BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

Probability

This topic has expert replies
Post new topic Post Reply
Previous Topic Next Topic
Senior | Next Rank: 100 Posts
Posts: 40
Joined: Sat Jan 30, 2010 10:38 am

Probability

by vijaynarayanan » Thu Jul 15, 2010 10:45 pm
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
Top
Source: — Data Sufficiency |
User avatar
Master | Next Rank: 500 Posts
Posts: 324
Joined: Mon Jul 05, 2010 6:44 am
Location: London
Thanked: 70 times
Followed by:3 members

by kmittal82 » Fri Jul 16, 2010 3:06 am
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)
Top
Senior | Next Rank: 100 Posts
Posts: 40
Joined: Sat Jan 30, 2010 10:38 am

by vijaynarayanan » Fri Jul 16, 2010 3:22 am
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.
Top
User avatar
Master | Next Rank: 500 Posts
Posts: 324
Joined: Mon Jul 05, 2010 6:44 am
Location: London
Thanked: 70 times
Followed by:3 members

by kmittal82 » Fri Jul 16, 2010 4:52 am
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 :)
Top
User avatar
GMAT Instructor
Posts: 3225
Joined: Tue Jan 08, 2008 2:40 pm
Location: Toronto
Thanked: 1710 times
Followed by:614 members
GMAT Score:800

by Stuart@KaplanGMAT » Fri Jul 16, 2010 8:44 pm
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.
Image

Stuart Kovinsky | Kaplan GMAT Faculty | Toronto

Kaplan Exclusive: The Official Test Day Experience | Ready to Take a Free Practice Test? | Kaplan/Beat the GMAT Member Discount
BTG100 for $100 off a full course
Top
User avatar
Legendary Member
Posts: 758
Joined: Sat Aug 29, 2009 9:32 pm
Location: Bangalore,India
Thanked: 67 times
Followed by:2 members

by sumanr84 » Mon Jul 19, 2010 1:02 am
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..
Top
User avatar
Senior | Next Rank: 100 Posts
Posts: 51
Joined: Mon May 10, 2010 11:55 pm
Thanked: 4 times

by surajgarg » Tue Jul 20, 2010 4:35 am
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? :?
Top
User avatar
GMAT Instructor
Posts: 3225
Joined: Tue Jan 08, 2008 2:40 pm
Location: Toronto
Thanked: 1710 times
Followed by:614 members
GMAT Score:800

by Stuart@KaplanGMAT » Tue Jul 20, 2010 8:56 am
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!).
Image

Stuart Kovinsky | Kaplan GMAT Faculty | Toronto

Kaplan Exclusive: The Official Test Day Experience | Ready to Take a Free Practice Test? | Kaplan/Beat the GMAT Member Discount
BTG100 for $100 off a full course
Top
Post new topic Post Reply
• Page 1 of 1