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Re: Probability of 2 actresses being chosen together from 5 [#permalink]
02 Dec 2010, 04:51

2

This post received KUDOS

Expert's post

ViG wrote:

5 a list actresses are vying for 3 spots. J, M, S, L and H. What is the probability that J and H will star in the film together?

\(Probability=\frac{# \ of \ favorable \ outcomes}{total \ # \ of \ outcomes}\).

Now, total # of outcomes will be \(C^3_5=10\), (# of ways to choose 3 different actresses out of 5 when order of selection doesn't matter);

As for the # of favorable outcomes: we want 2 places out of 3 to be taken be J and H: so {JH-?}, for the third spot we can choose ANY from the 3 actresses left M, S, L. So, there are 3 such favorable groups possible: {JH-M}, {JH-S}, {JH-L}. Or \(C^1_1*C^1_3\), where \(C^1_1=1\) is 1 way to choose J and H (as it's one group) and \(C^1_3=3\) is 3 ways to choose the third member;

Re: Probability of 2 actresses being chosen together from 5 [#permalink]
04 Dec 2010, 17:03

Expert's post

ViG wrote:

ok, I got it.

It's not 5c3 as order matters so it's in fact 5!/3! which yields 6/20 = 3/10

It's pretty simple. Think of it as an arrangement, really.

The ways in which you can get those two acting together can be done in three ways, depending on who the third actress is. The ways in which you can pick any three actresses is 5C3. Hence 3/5C3 = 3/10 is the answer.

Re: Probability of 2 actresses being chosen together from 5 [#permalink]
13 Feb 2011, 08:59

How about this:

since the order doesn't matter the probability of any actress to be selected on any spot is 3/5. (in other words ANY actress has 3/5 chance to be selected) Next, we have 2 spots with 4 actresses, which gives us 2/4=1/2 probability of any given actress to be selected. Now, we just multiply probabilities 3/5 * 1/2 = 3/10. this might be a shortcut.

Re: Probability of 2 actresses being chosen together from 5 [#permalink]
11 Sep 2011, 00:12

whiplash2411 wrote:

ViG wrote:

ok, I got it.

It's not 5c3 as order matters so it's in fact 5!/3! which yields 6/20 = 3/10

It's pretty simple. Think of it as an arrangement, really.

The ways in which you can get those two acting together can be done in three ways, depending on who the third actress is. The ways in which you can pick any three actresses is 5C3. Hence 3/5C3 = 3/10 is the answer.

i dont get why do we divide 3 / 10 ?? why is 10 in the denominator?

Re: Probability of 2 actresses being chosen together from 5 [#permalink]
11 Sep 2011, 00:20

siddhans wrote:

whiplash2411 wrote:

ViG wrote:

ok, I got it.

It's not 5c3 as order matters so it's in fact 5!/3! which yields 6/20 = 3/10

It's pretty simple. Think of it as an arrangement, really.

The ways in which you can get those two acting together can be done in three ways, depending on who the third actress is. The ways in which you can pick any three actresses is 5C3. Hence 3/5C3 = 3/10 is the answer.

i dont get why do we divide 3 / 10 ?? why is 10 in the denominator?

P=Favorable Outcome/Total outcome Total outcome= Number of ways to choose any 3 actresses out of 5= \(C^5_3=\frac{5!}{3!(5-3)!}=\frac{5!}{3!2!}=10\) Thus, denominator is 10.

Numerator=3, because there are only 3 favorable outcomes: JHM JHS JHL _________________

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