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Re M0531 [#permalink]
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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) \(1\frac{1}{28} = \frac{28}{28}\frac{1}{28} = \frac{27}{28}\) Answer: E
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Re: M0531 [#permalink]
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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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Re: M0531 [#permalink]
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02 Oct 2015, 07:58



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Re: M0531 [#permalink]
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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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Re: M0531 [#permalink]
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02 Oct 2015, 22:34
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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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Re: M0531 [#permalink]
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03 Oct 2015, 04:04



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Re: M0531 [#permalink]
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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) \(1\frac{1}{28} = \frac{28}{28}\frac{1}{28} = \frac{27}{28}\)
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: M0531 [#permalink]
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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)= 281/28=27/28



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Re: M0531 [#permalink]
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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?



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Re: M0531 [#permalink]
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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. 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: M0531 [#permalink]
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13 Feb 2018, 07:53










