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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, 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)
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New post 20 Jun 2004, 10:49
If the answer is 5/24, i will explain.

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New post 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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New post 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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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)


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

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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.


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

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New post 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.

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

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New post 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 !

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

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New post 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?

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

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

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New post 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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New post 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.

Please explain the steps (procedure)


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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