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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] New post 20 Jun 2004, 08:00
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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.

Please explain the steps (procedure)
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 [#permalink] New post 20 Jun 2004, 09:49
If the answer is 5/24, i will explain.
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 [#permalink] New post 20 Jun 2004, 11:28
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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 [#permalink] New post 20 Jun 2004, 20:52
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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Re: Combintions [#permalink] New post 20 Jun 2004, 23:17
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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.

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] New post 14 Jul 2014, 19:41
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] New post 14 Jul 2014, 20:58
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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.

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] New post 23 Sep 2015, 12:39
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Re: 10 t-shirts are available to choose. 5 of them are printed, [#permalink] New post 23 Sep 2015, 13:17
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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.


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] New post 25 Sep 2015, 07:58
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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Re: 10 t-shirts are available to choose. 5 of them are printed, [#permalink] New post 25 Sep 2015, 08:39
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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Re: 10 t-shirts are available to choose. 5 of them are printed, [#permalink] New post 25 Sep 2015, 08:44
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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Re: 10 t-shirts are available to choose. 5 of them are printed,   [#permalink] 25 Sep 2015, 08:44
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