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Women can be = 6, 7, 8, 9, 10 and Men respectively will be 4,3,2,1,0.

Say, W = 6 and M = 4

Probability that both representatives selected will be women, is p

p = 6C2/10C2 = 15/45 = 1/3

1/3 < 1/2

Say, W =10, M =0

p =1 > 1/2

So we can't answer.

Case ( ii )

Probability that both representatives selected will be men, is P < 1/10

Probability that None are men = 1-P = Probability that both are women

p<1/10

1-p = 1-1/10 = 9/10 > 1/2

Thus ( ii ) answers.

Answer should be "B"

you miss the case in which one is man and the other is woman.

this is the fastest way I have found to solve this one:

we must show whether wc2/10c2 >1/2

1. w>5 let's try with 6 or 8. with 6 we have 1/3, less than 1/2. with 8 we have 28/45, >1/2. not suff 2.mc2/10c2<1/10, mc2<4.5, m can be 3,2,1 or 0. if m=3, w=7, 7c2/10c2<1/2 ok. if m=2 and w=8, always 28/45>1/2 not suff

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

Last edited by dancinggeometry on 06 Sep 2008, 00:50, edited 1 time in total.

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

(1) More than 21 of the 10 employees are women.???

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

Last edited by vksunder on 17 Sep 2008, 12:13, edited 1 time in total.

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

w= # of women m= # of men

Probability that both selected are women:

\((w/10)*((w-1)/9) = (w^2-w)/90 > 1/2\), so the question is asking us whether \(w^2-w>45\)--->because \((1/2)*90 = 45\)

(1) So: if \(w=6\), then \(6^2-6 < 45\) False if \(w=9\), then \(9^2-9 > 45\) True --->NOT SUFF.

(w/10)*((w-1)/9) = (w^2-w)/90 > 1/2, so the question is asking us whether w^2-w>45--->because (1/2)*90 = 45

Quote:

Amitdgr... let me attempt to elaborate here:

Tarek's formula is just the words from the stem as an equation.. for example

From 1, say 6/10 are female... this means that when the first person is selceted the prob. is 6/10 (thus x/10) , when the next person is selceted, the prob. of Female becomes 5/9 (hence x-1/9), the total probability is 6/10 * 5/9 = 1/3 where

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

IMO D 1)INSUFFI ,say women=5 then 5*4/10*9 <1/2

when women=8 then p>1/2

2)INSUFFI thouh im not able to put down he calculation

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.

Fact 1: p(w) > 1/2

Let there are n women out of 10 employees. so n > 5.

number of ways of selecting 2 women out of n = nC2 = n*(n-1)/2.

total number of possible ways of selecting 2 employees out of 10 = 10C2 = 45.

probability that two selected employees are women = n(n-1)/90

n = 6, p=1/3 n = 7, p=42/90 n = 8, p=56/90.

1st is not sufficient.

Fact 2: let number of men in the sample is m. number of ways of selecting 2 men out of m = mC2 = m*(m-1)/2.

total number of ways = 45.

p(both men) = m*(m-1)/90

m*(m-1)/90 < 1/10

m*(m-1)<9 m = 3, 2, 1

second not sufficient.

combining both, we have two solutions 8 women and 2 men OR 7 women or 3 men.

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

A is insuff B is insuff as it provides info for not two men i.e. it could either be two men or one man and one woman together the prob of 2 women is between 1/3 and 9/10. therefore insuff.

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

Thanks in advance

for (1), suppose W=6, then both representatives are women would be 6C2, and total select ways are 10C2 p=6C2/10C2=(6*5)/(10*9)=1/3<1/2 therefore (1) is nsf.

for (2) suppose there are M mem, then Pman=MC2/10C2<1/10, so, M(M-1)<9, M=0, 1, 2, 3 for M=3, then W=7, p=7C2/10C2=(7*6)/(10*9)=42/90<45/90=1/2

The condition II states that the prob of selecting two men is less than 1/10. This provides information for "prob of not selecting two men" and this means that either both reps are women OR one rep is a man and one rep is a woman. This is insuff coz it does not help calculate the prob of two women reps >1/2. hope this helps!

gmatclubot

Re: Probability problems
[#permalink]
19 Aug 2009, 05:50