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# M05-31

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Math Expert
Joined: 02 Sep 2009
Posts: 52344

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15 Sep 2014, 23:26
00:00

Difficulty:

35% (medium)

Question Stats:

73% (00:53) correct 27% (01:38) wrong based on 130 sessions

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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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Math Expert
Joined: 02 Sep 2009
Posts: 52344

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15 Sep 2014, 23:26
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)
$$1-\frac{1}{28} = \frac{28}{28}-\frac{1}{28} = \frac{27}{28}$$

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Joined: 03 Apr 2015
Posts: 10

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02 Oct 2015, 05:01
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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Daniela

From 330 - Doing my best to beat the gmat!!

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Joined: 02 Sep 2009
Posts: 52344

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02 Oct 2015, 07:58
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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02 Oct 2015, 10:34
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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Daniela

From 330 - Doing my best to beat the gmat!!

Intern
Joined: 06 May 2014
Posts: 6

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02 Oct 2015, 22:34
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
Math Expert
Joined: 02 Sep 2009
Posts: 52344

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03 Oct 2015, 04:04
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

____________
Yes, that's right.
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Intern
Joined: 06 May 2014
Posts: 6

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03 Oct 2015, 08:47
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)
$$1-\frac{1}{28} = \frac{28}{28}-\frac{1}{28} = \frac{27}{28}$$

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.
Intern
Joined: 24 Nov 2014
Posts: 3
Schools: Alberta'18

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20 Jul 2016, 05:55
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
Intern
Joined: 23 Jan 2017
Posts: 12

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19 Jan 2018, 11:03
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?
Intern
Joined: 23 Jan 2017
Posts: 12

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13 Feb 2018, 07:46
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.

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?
Math Expert
Joined: 02 Sep 2009
Posts: 52344

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13 Feb 2018, 07:53
Neo94 wrote:

Hi Bunuel,

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

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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Re: M05-31 &nbs [#permalink] 13 Feb 2018, 07:53
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# M05-31

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