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Probability Question (m05q31)

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Probability Question (m05q31) [#permalink] New post 26 Oct 2006, 16:41
A flower shop has 2 tulips, 2 roses, 2 daisies, and 2 lilies. If two flowers are sold at random, what is the probability of not picking exactly two tulips?

(A) \frac{1}{8}
(B) \frac{1}{7}
(C) \frac{1}{2}
(D) \frac{7}{8}
(E) \frac{27}{28}

[Reveal] Spoiler: OA
E

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 [#permalink] New post 26 Oct 2006, 17:12
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Number of combinations for selecting 2 flowers from 8 flowers is 8C2

Numebr of ways 2 tulips can be selected is 2C2

Probability of selecting tulips is 2c2/8c2 =1/28

Probabiliy of not selecting=> 1-1/28 =27/28
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 [#permalink] New post 26 Oct 2006, 16:47
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The probability that both the flowers selected are tulip is

2/8 * 1/7 = 1/28

Therefore, the probability that both the flowers are not tulips is

1- 1/28 = 27/28
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Re: Please Explain [#permalink] New post 26 Oct 2006, 22:26
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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 flowers is 2/8, because there are two tulips and you have to take that into account....
Once one tulip is selected, only one is left from 7 total, so the probability goes down to 1/7....
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Re: Probability Question (m05q31) [#permalink] New post 02 Aug 2010, 06:46
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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}



Yup, that's the trick in the language of the posted question. Read the last line again - "what is the probability of not picking exactly two tulips"? The "exactly" 2 tulips means that one of them can be a tulip but both can't be tulips. So what happens when you do 6/8 * 5/7 is that possibilities in which only one of the flowers is a tulip do not get counted.

So that's why the right approach is to subtract the possibility of getting exactly two tulips from the "entire universe" of possible options, because that will include possibilities of one tulip as well.

Good thing is none of the answer choices is 15/28, so you can figure out that you have gone wrong somewhere, but knowing the makers of the GMAT, I am sure that's not a luxury which will offered to you on the test day (i.e. they will most certainly include 15/28 as an option) :)
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Re: Probability Question (m05q31) [#permalink] New post 02 Aug 2010, 18:30
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ZMAT wrote:
Most of the users are following below.Can someone pls. clearly explain?

(2/8)*(1/7)

Once 2/8 is there why is the need for 1/7?


ZMAT we have two events

1st pick, then 2nd pick

Probability of the first event (2 tulips in 8 flowers)
Probability of the second event (1 tulip in 7 flowers)
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Re: Probability Question (m05q31) [#permalink] New post 08 Aug 2013, 06:50
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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
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Please Explain [#permalink] New post 26 Oct 2006, 17:03
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
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 [#permalink] New post 26 Oct 2006, 22:33
E

Work out the probability of getting Tulip and take from 1

27/28
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Re: Probability Question (m05q31) [#permalink] New post 06 Jun 2010, 23:03
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}
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Re: Probability Question (m05q31) [#permalink] New post 02 Aug 2010, 04:48
IMO E...

Probability of picking 2 tulips = 2/8 * 1/7 = 2/56
Probability of not picking 2 tulips = 1-2/56 = 54/56 = 27/28
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Re: Probability Question (m05q31) [#permalink] New post 02 Aug 2010, 05:38
Ans : (8c2 - 1)/8c2 = 27/28
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Re: Probability Question (m05q31) [#permalink] New post 02 Aug 2010, 06:36
(6c2+2*(6c1))/(8c2)=27/28
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Re: Probability Question (m05q31) [#permalink] New post 02 Aug 2010, 06:45
The probability that tulip can be selected is

(2C1/8C1)*(1C1/7C1)
= (2/8)*(1/7)
Why ??

When you select the first Tulip , there are two tulips among 8 flowers. When you draw the second flower , you have only one tulip remaining among 7 flowers.

So , probability of not selected is
1-(1/28) = 27/28
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Re: Probability Question (m05q31) [#permalink] New post 02 Aug 2010, 07:17
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
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Re: Probability Question (m05q31) [#permalink] New post 02 Aug 2010, 08:23
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?
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Re: Probability Question (m05q31) [#permalink] New post 02 Aug 2010, 09:40
Most of the users are following below.Can someone pls. clearly explain?

(2/8)*(1/7)

Once 2/8 is there why is the need for 1/7?
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Re: Probability Question (m05q31) [#permalink] New post 02 Aug 2010, 20:43
Thanks TallJTinChina.
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Re: Probability Question (m05q31) [#permalink] New post 03 Aug 2011, 08:37
Can someone please explain where patilkumar's approach is wrong?
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Re: Probability Question (m05q31) [#permalink] New post 03 Aug 2011, 11:07
easy one ..naswer is E
Re: Probability Question (m05q31)   [#permalink] 03 Aug 2011, 11:07
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