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M05-31

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Bunuel
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M05-31 [#permalink] 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 chosen at random one after the other without replacement, what is the probability of not selecting 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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Bunuel
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Re M05-31 [#permalink] 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 chosen at random one after the other without replacement, what is the probability of not selecting exactly two tulips?

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


Let's find the probability of the opposite event, which is choosing two tulips, and subtract that from 1. The probability of choosing two tulips is \(P(both \ tulips) = \frac{2}{8}*\frac{1}{7} = \frac{1}{28}\).

Therefore, the probability that not both selected flowers are tulips is \(P(not \ both \ are \ tulips) = 1 - P(both \ tulips) = 1 - \frac{1}{28} = \frac{27}{28}\).


Answer: E
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dmatinho
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Re: M05-31 [#permalink] 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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Bunuel
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Re: M05-31 [#permalink] New post   02 Oct 2015, 08:58
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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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vbbvaibhav
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Re: M05-31 [#permalink] New post   20 Jul 2016, 06:55
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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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swapnils10
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Re: M05-31 [#permalink] New post   18 Sep 2021, 01:39
Hi Bunnel,

Can we take here for non relevant event (Not selecting the tulip) 6C6

and then find the P = 1- (Non relevant event/relevant event) = 1-(6C6/8C2) = 1-(1/28). Is it correct approach ?

Thanks,
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Bunuel
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Re: M05-31 [#permalink] New post   18 Sep 2021, 02:17
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swapnils10 wrote:
Hi Bunnel,

Can we take here for non relevant event (Not selecting the tulip) 6C6

and then find the P = 1- (Non relevant event/relevant event) = 1-(6C6/8C2) = 1-(1/28). Is it correct approach ?

Thanks,


It should be:

\(P(no \ two \ tulips) = 1 - P(two \ tulips) = 1 - \frac{the \ number \ of \ ways \ to \ pick \ two \ tulips \ out \ of \ two}{the \ number \ of \ ways \ to \ pick \ two \ flowers \ out \ of \ eight}= 1 - \frac{C^2_2}{C^2_8}=\frac{27}{28}\).
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Bunuel
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Re: M05-31 [#permalink] New post   20 Apr 2023, 05:24
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I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.
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Re: M05-31 [#permalink]
20 Apr 2023, 05:24
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Bunuel
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