A firm has at least 5 partners, of whom at least 2 are male and at

Yes , i forgot this case , my bad


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Adding to my solution stated above, one more case , which i forgot to consider

With MMMFFF ->
3C2.3C2 / 6C4
3x3/15
3/5
Therefore C

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Yep.

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.)

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.

__


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.

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

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

To find the probability of selecting equal number , we need to know the gender of 5th member !!

Given:-
Total of Male + Female is atleast 5, It can be more as well !!
2 Male & 2 Female confirmed !! No idea about the 5th member or other members (if any) ?

Statement(1) :-
The firm has no more than 3 male partners.
This statement only tells us, that Male <= 3 , Now male can be 2 or 3 !!
Statement(1) is insufficient !!

Statement(2) :-
The firm has no more than 3 female partners.
This statement only tells us, that Female <= 3 , Now Female can be 2 or 3 !!
Statement(2) is again insufficient !!

Combining both statements :-
Male can be either 2 or 3.
Female can be either 2 or 3 .

Case (1) :- Male 2 , Female 3
Case (2) :- Male 3 , Female 2
Case(3) :- Male 3 , Female 3

The probabilities of selecting 4(2 Male / 2 Female) or 6 (3 Male/ 3 Female) equal number of male and female partner will differ for the above cases !!

Hence Statement(1) & (2) together are insufficient !!
Correct Option :- (E)

Merging topics. Please search before posting a question.

IMO C

1) possibilities:

3M, 4F
2M,3F

In each case, probability is different

2)possibilities:

3F, 2M
2F, 3M

In each case probability is different

Combining:
There are 3 possibilities:
a) 3M, 2F b) 2M, 3F c) 3M,3F

In each of the case, probability is 3/5. Hence C is the answer.

I received a PM requesting that I comment.

Statement 1:
Case 1: Exactly 2 female partners
Case 2: 1,000,000 female partners
Clearly, the probability that exactly 2 women will be selected is higher in Case 2 than in Case 1.
INSUFFICIENT.

Statement 2:
Case 1: Exactly 2 male partners
Case 2: 1,000,000 male partners
Clearly, the probability that exactly 2 men will be selected is higher in Case 2 than in Case 1.
INSUFFICIENT.

Statements combined:
A favorable outcome will be yielded if 2 men and 2 women are selected:
MMWW
The 2 men and 2 women can be selected in ANY ORDER.
For this reason, the last step in each of the 3 cases below is to multiply by the number of orderings for the 4 letters MMWW.
Number of ways to arrange 4 elements = 4!.
But when an arrangement includes IDENTICAL elements, we must divide by the number of ways each set of identical elements can be ARRANGED.
The reason:
When the identical elements swap positions, the arrangement doesn't change.
Here, we must divide by 2! to account for the two identical M's and by another 2! to account for the two identical W's.
Thus:
Number of ways to arrange the 4 letters MMWW = 4!/(2!2!) = 6.

Case 1: The partners are composed of 2 men and 3 women, for a total of 5 partners
P(MMWW) \(= \frac{2}{5} * \frac{1}{4} * \frac{3}{3} * \frac{2}{2} = \frac{1}{10}\)
Multiplying the result above by 6, we get:
\(\frac{1}{10} * 6 = \frac{6}{10} = \frac{3}{5}\)

Case 2: The partners are composed of 3 men and 2 women, for a total of 5 partners
P(MMWW) \(= \frac{3}{5} * \frac{2}{4} * \frac{2}{3} * \frac{1}{2} = \frac{1}{10}\)
Multiplying the result above by 6, we get:
\(\frac{1}{10} * 6 = \frac{6}{10} = \frac{3}{5}\)

Case 3: The partners are composed of 3 men and 3 women, for a total of 6 partners
P(MMWW) \(= \frac{3}{6} * \frac{2}{5} * \frac{3}{4} * \frac{2}{3} = \frac{1}{10}\)
Multiplying the result above by 6, we get:
\(\frac{1}{10} * 6 = \frac{6}{10} = \frac{3}{5}\)


Since the resulting probability in each case is THE SAME -- \(\frac{3}{5}\)-- the two statements combined are SUFFICIENT.

Firm has at least 5 partners...then why we can't take 3 male and 3 female ; then 2 and 3 from each category? At that time we get different values...hence E

Please correct me if I'm wrong