Probability Question (m05q31) : Retired Discussions [Locked] - Page 2
Check GMAT Club Decision Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack

 It is currently 21 Jan 2017, 10:37

### GMAT Club Daily Prep

#### 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

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

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# Probability Question (m05q31)

Author Message
Senior Manager
Joined: 19 Oct 2010
Posts: 271
Location: India
GMAT 1: 560 Q36 V31
GPA: 3
Followers: 7

Kudos [?]: 75 [0], given: 27

### Show Tags

04 Sep 2011, 01:28
I worked it out this way:

I firstly assumed that we are not going to pick them one-by-one, and that we are going to together.

Then, the probability that both flowers will be tulips is 2/16, viz., 1/8.

Therefore, the probability that both will NOT be tulips = 1 - 1,8
= 7/8

Do feel free to chip in.
_________________

petrifiedbutstanding

Director
Status: Final Countdown
Joined: 17 Mar 2010
Posts: 563
Location: India
GPA: 3.82
WE: Account Management (Retail Banking)
Followers: 17

Kudos [?]: 278 [0], given: 75

### Show Tags

07 Aug 2012, 04:41
prob. of not selecting 2 tulips = 1- prob.of selecting two tulips
lets calculate ; prob.of selecting two tulips= (2/8)*(1/7)
selection without replacement, one by one
=1/28
therefore;prob. of not selecting 2 tulips={1-(1/28)}=27/28
_________________

" Make more efforts "
Press Kudos if you liked my post

Intern
Joined: 25 Jan 2011
Posts: 1
Location: New Delhi, India
Followers: 0

Kudos [?]: 0 [0], given: 2

### Show Tags

07 Aug 2012, 11:51
rpgmat2010 wrote:
Probability of

Probability of finding tulip on both attempts is 2/8 x 1/7 = 1/28
Probability of NOT finding tulip is 1 - 1/28 = 27/28.

-----------------------
But, in what way is this incorrect?
Probability of not finding tulip on 1st and 2nd try =
Probability of NOT finding tulip on 1st try x Probability of NOT finding tulip on 2nd try =
( 1 - Probability of finding tulip on 1st try ) x (1 - Probability of finding tulip on 2nd try) =
(1-2/8) x ( 1 - 1/7) = 18/28
Where am i going wrong?

Your mentioned approach is wrong because it considers only the case where both the attempts do not select any tulip. But in fact we have to consider the cases where either of the attempts can have a tulip but not both. Moreover, your calculation is also wrong because:
It should be (1-2/8) X (1-2/7) because if the first attempt had no tulip then in second attempt we will have not consider 2 tulips, not one.

The detailed other way round solution is as follows:
1st attempt - TULIP 2nd attempt - NOT TULIP (Probability = 2/8 X 6/7 = 6/28)
1st attempt - NOT TULIP 2nd attempt - TULIP (Probability = 6/8 X 2/7 = 6/28)
1st attempt - NOT TULIP 2nd attempt - NOT TULIP (Probability = 6/8 X 5/7 = 15/28) (The only case you considered)

Add all the probabilities and the ans is 27/28.

Hope it helps.
Manager
Joined: 14 Jun 2012
Posts: 66
Followers: 0

Kudos [?]: 13 [0], given: 1

### Show Tags

08 Aug 2012, 06:11
There are two ways to approach the solution.

1. Approach 1 : Sum of probability of choosing no tulips, choosing 1st as tulip and 2nd as something else and choosing 1st as something else and 2nd as tulip. This would be the longer approach.

2. Approach 2 : 1 - probability that both the flowers chosen are tulips.

probability that the first flower can be a tulip = 2/8

probability that the second flower can be a tulip = 1/7 (because one tulip is already taken hence total flowers are now 7)

Probability that 1st AND 2nd are chosen as tulips = (2/8)*(1/7) = 1/28.

Hence answer is 1 - (1/28) = 27/28

I am feeling happy that the practice on probability is paying off.
_________________

My attempt to capture my B-School Journey in a Blog : tranquilnomadgmat.blogspot.com

There are no shortcuts to any place worth going.

Senior Manager
Joined: 14 Jul 2013
Posts: 295
Location: India
Concentration: Marketing, Strategy
GMAT 1: 690 Q49 V34
GMAT 2: 670 Q49 V33
GPA: 3.6
WE: Brand Management (Retail)
Followers: 13

Kudos [?]: 186 [0], given: 130

### Show Tags

07 Aug 2013, 12:49
study wrote:
can someone explain how to solve this problem the direct way.

I did the following, but didnt get to the right answer. What is wrong in my method?

6/8 * 5/7 = 30/56 = 15/28

The problem below is similar. And the answer is 8/10. So why cant the answer to the above be 6/8*5/7 ?

10 applicants are interviewed for a position. Among them are Paul and Jen. If a randomly chosen applicant is invited to interview first, what is the probability to have neither Paul nor Jen at the first interview?

* $$\frac{1}{10}$$
* $$\frac{1}{9}$$
* $$\frac{1}{2}$$
* $$\frac{8}{10}$$
* $$\frac{9}{10}$$

Neither P nor J is different from atleast both p and J. In this case remember p can be selected or j can be selected but both p and j cant be selected. If you have to take the long route find the probability of getting 0 tulips that will be 6C2/8C2 15/28 and add it to the probability of 1 tulip and 1 other flower to be selected (2C1x6C1)/8C2 which will give you 12/28. Sum is 27/28.

Cheers

Posted from my mobile device
_________________

Cheers
Farhan

My Blog - Student for Life ( Oxford MBA)

Intern
Joined: 05 Oct 2011
Posts: 2
Location: United States
GMAT 1: 740 Q51 V38
GPA: 3.5
Followers: 0

Kudos [?]: 0 [0], given: 5

### Show Tags

07 Aug 2013, 13:12
patilkunal wrote:
probability of not picking exactly two tulips = probability of picking exactly one tulip and one other flower + probability of picking no tulips at all

probability of picking exactly one tulip and other flower = 2/8 * 6/7 = 12/56
probability of picking no tulips at all= 6/8 * 5/7 = 30/56

I did the same way but we never considered that there are 2 ways we can get exactly one tulip either in the first attempt or 2nd.
so it will be 12/56 + 30/56 + 12/56= 27/28
Intern
Joined: 20 Jun 2013
Posts: 9
Followers: 0

Kudos [?]: 0 [0], given: 10

### Show Tags

08 Aug 2013, 04:42
patilkunal wrote:
probability of not picking exactly two tulips = probability of picking exactly one tulip and one other flower + probability of picking no tulips at all

probability of picking exactly one tulip and other flower = 2/8 * 6/7 = 12/56
probability of picking no tulips at all= 6/8 * 5/7 = 30/56

Could anyone explain this method. The steps seem completely correct but the final answer is coming out 3/4... please explain.
VP
Status: Far, far away!
Joined: 02 Sep 2012
Posts: 1123
Location: Italy
Concentration: Finance, Entrepreneurship
GPA: 3.8
Followers: 181

Kudos [?]: 1965 [0], given: 219

### Show Tags

08 Aug 2013, 04:49
hsb91 wrote:
patilkunal wrote:
probability of not picking exactly two tulips = probability of picking exactly one tulip and one other flower + probability of picking no tulips at all

probability of picking exactly one tulip and other flower = 2/8 * 6/7 = 12/56
probability of picking no tulips at all= 6/8 * 5/7 = 30/56

Could anyone explain this method. The steps seem completely correct but the final answer is coming out 3/4... please explain.

That method is not complete. The probability of picking a tulip and not a tulip is not 12/56, but it's 24/56.

The selection can happen in this order $$T,NT$$(with probability 12/56) and in this order $$NT,T$$(with probability 12/56).

So the answer is $$\frac{12+12+30}{56}=\frac{27}{28}$$.

Hope it's clear
_________________

It is beyond a doubt that all our knowledge that begins with experience.

Kant , Critique of Pure Reason

Tips and tricks: Inequalities , Mixture | Review: MGMAT workshop
Strategy: SmartGMAT v1.0 | Questions: Verbal challenge SC I-II- CR New SC set out !! , My Quant

Rules for Posting in the Verbal Forum - Rules for Posting in the Quant Forum[/size][/color][/b]

Intern
Joined: 20 Jun 2013
Posts: 9
Followers: 0

Kudos [?]: 0 [0], given: 10

### Show Tags

08 Aug 2013, 05:02
Zarrolou wrote:
hsb91 wrote:
patilkunal wrote:
probability of not picking exactly two tulips = probability of picking exactly one tulip and one other flower + probability of picking no tulips at all

probability of picking exactly one tulip and other flower = 2/8 * 6/7 = 12/56
probability of picking no tulips at all= 6/8 * 5/7 = 30/56

Could anyone explain this method. The steps seem completely correct but the final answer is coming out 3/4... please explain.

That method is not complete. The probability of picking a tulip and not a tulip is not 12/56, but it's 24/56.

The selection can happen in this order $$T,NT$$(with probability 12/56) and in this order $$NT,T$$(with probability 12/56).

So the answer is $$\frac{12+12+30}{56}=\frac{27}{28}$$.

Hope it's clear

So it means that arrangements are used here and not combination/ selections ?? Are we treating tulip and non tulip as two different sources/ groups ?
VP
Status: Far, far away!
Joined: 02 Sep 2012
Posts: 1123
Location: Italy
Concentration: Finance, Entrepreneurship
GPA: 3.8
Followers: 181

Kudos [?]: 1965 [0], given: 219

### Show Tags

08 Aug 2013, 05:07
Here we are using "probability" to solve this question, but you can use combinations as well. (I do not know what you mean by "arrangements")

And yes, we are considering tulips and Non tulips as two different groups.
_________________

It is beyond a doubt that all our knowledge that begins with experience.

Kant , Critique of Pure Reason

Tips and tricks: Inequalities , Mixture | Review: MGMAT workshop
Strategy: SmartGMAT v1.0 | Questions: Verbal challenge SC I-II- CR New SC set out !! , My Quant

Rules for Posting in the Verbal Forum - Rules for Posting in the Quant Forum[/size][/color][/b]

Intern
Joined: 20 Jun 2013
Posts: 9
Followers: 0

Kudos [?]: 0 [0], given: 10

### Show Tags

08 Aug 2013, 06:25
By combinations, total possibilities = 8c2 = 28
NO Tulip -- 6c2 = 15
Thus, prob of no tulip = 15/28

For same case, using probability = 6/8* 5/7 = 15/28

When 1 tulip = 2c1* 6c1 = 2*6 = 12 .... NOT 24 ( i.e. only selections and not arrangements or order does not matter TNT is same as NT T)
Thus, Prob. = 12/28
For same case, Using prob = 2/8* 6/7 =6/28

Why we have to multiply with 2 when using prob. and no multiplication with 2 when using combinations ?

I hope you are getting what i mean to say
VP
Status: Far, far away!
Joined: 02 Sep 2012
Posts: 1123
Location: Italy
Concentration: Finance, Entrepreneurship
GPA: 3.8
Followers: 181

Kudos [?]: 1965 [1] , given: 219

### Show Tags

08 Aug 2013, 06:50
1
KUDOS
hsb91 wrote:
By combinations, total possibilities = 8c2 = 28
NO Tulip -- 6c2 = 15
Thus, prob of no tulip = 15/28

For same case, using probability = 6/8* 5/7 = 15/28

When 1 tulip = 2c1* 6c1 = 2*6 = 12 .... NOT 24 ( i.e. only selections and not arrangements or order does not matter TNT is same as NT T)
Thus, Prob. = 12/28
For same case, Using prob = 2/8* 6/7 =6/28

Why we have to multiply with 2 when using prob. and no multiplication with 2 when using combinations ?

I hope you are getting what i mean to say

In combinations you do not have to multiply by two T,NT as you already get both scenarios (T,NT and NT,T) with the formula (2C1)*(6C1).

In the other method however you do have to multiply: with this $$\frac{2}{8}* \frac{6}{7}=\frac{6}{28}$$ you are calculating the probability of getting a Tulip FIRST and a N-Tulip SECOND, that's it => this DOES NOT include the other way round(NT,T).

To get NT, T you have to consider $$\frac{6}{8}*\frac{2}{7}=NT*T$$ separately.

So combinations include both cases, probability does not. Hope it's clear
_________________

It is beyond a doubt that all our knowledge that begins with experience.

Kant , Critique of Pure Reason

Tips and tricks: Inequalities , Mixture | Review: MGMAT workshop
Strategy: SmartGMAT v1.0 | Questions: Verbal challenge SC I-II- CR New SC set out !! , My Quant

Rules for Posting in the Verbal Forum - Rules for Posting in the Quant Forum[/size][/color][/b]

Intern
Joined: 20 Jun 2013
Posts: 9
Followers: 0

Kudos [?]: 0 [0], given: 10

### Show Tags

08 Aug 2013, 08:11
Zarrolou wrote:
hsb91 wrote:
By combinations, total possibilities = 8c2 = 28
NO Tulip -- 6c2 = 15
Thus, prob of no tulip = 15/28

For same case, using probability = 6/8* 5/7 = 15/28

When 1 tulip = 2c1* 6c1 = 2*6 = 12 .... NOT 24 ( i.e. only selections and not arrangements or order does not matter TNT is same as NT T)
Thus, Prob. = 12/28
For same case, Using prob = 2/8* 6/7 =6/28

Why we have to multiply with 2 when using prob. and no multiplication with 2 when using combinations ?

I hope you are getting what i mean to say

In combinations you do not have to multiply by two T,NT as you already get both scenarios (T,NT and NT,T) with the formula (2C1)*(6C1).

In the other method however you do have to multiply: with this $$\frac{2}{8}* \frac{6}{7}=\frac{6}{28}$$ you are calculating the probability of getting a Tulip FIRST and a N-Tulip SECOND, that's it => this DOES NOT include the other way round(NT,T).

To get NT, T you have to consider $$\frac{6}{8}*\frac{2}{7}=NT*T$$ separately.

So combinations include both cases, probability does not. Hope it's clear

Thanks for clearing my doubt ...
Intern
Joined: 24 Jan 2013
Posts: 9
Followers: 0

Kudos [?]: 9 [0], given: 5

### Show Tags

23 Jul 2014, 01:54
mitul wrote:
Prashant,

Why do we have to have to multiply 2/8x 1/7

According to me,

The probability that one tulip is selected from 8 = 1/8
So there are 7 remaining
The probability that 1 tulip is selected from 7 = 1/7
Therefore selectng 2 tulips
= 1/8*1/7
= 1/56

Probability that both flowers are not tulip = 1-1/56
= 55/56

The probability that one tulip is selected from 8 is not 1/8 but 2/8 as out of 8 no of tulips is 2 not 1.
Re: Probability Question (m05q31)   [#permalink] 23 Jul 2014, 01:54

Go to page   Previous    1   2   [ 34 posts ]

Similar topics Replies Last post
Similar
Topics:
GMAT Club Hardest Questions: Probability and Combinatorics 1 26 Mar 2011, 06:17
Probability Question 3 06 Sep 2010, 17:50
Probability Die 2 06 Jun 2010, 23:10
1 Probability 2 22 Jan 2010, 09:30
1 Event A and Event B are independent. Is the probability that 6 23 Oct 2007, 07:13
Display posts from previous: Sort by

# Probability Question (m05q31)

Moderator: Bunuel

 Powered by phpBB © phpBB Group and phpBB SEO Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.