily, we could have just calculated the probability of “apple then banana,” then multiplied by 2! = 2 to get our result. This is much faster.

Let’s now figure out the other scenarios:

3 apples and 5 bananas:

The probability of picking 1 of each is (3/8)(5/7)×2 = 30/56, also bigger than 1/3. (Notice that we have not wasted time simplifying these fractions.)

By symmetry, the probability must be the same for 5 apples and 3 bananas.

2 apples and 6 bananas:

The probability of picking 1 of each is (2/8)(6/7)×2 = 24/56, also bigger than 1/3. (We can compare 24/56 to 20/60 by eye.)

The same again is true for 6 apples and 2 bananas.

1 apple and 7 bananas:

The probability of picking 1 of each is (1/8)(7/7)×2 = 14/56, which is ¼. Too small.

Thus, this statement says that the number of bananas is 6, 5, 4, 3, or 2.

Statements (1) and (2) together: SUFFICIENT. The number of bananas must be 2.

Re: MGMAT Challenge Problem Showdown [#permalink]
29 Sep 2010, 14:39

zisis wrote:

Two pieces of fruit are selected out of a group of 8 pieces of fruit consisting only of apples and bananas. What is the probability of selecting exactly 2 bananas?

(1) The probability of selecting exactly 2 apples is greater than ½.

(2) The probability of selecting 1 apple and 1 banana in either order is greater than 1/3.

IMO A

no OA yet - will post as soon as it is provided along with OE

Re: MGMAT Challenge Problem Showdown [#permalink]
29 Sep 2010, 14:53

zisis wrote:

Two pieces of fruit are selected out of a group of 8 pieces of fruit consisting only of apples and bananas. What is the probability of selecting exactly 2 bananas?

(1) The probability of selecting exactly 2 apples is greater than ½.

(2) The probability of selecting 1 apple and 1 banana in either order is greater than 1/3.

Re: MGMAT Challenge Problem Showdown [#permalink]
29 Sep 2010, 17:34

zisis wrote:

Two pieces of fruit are selected out of a group of 8 pieces of fruit consisting only of apples and bananas. What is the probability of selecting exactly 2 bananas?

(1) The probability of selecting exactly 2 apples is greater than ½.

(2) The probability of selecting 1 apple and 1 banana in either order is greater than 1/3.

I think it's (C)

1) Testing a few numbers of apples: the probability of selecting 2 apples if there are 5 is (5/8)(4/7) = 20/56, which is less than 1/2. If there are 6 apples, the probability is (6/8)(5/7) = 30/56, which is greater than 1/2. So there are 6, 7, or 8 apples. If there are 6 apples, then the probability of selecting 2 bananas is (2/8)(1/7) = 1/28. But if there are 7 or 8 apples, the probability of selecting 2 bananas is 0. So insufficient.

2) The probability of selecting 1 apple and 1 banana is (a/8)(b/7) + (b/8)(a/7) = (ab/56) + (ab/56) = ab/28. Since (ab/28) = (1/3), we know that ab > (28/3) = 9.333... This doesn't tell us how many of each fruit there is, though.

Together: We know there are 6, 7, or 8 apples, and that a*b > 9.333... If there are 8 apples, then there are 0 bananas, and 8*0 < 9.333... Likewise, if there are 7 apples and 1 banana, 7*1 < 9.333... However, if there are 6 apples and 2 bananas, 6*2 > 9.333...

So there are 6 apples and 2 bananas, and the probability of selecting both bananas is 1/28.

Re: MGMAT Challenge Problem Showdown [#permalink]
14 May 2011, 08:50

C is 2:55 min, without any writing. i think this could have been faster, it is clear that there are either 7 apples or 6 apples, so either there is 1 banana or 2 bananas. now plugging in the values and checking out for C is easier. _________________

What is of supreme importance in war is to attack the enemy's strategy.

Re: MGMAT Challenge Problem Showdown [#permalink]
14 May 2011, 09:13

Statment 1: Tells that there are either 3 or 2 or 1 or 0 bananas ==> Not sufficient alone Statment 2: Tells that A and B are in a fixed particular ratio (what ever it is )but it can be ether way round. ==> Not sufficient alone

Both together tells that the apples are more the given ratio. So gives counts of A and B. Hence sufficient find all kind of probabilities. _________________

If you liked my post, please consider a Kudos for me. Thanks!

Re: MGMAT Challenge Problem Showdown [#permalink]
17 May 2011, 01:15

subhashghosh wrote:

Bananas - x

Apples - 8-x

P(Exactly 2 Bananas) - x/8 * (x-1)/7

(1)

Exactly 2 Apples

(8-x)/8 * (8-x-1)/7 > 1/2

=> (8-x)/8 * (7-x)/7 > 1/2

=> (8-x)(7-x) > 28

X can be 2 or 1 or 0

Insufficient

(2)

x/8 * (8-x)/7 + (8-x)/8 * x/7 > 1/3

2 * x/8 * (8-x)/7 > 1/3

x/8 * (8-x)/7 > 1/6

x(8-x) > 28/3 > 9.something

=> x = 2, 3, 4 etc.

Not Sufficient

(1) + (2)

x = 2

Answer - C

That I understood. See according to your explanation of statement 1: X can be 2, 1 or 0. Now the question says the group has bananas and apples, so bananas cannot be 0 from what I understand.

Leaving X = 2 or 1 Now the question is probability of getting exactly 2 bananas - doesn't that imply that the group has atleast 2 bananas?

If that is the case then X has to be 2, so A would become sufficient. If that is not the case then your explanation is correct.

I needed a clarification on the above! _________________

GMAT done - a mediocre score but I still have a lot of grit in me

On September 6, 2015, I started my MBA journey at London Business School. I took some pictures on my way from the airport to school, and uploaded them on...

When I was growing up, I read a story about a piccolo player. A master orchestra conductor came to town and he decided to practice with the largest orchestra...

Although I have taken many lessons from Field Foundations that can be leveraged later, the lessons that will stick with me the strongest have been the emotional intelligence lessons...

Tick, tock, tick...the countdown to January 7, 2016 when orientation week kicks off. Been a tiring but rewarding journey so far and I really can’t wait to...