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

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

If the answer is 5/24, i will explain.

P(all non-printed) = 5C3/10C3=5*4*3/10*9*8=1/12
P(atleast one printed)=1-1/12=11/12

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.

do you mean how many methods or anything other?

Nina from forest and lakes

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.

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 !

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: https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2016/01 ... mbination/

P(all non printed)= 5C3/10C3 = 10/120 = 1/12
P(atleast one printed) = 1-1/12 = 11/12

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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