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# 10 t-shirts are available to choose. 5 of them are printed,

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10 t-shirts are available to choose. 5 of them are printed, [#permalink]

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20 Jun 2004, 09:00
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10 t-shirts are available to choose. 5 of them are printed, and 5 of them are plain t-shirts. If 3 t-shirts are to be selected at random from the 10, what is the probability that at least one of them is a printed t-shirt.

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20 Jun 2004, 10:49
If the answer is 5/24, i will explain.
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20 Jun 2004, 12:28
P(all non-printed) = 5C3/10C3=5*4*3/10*9*8=1/12
P(atleast one printed)=1-1/12=11/12
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20 Jun 2004, 21:52
Pardon me GUEST ...

I can let know the answer only if you explain. Sorry about that.

Goal Stanford...

I understand the first step - Desired/Total combinations.

But

P(atleast one printed)=1-1/12=11/12 - Can you explain me this step?.

Thanks bro.
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21 Jun 2004, 00:17
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carsen wrote:
10 t-shirts are available to choose. 5 of them are printed, and 5 of them are plain t-shirts. If 3 t-shirts are to be selected at random from the 10, what is the probability that at least one of them is a printed t-shirt.

1. Probability of at least 1 being a printed = 1 - P(none of them is printed).

2. P(all 3 are plain) = 5/10*4/9*3/8 = 1/12!

3. The answer is 1 - 1/12 = 11/12.
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Re: 10 t-shirts are available to choose. 5 of them are printed, [#permalink]

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14 Jul 2014, 20:41
do you mean how many methods or anything other?

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Re: 10 t-shirts are available to choose. 5 of them are printed, [#permalink]

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14 Jul 2014, 21:58
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carsen wrote:
10 t-shirts are available to choose. 5 of them are printed, and 5 of them are plain t-shirts. If 3 t-shirts are to be selected at random from the 10, what is the probability that at least one of them is a printed t-shirt.

You have 10 t-shirts. You need to choose 3 of them. You can do this in 10C3 = 10*9*8/3*2*1 = 120 ways.

You have 5 plain t-shirts. You can choose 3 out of them in 5C3 = 5*4/2 = 10 ways.

So you have total 120 ways of picking 3 t-shirts out of 10. In 10 of those ways, you pick all plain t-shirts. So what happens in the rest of the 110 ways? In those, you must be picking up at least one printed t-shirt.

So probability of picking at least one printed t-shirt = 110/120 = 11/12
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews GMAT Club Legend Joined: 09 Sep 2013 Posts: 11062 Followers: 511 Kudos [?]: 134 [0], given: 0 Re: 10 t-shirts are available to choose. 5 of them are printed, [#permalink] ### Show Tags 23 Sep 2015, 13:39 Hello from the GMAT Club BumpBot! Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos). Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________ Director Joined: 12 Sep 2015 Posts: 545 Location: Canada Followers: 42 Kudos [?]: 383 [1] , given: 12 Re: 10 t-shirts are available to choose. 5 of them are printed, [#permalink] ### Show Tags 23 Sep 2015, 14:17 1 This post received KUDOS carsen wrote: 10 t-shirts are available to choose. 5 of them are printed, and 5 of them are plain t-shirts. If 3 t-shirts are to be selected at random from the 10, what is the probability that at least one of them is a printed t-shirt. When it comes to probability questions involving "at least," it's best to try using the complement, as the above solutions have done. However, what if you didn't spot that shortcut? No problem, it will just take us a little bit longer. P(at least 1 printed shirt) = (# of outcomes with at least 1 printed shirt)/(TOTAL # of possible outcomes) Always start with the denominator. TOTAL # of possible outcomes We have 10 shirts and we must select 3. Since the order in which we select the shirts doesn't matter, we can use combinations. We can select 3 shirts from 10 shirt in 10C3 ways (= 120 outcomes) # of outcomes with at least 1 printed shirt We have 3 cases: Case 1: 1 printed shirt and 2 plain. So, select 1 of the 5 printed, and select 2 of the 5 plain. # of outcomes = (5C1)(5C2) = (5)(10) = 50 Case 2: 2 printed shirts and 1 plain. So, select 2 of the 5 printed, and select 1 of the 5 plain. # of outcomes = (5C2)(5C1) = (10)(5) = 50 Case 3: 3 printed shirts and 0 plain. So, select 3 of the 5 printed, and select 0 of the 5 plain. # of outcomes = (5C3)(5C0) = (10)(1) = 10 Total number of outcomes = 50 + 50 + 10 = 110 So, P(at least 1 printed shirt) = 110/120 = 11/12 Cheers, Brent _________________ Brent Hanneson – Founder of gmatprepnow.com Brent also tutors students for the GMAT Manager Joined: 13 Apr 2013 Posts: 98 Followers: 2 Kudos [?]: 55 [0], given: 577 Re: 10 t-shirts are available to choose. 5 of them are printed, [#permalink] ### Show Tags 25 Sep 2015, 08:58 VeritasPrepKarishma wrote: carsen wrote: 10 t-shirts are available to choose. 5 of them are printed, and 5 of them are plain t-shirts. If 3 t-shirts are to be selected at random from the 10, what is the probability that at least one of them is a printed t-shirt. Please explain the steps (procedure) You have 10 t-shirts. You need to choose 3 of them. You can do this in 10C3 = 10*9*8/3*2*1 = 120 ways. You have 5 plain t-shirts. You can choose 3 out of them in 5C3 = 5*4/2 = 10 ways. So you have total 120 ways of picking 3 t-shirts out of 10. In 10 of those ways, you pick all plain t-shirts. So what happens in the rest of the 110 ways? In those, you must be picking up at least one printed t-shirt. So probability of picking at least one printed t-shirt = 110/120 = 11/12 Karishma, This is the combinatorics approach if I am not wrong. Can you tell me how to solve this with probability approach. _________________ Freedom is not a gift...It is a responsibility to pass on..... Director Joined: 12 Sep 2015 Posts: 545 Location: Canada Followers: 42 Kudos [?]: 383 [0], given: 12 Re: 10 t-shirts are available to choose. 5 of them are printed, [#permalink] ### Show Tags 25 Sep 2015, 09:39 Learning4mU wrote: Karishma, This is the combinatorics approach if I am not wrong. Can you tell me how to solve this with probability approach. Emmanuel posted a nice probability approach above. Cheers, Brent _________________ Brent Hanneson – Founder of gmatprepnow.com Brent also tutors students for the GMAT Manager Joined: 13 Apr 2013 Posts: 98 Followers: 2 Kudos [?]: 55 [0], given: 577 Re: 10 t-shirts are available to choose. 5 of them are printed, [#permalink] ### Show Tags 25 Sep 2015, 09:44 GMATPrepNow wrote: Learning4mU wrote: Karishma, This is the combinatorics approach if I am not wrong. Can you tell me how to solve this with probability approach. Emmanuel posted a nice probability approach above. Cheers, Brent Thanks a lot Brent. _________________ Freedom is not a gift...It is a responsibility to pass on..... Intern Joined: 07 Mar 2016 Posts: 9 Followers: 1 Kudos [?]: 1 [0], given: 2 10 t-shirts are available to choose. 5 of them are printed, [#permalink] ### Show Tags 27 Mar 2016, 02:37 Hi There, Allow me to give u all the approach i did with this question Firstly, there are 10 Shirts, which is 5 Printed (Pr) and 5 Plain (Pl) and we need to find the probability for at least one of them 3 shirts is Printed (Pr) and this can be done with the following event : 1 of 3 shirts is Pr (Pr Pl Pl) : there are 5C1 combination of this event or 2 of 3 shirts is Pr (Pr Pr Pl) : there are 5C2 combination of this event or all of the shirts is Pr (Pr Pr Pr) : there are 5C3 combination of this event (please note that order does not matter so we can use Combination in here) (also notice the "or" above so that means we must use + (plus) for the total event) and for the total probability of all shirts, there is total 3 shirts out of 10 we can use 10C3 So in the end we can calculate the probability with the following equation : (5C1 + 5C2 + 5C3) / 10C3 = 5/24 there you have it ! Manager Joined: 01 Mar 2014 Posts: 117 Followers: 2 Kudos [?]: 6 [0], given: 616 Re: 10 t-shirts are available to choose. 5 of them are printed, [#permalink] ### Show Tags 27 Mar 2016, 06:52 VeritasPrepKarishma wrote: carsen wrote: 10 t-shirts are available to choose. 5 of them are printed, and 5 of them are plain t-shirts. If 3 t-shirts are to be selected at random from the 10, what is the probability that at least one of them is a printed t-shirt. Please explain the steps (procedure) You have 10 t-shirts. You need to choose 3 of them. You can do this in 10C3 = 10*9*8/3*2*1 = 120 ways. You have 5 plain t-shirts. You can choose 3 out of them in 5C3 = 5*4/2 = 10 ways. So you have total 120 ways of picking 3 t-shirts out of 10. In 10 of those ways, you pick all plain t-shirts. So what happens in the rest of the 110 ways? In those, you must be picking up at least one printed t-shirt. So probability of picking at least one printed t-shirt = 110/120 = 11/12 Do we not need to consider the sequence in this - how do we decide? Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 6829 Location: Pune, India Followers: 1922 Kudos [?]: 11931 [0], given: 221 Re: 10 t-shirts are available to choose. 5 of them are printed, [#permalink] ### Show Tags 29 Mar 2016, 22:51 MeghaP wrote: VeritasPrepKarishma wrote: carsen wrote: 10 t-shirts are available to choose. 5 of them are printed, and 5 of them are plain t-shirts. If 3 t-shirts are to be selected at random from the 10, what is the probability that at least one of them is a printed t-shirt. Please explain the steps (procedure) You have 10 t-shirts. You need to choose 3 of them. You can do this in 10C3 = 10*9*8/3*2*1 = 120 ways. You have 5 plain t-shirts. You can choose 3 out of them in 5C3 = 5*4/2 = 10 ways. So you have total 120 ways of picking 3 t-shirts out of 10. In 10 of those ways, you pick all plain t-shirts. So what happens in the rest of the 110 ways? In those, you must be picking up at least one printed t-shirt. So probability of picking at least one printed t-shirt = 110/120 = 11/12 Do we not need to consider the sequence in this - how do we decide? No. Here you are just "selecting" 3 t-shirts. If you had to decide the sequence in which you could wear them over 3 days, then you would need to arrange them too. In this post, I have tried to explain the difference: http://www.veritasprep.com/blog/2016/01 ... mbination/ _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

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Re: 10 t-shirts are available to choose. 5 of them are printed, [#permalink]

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04 Apr 2016, 03:38
P(all non printed)= 5C3/10C3 = 10/120 = 1/12
P(atleast one printed) = 1-1/12 = 11/12
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Re: 10 t-shirts are available to choose. 5 of them are printed, [#permalink]

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09 Apr 2016, 02:01
carsen wrote:
10 t-shirts are available to choose. 5 of them are printed, and 5 of them are plain t-shirts. If 3 t-shirts are to be selected at random from the 10, what is the probability that at least one of them is a printed t-shirt.

probability of At least one is probability of (1 - none)

there are 5 non painted t-shirts that comes in 'none' category. choose 3 out of these 5.

Probability of at least one of them is a printed t-shirt = 1 - (5choose3/10choose3) = 1 - 1/12 = 11/12.
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Re: 10 t-shirts are available to choose. 5 of them are printed,   [#permalink] 09 Apr 2016, 02:01
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