Data Sufficiency

If 2 different representative are to be selected at random from a group of 10 employees and if p is the probability that both representatives selected will be women, is p>1/2 ?

(1) more than 1/2 of the 10 employees are women.
(2) The probability that both representatives selected will be men is less than 1/10.


OA: [color=white]E[/color]

Ans is E

Lets say women = w

then the prob = p = wC2/10C2 = wC2 / 45

from (A)
w>5 , lest say w=6
p = 6C2/45 = 15/45 = 1/3 < 1/2
lets say w=10 then p = 1
so insufficient

from (B)

prob both are M = mC2/45 where m=number of men
mC2/45 < 1/10
mC2 < 45/10
m x (m-1) < 9
or m <=3 which means women >= 7

when w = 7 then prob for women = 7C2/45 = 21/45 < 1/2
so again insufficient

yikes, thats a hard one.

what does C stand for in your explanation netigen? also, how did you get the 45 value?

Yes, please reply - netigen.

You are much better than me at combinations.. trying to sponge knowledge quickly here : )

C = combinations

10C2 = 10!/[(10-2)! x 2!] = 45

Read up more here - https://www.mathwords.com/c/combination_formula.htm

Hey,

How did you get this bit:

mC2 < 45/10
m x (m-1) < 9

Could you explain?

mC2 < 45/10

--> m!/2!(m-2)! <45/10
--> m!/(m-2)! < 90/10
--> m(m-1) <9

Thanks!

Sorry to be so silly but I don't get the next bit either:

m(m-1)<9
m<=3?

how!

i am not quite sure but i think its because

3x3=9

and if m=4, then m(m-1)= 4*3 = 12 which is bigger than 9, so m must be smaller or equal to 3.

silvia928 wrote:If 2 different representative are to be selected at random from a group of 10 employees and if p is the probability that both representatives selected will be women, is p>1/2 ?

(1) more than 1/2 of the 10 employees are women.
(2) The probability that both representatives selected will be men is less than 1/10.

All that math looks way too complicated to me - let's use probability instead of combinations.

(1) If 6 are women, then the chance of 2 women is (6/10)(5/9) = 30/90.

Is 1/3 > 1/2? NO.

If all 10 are women, then the chance of 2 women is 100%. Is 100% > 50%? YES.

(2) Let's see the maximum number of men that would make this true:

If there were 2 men, prob of MM would be (2/10)(1/9) = 2/90... allowed.

If there were 3 men, prob of MM would be (3/10)(2/9) = 6/90... allowed.

If there were 4 men, prob of MM would be (4/10)(3/9) = 12/90 which is more than 1/10, so not allowed!

So, the maximum # of men is 3, which means that the minumum # of women is 7.

If we had 7 women, the chance of WW would be (7/10)(6/9) = 42/90. That's less than half, so "NO" answer.

Again, if we had 10 women, chance would be 100%, which gives us a "YES" answer.

So, each statement by itself can give us a "yes" or "no". Since the scenarios we examined for statement (2) also fit statement (1), we're no better off after combination: choose (E).

Thanks!