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Probability Question (m05q31) [#permalink]
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26 Oct 2006, 17:41
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This topic is locked. If you want to discuss this question please repost it in the respective forum. 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) \(\frac{27}{28}\) Source: GMAT Club Tests  hardest GMAT questions



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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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Please Explain [#permalink]
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26 Oct 2006, 18:03
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 = 11/56
= 55/56



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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=> 11/28 =27/28



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Re: Please Explain [#permalink]
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26 Oct 2006, 23:26
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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 = 11/56 = 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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E
Work out the probability of getting Tulip and take from 1
27/28



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Re: Probability Question (m05q31) [#permalink]
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07 Jun 2010, 00:03
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]
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02 Aug 2010, 05:48
IMO E... Probability of picking 2 tulips = 2/8 * 1/7 = 2/56 Probability of not picking 2 tulips = 12/56 = 54/56 = 27/28
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Re: Probability Question (m05q31) [#permalink]
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02 Aug 2010, 06:38
Ans : (8c2  1)/8c2 = 27/28
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Re: Probability Question (m05q31) [#permalink]
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02 Aug 2010, 07:36
(6c2+2*(6c1))/(8c2)=27/28



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Re: Probability Question (m05q31) [#permalink]
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02 Aug 2010, 07:45
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]
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02 Aug 2010, 07:46
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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]
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02 Aug 2010, 08:17
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]
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02 Aug 2010, 09:23
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) = (12/8) x ( 1  1/7) = 18/28 Where am i going wrong?



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Re: Probability Question (m05q31) [#permalink]
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02 Aug 2010, 10:40
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]
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02 Aug 2010, 19:30
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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]
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02 Aug 2010, 21:43
Thanks TallJTinChina.



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Re: Probability Question (m05q31) [#permalink]
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03 Aug 2011, 09:37
Can someone please explain where patilkumar's approach is wrong?



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Re: Probability Question (m05q31) [#permalink]
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03 Aug 2011, 12:07
easy one ..naswer is E



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Re: Probability Question (m05q31) [#permalink]
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10 Aug 2011, 23:46
Correct Answer Getting both tulips 2/8*1/7 (because two event happen one after another) = 1/28 Not getting the both tulips = 11/28 = 27/28
Now come study question
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?
Here paul nor jen .. let calculate
if paul get selected P1= 1/10 Jen get selected P2= 1/10 if jen or pual get selected = 1/10+1/10 = 2/10
Now jen or paul not getting selected = 12/10 = 8/10
Correct me if I am wrong




Re: Probability Question (m05q31)
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