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M05-31

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M05-31  [#permalink]

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New post 16 Sep 2014, 00:26
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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. \(\frac{27}{28}\)

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Re M05-31  [#permalink]

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New post 16 Sep 2014, 00:26
Official Solution:

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. \(\frac{27}{28}\)


The probability of not picking exactly two tulips is 1 minus the probability of picking 2 tulips. Out of 8 flowers, there is a 2 out of 8 chance of picking a tulip: \(\frac{2}{8} = \frac{1}{4}\). Out of the 7 remaining flowers, there is a 1 out of 7 or \(\frac{1}{7}\) chance of picking a tulip. Multiply the two fractions together to get the probability of picking both tulips:
\(\frac{1}{4}*\frac{1}{7} = \frac{1}{28}\)

Find probability of not picking exactly two tulips using the following equation:

P (both not Tulips) = 1 - P(both Tulips)
\(1-\frac{1}{28} = \frac{28}{28}-\frac{1}{28} = \frac{27}{28}\)


Answer: E
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Re: M05-31  [#permalink]

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New post 02 Oct 2015, 06:01
Hi Bunel,

I thought :
Probability of not picking two tulips :
6/8*5/7=5/12

I know I didn't get the correct answer but why do I have to find first the probability of getting both tulips (as you suggest)?
Could you please explain me ?

Thanks,
Daniela
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New post 02 Oct 2015, 08:58
dmatinho wrote:
Hi Bunel,

I thought :
Probability of not picking two tulips :
6/8*5/7=5/12

I know I didn't get the correct answer but why do I have to find first the probability of getting both tulips (as you suggest)?
Could you please explain me ?

Thanks,
Daniela


Notice that we are allowed to have one tulip, while you are calculating the probability of 0 tulips.

P(not 2 tulips) = P(1 tulip) + P(0 tulips) = 2*2/8*6/7 + 6/8*5/7 = 27/28.
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Re: M05-31  [#permalink]

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New post 02 Oct 2015, 11:34
Thank you very much, Bunel.

But why are you multiplying the probability of getting tulips (2/8) by 2?
2*2/8 (doesn't make sense in my head). :S

Thanks
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Re: M05-31  [#permalink]

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New post 02 Oct 2015, 23:34
1
dmatinho wrote:
Thank you very much, Bunel.

But why are you multiplying the probability of getting tulips (2/8) by 2?
2*2/8 (doesn't make sense in my head). :S

Thanks


2 because order matters, first can be a tulip or second can be a tulip
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New post 03 Oct 2015, 05:04
gillm wrote:
dmatinho wrote:
Thank you very much, Bunel.

But why are you multiplying the probability of getting tulips (2/8) by 2?
2*2/8 (doesn't make sense in my head). :S

Thanks


2 because order matters, first can be a tulip or second can be a tulip

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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: M05-31  [#permalink]

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New post 03 Oct 2015, 09:47
Bunuel wrote:
Official Solution:

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. \(\frac{27}{28}\)


The probability of not picking exactly two tulips is 1 minus the probability of picking 2 tulips. Out of 8 flowers, there is a 2 out of 8 chance of picking a tulip: \(\frac{2}{8} = \frac{1}{4}\). Out of the 7 remaining flowers, there is a 1 out of 7 or \(\frac{1}{7}\) chance of picking a tulip. Multiply the two fractions together to get the probability of picking both tulips:
\(\frac{1}{4}*\frac{1}{7} = \frac{1}{28}\)


Find probability of not picking exactly two tulips using the following equation:

P (both not Tulips) = 1 - P(both Tulips)
\(1-\frac{1}{28} = \frac{28}{28}-\frac{1}{28} = \frac{27}{28}\)


Answer: E


Can this be solved by permutation/combination also? Total num of ways two tulips be selected/Total num of ways any two flowers be selected?
And then you subtract that from 1? I would like to understand the explanation if anyone can provide please.
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Re: M05-31  [#permalink]

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New post 20 Jul 2016, 06:55
2 tulips out of 8 flowers can be selected by 1 way.

Total No. of ways 2 flowers can be selected out of 8 flowers is 8c2= 7*8/2= 28

Probability of selecting 2 tulips out of 8 flowers is 1/28.

Therefor P(Not Selecting a Tulip)= 1-(1/28)= 28-1/28=27/28
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Re: M05-31  [#permalink]

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New post 19 Jan 2018, 12:03
Why wouldn't the answer be 1 - (2/8) squared?

I got the answer correct, but I'm just trying to conceptualize why, if we are selling/picking two tulips simultaneously, why the chances of picking the second tulip would be reduced to (1/7).

Or am I just implicitly assuming that the tulips are sold one after the other?
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Re: M05-31  [#permalink]

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New post 13 Feb 2018, 08:46
Bunuel wrote:
gillm wrote:
dmatinho wrote:
Thank you very much, Bunel.

But why are you multiplying the probability of getting tulips (2/8) by 2?
2*2/8 (doesn't make sense in my head). :S

Thanks


2 because order matters, first can be a tulip or second can be a tulip

____________
Yes, that's right.


Hi Bunuel,

I am quoting you hoping that you will get a notification.

Can you please help? -->

Why wouldn't the answer be 1 - (2/8) squared?

I got the answer correct, but I'm just trying to conceptualize why, if we are selling/picking two tulips simultaneously, why the chances of picking the second tulip would be reduced to (1/7).

Or am I just implicitly assuming that the tulips are sold one after the other?
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Re: M05-31  [#permalink]

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New post 13 Feb 2018, 08:53
Neo94 wrote:

Hi Bunuel,

I am quoting you hoping that you will get a notification.

Can you please help? -->

Why wouldn't the answer be 1 - (2/8) squared?

I got the answer correct, but I'm just trying to conceptualize why, if we are selling/picking two tulips simultaneously, why the chances of picking the second tulip would be reduced to (1/7).

Or am I just implicitly assuming that the tulips are sold one after the other?


Both tulips cannot have the probability of 2/8. In any case one should be considered picked after another. Mathematically the probability of picking two tulips simultaneously, or picking them one at a time (without replacement) is the same.
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New to the Math Forum?
Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: M05-31 &nbs [#permalink] 13 Feb 2018, 08:53
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