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Re: Probability Question (m05q31) [#permalink]
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 _________________

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Re: Probability Question (m05q31) [#permalink]
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)

Re: Probability Question (m05q31) [#permalink]
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

Re: Probability Question (m05q31) [#permalink]
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?

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.

Re: Probability Question (m05q31) [#permalink]
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

addition = 42/56 = 3/4.

so the answer is 3/4

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

Re: Probability Question (m05q31) [#permalink]
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

addition = 42/56 = 3/4.

so the answer is 3/4

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

Re: Probability Question (m05q31) [#permalink]
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

addition = 42/56 = 3/4.

so the answer is 3/4

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.

Re: Probability Question (m05q31) [#permalink]
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

addition = 42/56 = 3/4.

so the answer is 3/4

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 ?

Re: Probability Question (m05q31) [#permalink]
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 ?

Re: Probability Question (m05q31) [#permalink]
08 Aug 2013, 06:50

1

This post received 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.

Re: Probability Question (m05q31) [#permalink]
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

Re: Probability Question (m05q31) [#permalink]
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.

gmatclubot

Re: Probability Question (m05q31)
[#permalink]
23 Jul 2014, 01:54