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# Probability of 2 actresses being chosen together from 5

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Probability of 2 actresses being chosen together from 5 [#permalink]

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01 Dec 2010, 20:13
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67% (01:39) correct 33% (02:34) wrong based on 6 sessions

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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?

Here are my workings:

There are 6 ways for them to be chosen together:

JHX
HJX
JXH
HXJ
XJH
XHJ

The total possibilities in my book are 5c3 = 10 so I got the answer to be 6/10 and thus 3/5 which is wrong.

Anyone care to explain how they come to this answer?

Thanks!
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Re: Probability of 2 actresses being chosen together from 5 [#permalink]

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01 Dec 2010, 20:15
ok, I got it.

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

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Re: Probability of 2 actresses being chosen together from 5 [#permalink]

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01 Dec 2010, 23:35
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Order doesn't matter.
No. Of ways in which 3 actress can be chosen out of 5 = 5C3=10

No. Of ways in which the two actresses (J and H) would work together = 1*no. Of ways in which the 3rd actress can be selected = 1*3 = 3

hence the required probability = 3/10

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Re: Probability of 2 actresses being chosen together from 5 [#permalink]

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02 Dec 2010, 05:24
Thanks but I'm struggling to see why it would be 1 * 3 at the end?
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Re: Probability of 2 actresses being chosen together from 5 [#permalink]

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02 Dec 2010, 05:51
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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;

Thus $$P=\frac{3}{10}$$.

For more on probability and combinatorics check: math-probability-87244.html and math-combinatorics-87345.html

Hope it helps.
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Re: Probability of 2 actresses being chosen together from 5 [#permalink]

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02 Dec 2010, 06:28
That totally makes sense. Thanks.

I'm usually pretty good at combos and probability but for some reason this one stumped me.
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Re: Probability of 2 actresses being chosen together from 5 [#permalink]

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04 Dec 2010, 10:59
ViG wrote:
That totally makes sense. Thanks.

I'm usually pretty good at combos and probability but for some reason this one stumped me.

good now go try this one: probability-mgmat-test-102027.html?hilit=anthony#p824969

a little bit tougher, but same deal.
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Re: Probability of 2 actresses being chosen together from 5 [#permalink]

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04 Dec 2010, 18:03
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.
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Re: Probability of 2 actresses being chosen together from 5 [#permalink]

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13 Feb 2011, 09:59

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.
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Re: Probability of 2 actresses being chosen together from 5 [#permalink]

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25 Apr 2011, 04:24
3C1 * 1 * 1/5C3

= 3!/2!/(5*4*3!)/(3!2!)

= 3 * 2/(5*4)

= 3/10
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Re: Probability of 2 actresses being chosen together from 5 [#permalink]

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11 Sep 2011, 01:12
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?
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Re: Probability of 2 actresses being chosen together from 5 [#permalink]

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11 Sep 2011, 01:20
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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Re: Probability of 2 actresses being chosen together from 5 [#permalink]

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11 Sep 2011, 10:34
total outcomes = 5c3 = 10

favorable outcomes = 2c2*3c1 = 3

=> probability = 3/10
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Re: Probability of 2 actresses being chosen together from 5 [#permalink]

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Re: Probability of 2 actresses being chosen together from 5   [#permalink] 19 Dec 2015, 04:57
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