total no of ways = 10C2 = 45.
Let there be x women, thus p = XC2/45.
p>1/2 means XC2/45> 1/2 or (x)(x-1)>45...
a) x>5. if x=5, 20>45 false.
if x=9, 72>45 true. insufficient
b) let men be y. thus YC2/45 < 1/10 or (y)(y-1)<9.
y=0,1,2,3 is possible. thus x can be 7,8,9,10.
42>45 not true. 72>45 true
Insufficient.
combined also means x=7,8,9,10...
Insufficient
IMO E

mehrasa wrote:If 2 different representatives 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

i followed a diff method...

stmnt 1) basically says W (# of women) = 6,7,8,9,10
if W = 10, then p > 0.5 (since p = 1)
if W = 6, then p = (6/10) x (5/9) = 1/3 = 0.33 .. so p < 0.5
INSUFF

stmnt 2) is saying P(M,M, -> of getting 2 men) < 0.1
so here we just try numbers one by one...
if M (# of men) = 0 or 1, then P(M,M) = 0
if M = 2, then P(M,M) = (2/10) x (1/9) = 1/45 < 0.1 so lets try a bigger #
M = 3, then P(M,M) = (3/10) x (2/9) = 1/15 < 0.1 ... this is so close to 1/10 so we can stop here and we know tht stmnt 2 is trying to say that the largest possible value of M = 3
so the smallest possible value of W = 7
then p = (7/10) x (6/9) = 14/30 which is a bit short of 0.5 (15/30)

but of course if M=0 & W =10 then p> 0.5
INSUFF

stmnt 1,2)
1) says W = 6,7,8,9,10 & 2) says W = 7,8,9,10
basically we get W = 7,8,9,10 by combining the 2.. so its the same as stmnt 2..
INSUFF

Answer is E

imo E

I posted a solution here:

https://www.beatthegmat.com/if-2-differe ... 59122.html