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Time for a brand-new quant problem, ladies and gentlemen!

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A firm has at least 5 partners, of whom at least 2 are male and at least 2 are female. If four of this firm’s partners are to be selected at random for an audit, what is the probability that equal numbers of male and female partners will be selected?

(1) The firm has no more than 3 male partners.

(2) The firm has no more than 3 female partners.

__

It appears that the "delayed OA" option has disappeared, so, the OA for this problem is available immediately. I'll post a detailed solution in a day or two.

Official Answer and Stats are available only to registered users. Register/Login.

_________________

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A firm has at least 5 partners, of whom at least 2 are male and at [#permalink]

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12 Nov 2017, 10:18

Hello ron I am feeling happy to reply on your post in first comment My solution :

1) consider MM FFF 5 members and MMFFFF 6 members MMFFF probability equal MMFF => 2C2.3C2 / 5C4 = 3/5 MMFFFF probability equal MMFF => 2C2.4C2/ 6C4 = 2/5

A and D are out

2) Same case apply for 2nd point The firm has no more than 3 female partners. just replace M by F and F by M in part 1) so B is also out

Combine 1 and 2

Possibilities

MMFFF and FFMMM case a MMMFF and FFFMM case b

consider both case now case a MMFFF => probability is 3/5 calculated above , holds good for FFMMM

case b MMMFF => its same case as later part of case a , which gives 3/5 and same is FFFMM which is first part of case a

so combination of 1 and 2 gives => probability 3/5.

IMO C
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Give Kudos for correct answer and/or if you like the solution.

sahilvijay — it seems you've forgotten to consider "MMMFFF" — i.e., 3 men and 3 women — as a pool from which to choose. This, too, is a possibility under statements 1 and 2 together — and so you can't conclude C without verifying that it gives the same probability as the cases you actually have considered.

(You've considered only "2 men, 3 women" and "3 men, 2 women" as possible pools for the two statements together.)
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You should ALSO work this problem by LISTING OUT THE CASES for the "MMFFF" and "MMMFFF" pools — and verifying that exactly three-fifths of the possibilities in each list comprise 2 men and 2 women. (Having made the list for "MMFFF", you don't need to make the list for "MMMFF"; that would just be the same list, with all the M's and F's reversed. Since the desired outcome is symmetric with regard to M's and F's, the probability will definitely be the same if all the M's and F's are switched.)
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Pueden hacerle preguntas a Ron en castellano Potete fare domande a Ron in italiano On peut poser des questions à Ron en français Voit esittää kysymyksiä Ron:lle myös suomeksi __

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Time for a brand-new quant problem, ladies and gentlemen!

__

A firm has at least 5 partners, of whom at least 2 are male and at least 2 are female. If four of this firm’s partners are to be selected at random for an audit, what is the probability that equal numbers of male and female partners will be selected?

(1) The firm has no more than 3 male partners.

(2) The firm has no more than 3 female partners.

__

It appears that the "delayed OA" option has disappeared, so, the OA for this problem is available immediately. I'll post a detailed solution in a day or two.

individually statements are not sufficient...

so lets look at it when combined.... M and F can be 2 or 3 so total can be 5 or 6

possibilities- 1) MMFFF 2)MMMFF 3) MMMFFF

Now generally ans would be E as different possibilities are there But can be SUFFICIENT if all have same probability

MMFFF and FFMMM will be same equal :- 2 M are chosen and 2 are to be picked up from 3 F = \(3C2 = 3\) total :- 5C4 = 5 \(P = \frac{3}{5}\)

lets check MMMFFF choosing 4 :- 2 out of 3 M and 2 out of 3 F = \(3C2*3C2 = 3*3=9\) total ways :- 4 out of 6 = \(6C4 = \frac{6!}{4!2!} = \frac{6*5}{2}=15\)

\(P = \frac{9}{15} = \frac{3}{5}\)

so ans is \(\frac{3}{5}\) in all cases sufficient
_________________

Here's a KEY that uses EXHAUSTIVE LISTS. Even if you solved this problem with combinatorial formulas, you should ALSO practice MAKING EXHAUSTIVE LISTS when you REVIEW it!

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Statement (1)

Together with the prompt, this statement tells us that the firm has either 2 or 3 male partners. There is no maximum, however, on the number of female partners; we know only that there are at least 2 female partners (and there can’t be just 2 of each, since the firm has at least 5 partners).

Let’s test cases:

• If there are 3 female partners (call them ‘1’, ‘2’, ‘3’) and 2 male partners (‘A’, ‘B’), then there are five ways to select 4 of them all. The selections consisting of 2 women and 2 men are boldfaced and asterisked: 1, 2, 3, A 1, 2, 3, B 1, 2, A, B* 1, 3, A, B* 2, 3, A, B* Exactly three of these selections (the last three listed) are made up of 2 women and 2 men. So in this case the desired probability is 3/5.

• If there are 4 female partners (‘1’, ‘2’, ‘3’, ‘4’) and 2 male partners (‘A’, ‘B’), then there are 15 ways to select four of the six. The selections consisting of 2 women and 2 men are boldfaced and asterisked: 1, 2, 3, 4 1, 2, 3, A 1, 2, 3, B 1, 2, 4, A 1, 2, 4, B 1, 2, A, B* 1, 3, 4, A 1, 3, 4, B 1, 3, A, B* 1, 4, A, B* 2, 3, 4, A 2, 3, 4, B 2, 3, A, B* 2, 4, A, B* 3, 4, A, B* Of the 15 possible selections, exactly 6 contain equal numbers of women and men (two of each). The desired probability in this case is thus 6/15 = 2/5.

These two cases produce two different answers to the problem, so, Statement (1) is NOT SUFFICIENT.

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Statement (2)

This time, we know that there are either 2 or 3 female partners, while the number of male partners can be any integer from 2 onward.

These are precisely the same numerical restrictions as in Statement (1), except with men and women reversed. This statement is thus functionally identical to Statement (1) and is therefore NOT SUFFICIENT as well.

(Just switch men and women in the workup above. With 3 male and 2 female partners, the probability is 3/5; with 4 male and 2 female partners, the probability is 2/5.)

__

Statements (1) and (2) TOGETHER

With both statements in force, there are only three possibilities for the partners overall: • 3 women and 2 men • 2 women and 3 men • 3 women and 3 men

In either of the first two cases — as demonstrated under Statement (1) — the desired probability is 3/5.

The only case left to consider is that involving 3 women (‘1’, ‘2’, ‘3’) and 3 men (‘A’, ‘B’, ‘C’). There are, again, 15 ways to choose four of the six. (Note that you don’t have to do this work again from scratch! You can simply copy the list made above for 4 women and 2 men, substituting ‘C’ for ‘4’.) 1, 2, 3, C 1, 2, 3, A 1, 2, 3, B 1, 2, C, A* 1, 2, C, B* 1, 2, A, B* 1, 3, C, A* 1, 3, C, B* 1, 3, A, B* 1, C, A, B 2, 3, C, A* 2, 3, C, B* 2, 3, A, B* 2, C, A, B 3, C, A, B Of the 15 possible selections, exactly 9 contain equal numbers of women and men (two of each). The desired probability in this case is thus 9/15 = 3/5.

The probability is thus 3/5 for all cases possible under both statements. The two statements TOGETHER are thus SUFFICIENT.

The correct answer is C.
_________________

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Pueden hacerle preguntas a Ron en castellano Potete fare domande a Ron in italiano On peut poser des questions à Ron en français Voit esittää kysymyksiä Ron:lle myös suomeksi __

Un bon vêtement, c'est un passeport pour le bonheur. —Yves Saint-Laurent