Analysis of each case:

1) More than half the employees are women

Consider 2 scenarios:
a) #Women = 6
Then, probability of selecting 2 women = 6c2/10c2 = 1/3 < 1/2

b) #Women = 9
Then, probability of selecting 2 women = 9c2/10c2 = 4/5 > 1/2

Hence, (1) is not sufficient

2) Let #Men = x
Then, probability of selecting 2 men = xc2/10c2
Numerator is -> x(x-1)(x-2)!/(x-2)!2! = x(x-1)/2

Thus, probability = [x(x-1)/2] / 10c2 = x(x-1)/90 < 1/10 => x(x-1) < 9
Only x <= 3 satisfies this, which means #Women >= 7

Now, if #women = 7, then probability of selecting 2 women = 7c2/10c2 = 7/15 < 1/2 and if #Women >=8, then probability is > 1/2

Hence, (2) is also not sufficient.


Combining statement (1) and (2) doesn't tell us anything new, since we already established #Women >= 7 if case (2) is true. hence correct answer is (E)

kmittal82 wrote:Analysis of each case:

1) More than half the employees are women

Consider 2 scenarios:
a) #Women = 6
Then, probability of selecting 2 women = 6c2/10c2 = 1/3 < 1/2

b) #Women = 9
Then, probability of selecting 2 women = 9c2/10c2 = 4/5 > 1/2

Hence, (1) is not sufficient

2) Let #Men = x
Then, probability of selecting 2 men = xc2/10c2
Numerator is -> x(x-1)(x-2)!/(x-2)!2! = x(x-1)/2

Thus, probability = [x(x-1)/2] / 10c2 = x(x-1)/90 < 1/10 => x(x-1) < 9
Only x <= 3 satisfies this, which means #Women >= 7

Now, if #women = 7, then probability of selecting 2 women = 7c2/10c2 = 7/15 < 1/2 and if #Women >=8, then probability is > 1/2

Hence, (2) is also not sufficient.


Combining statement (1) and (2) doesn't tell us anything new, since we already established #Women >= 7 if case (2) is true. hence correct answer is (E)

thanks for the lovely explanation.

vijaynarayanan wrote:

kmittal82 wrote:Analysis of each case:

1) More than half the employees are women

Consider 2 scenarios:
a) #Women = 6
Then, probability of selecting 2 women = 6c2/10c2 = 1/3 < 1/2

b) #Women = 9
Then, probability of selecting 2 women = 9c2/10c2 = 4/5 > 1/2

Hence, (1) is not sufficient

2) Let #Men = x
Then, probability of selecting 2 men = xc2/10c2
Numerator is -> x(x-1)(x-2)!/(x-2)!2! = x(x-1)/2

Thus, probability = [x(x-1)/2] / 10c2 = x(x-1)/90 < 1/10 => x(x-1) < 9
Only x <= 3 satisfies this, which means #Women >= 7

Now, if #women = 7, then probability of selecting 2 women = 7c2/10c2 = 7/15 < 1/2 and if #Women >=8, then probability is > 1/2

Hence, (2) is also not sufficient.


Combining statement (1) and (2) doesn't tell us anything new, since we already established #Women >= 7 if case (2) is true. hence correct answer is (E)

thanks for the lovely explanation.

No problem, I enjoyed doing that one

vijaynarayanan 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.

For the algebraically inclined, we can rephrase the question to make the statements go quickly.

If there are w women and 10 people total, then the probability that we choose 2 women is:

(w/10) *((w-1)/9) = (w^2 - 1)/90

(I'm just applying the general probability formula:

prob = (# of desired outcomes)/(total # of possibilities).)

So, the question becomes:

Is (w^2 - 1)/90 > 1/2
Is (w^2 - 1) > 45
Is w^2 > 46
Is w > 6?

(We know that w is a non-negative integer, so we can do that final simplification.)

(1) w > 5

If w = 6, "NO"... if w>6, "YES"... insufficient.

(2) (m/10)(m-1/9) < 1/10

(m^2 - 1)/90 < 1/10
m^2 - 1 < 9
m^2 < 10
m < 4

and, accordingly,

w > 6... sufficient.

Now, I really just did that because it was fun - that's likely way too time consuming for test day.

vijaynarayanan 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.

A. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
B. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
C. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
D. EACH statement ALONE is sufficient.
E. Statements (1) and (2) TOGETHER are NOT sufficient.

Correct Answer: E

Quite a few questions I have seen in DS related to probability. I really liked this one. Also, answer is intelligently hidden by vijay in White color
Great solution by Stuart and kmittal82..

Stuart Kovinsky wrote:

vijaynarayanan 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.

For the algebraically inclined, we can rephrase the question to make the statements go quickly.

If there are w women and 10 people total, then the probability that we choose 2 women is:

(w/10) *((w-1)/9) = (w^2 - 1)/90

(I'm just applying the general probability formula:

prob = (# of desired outcomes)/(total # of possibilities).)

So, the question becomes:

Is (w^2 - 1)/90 > 1/2
Is (w^2 - 1) > 45
Is w^2 > 46
Is w > 6?

(We know that w is a non-negative integer, so we can do that final simplification.)

(1) w > 5

If w = 6, "NO"... if w>6, "YES"... insufficient.

(2) (m/10)(m-1/9) < 1/10

(m^2 - 1)/90 < 1/10
m^2 - 1 < 9
m^2 < 10
m < 4

and, accordingly,

w > 6... sufficient.

Now, I really just did that because it was fun - that's likely way too time consuming for test day.

Hey Stuart,

Thats a nice way to approach the problem, but I think I see a mistake in the calculation

(w/10) *((w-1)/9) should be (w^2 - w)/(90)

Similarly for the second point.

Or am I misreading something here?

surajgarg wrote:

Stuart Kovinsky wrote:

vijaynarayanan 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.

For the algebraically inclined, we can rephrase the question to make the statements go quickly.

If there are w women and 10 people total, then the probability that we choose 2 women is:

(w/10) *((w-1)/9) = (w^2 - 1)/90

(I'm just applying the general probability formula:

prob = (# of desired outcomes)/(total # of possibilities).)

So, the question becomes:

Is (w^2 - 1)/90 > 1/2
Is (w^2 - 1) > 45
Is w^2 > 46
Is w > 6?

(We know that w is a non-negative integer, so we can do that final simplification.)

(1) w > 5

If w = 6, "NO"... if w>6, "YES"... insufficient.

(2) (m/10)(m-1/9) < 1/10

(m^2 - 1)/90 < 1/10
m^2 - 1 < 9
m^2 < 10
m < 4

and, accordingly,

w > 6... sufficient.

Now, I really just did that because it was fun - that's likely way too time consuming for test day.

Hey Stuart,

Thats a nice way to approach the problem, but I think I see a mistake in the calculation

(w/10) *((w-1)/9) should be (w^2 - w)/(90)

Similarly for the second point.

Or am I misreading something here?

You are correct! Silly Ws!

That makes the math quite a bit more complicated - we'll have to solve some quadratic inequalities (boo!).