maihuna wrote:Out of 100 students, two sections of 40 and 60 are formed. If you and your friend are among the 100 students, what is the probability that
(a) you both enter the same section?
(b) you both enter the different sections?

Ans to follow,

Let section I have 40 students and section II have 60 students
If me and my friend enter section I, then there are 38 more students to be selected from I. But since 2 gone already, we have 98 students to chose from. 98C38. If we went to section II, 58 students would have to be chosen from 98. Since either is possible

98C38+ 98C58. The total way to choose is either 100C40 or 100C60. But note ACB = ACD if B+D=A.

So P(a)= (98C38+ 98C58)/100C40

After tedious and boring calculations you end up with
17/33.
GMAT will never ask this question. Too tedious for the test.

Use similar logic for b but after the above calculation, I leave that one for now.

dtweah wrote:

maihuna wrote:Out of 100 students, two sections of 40 and 60 are formed. If you and your friend are among the 100 students, what is the probability that
(a) you both enter the same section?
(b) you both enter the different sections?

Ans to follow,

Let section I have 40 students and section II have 60 students
If me and my friend enter section I, then there are 38 more students to be selected from I. But since 2 gone already, we have 98 students to chose from. 98C38. If we went to section II, 58 students would have to be chosen from 98. Since either is possible

98C38+ 98C58. The total way to choose is either 100C40 or 100C60. But note ACB = ACD if B+D=A.

So P(a)= (98C38+ 98C58)/100C40

After tedious and boring calculations you end up with
17/33.
GMAT will never ask this question. Too tedious for the test.

Use similar logic for b but after the above calculation, I leave that one for now.

The question is not as boring and tedious as is your approach, its like catching your ear with farther hand when nearer one will do the job.

as for as relevance with gmat is concerned see the BIRTHDAY problem given in OG which is very much on similar line as these one..

as for as solving it within a minute is concerned, here goes the logic...

Let us say the groups are color so people are balls so there are 40 Red ball and 60 Green ball, one is choosing two balls, and needs to find out are they of similar or different color,

case 1: When they need to fall in one section, it can happen in following two ways: either the two selections are from Red or from Green i.e. probability will be:
(40C2+60C2)/100C2
= (40*39+60*59)/100*99
= 20(78+177)/100*99
= 20*255/100*99
= 20*5*17*3/20*5*3*33
= 17/33

Where is the calulation? just a few addition single digit multiplication or cancelling the common factor...

case 2: The selection is from both section, one from Red another from Green:
40C1*60C1/100C2 = 40*60*2/100*99
= 20*2*3*20*2/20*5*3*33
= 16/33

Where is the long tedious calculation, if one has idea can be solved in a minute or so, laughingly some people are self declared authority on what should be asked in gmat what not...huh..

maihuna wrote:

dtweah wrote:

maihuna wrote:Out of 100 students, two sections of 40 and 60 are formed. If you and your friend are among the 100 students, what is the probability that
(a) you both enter the same section?
(b) you both enter the different sections?

Ans to follow,

Let section I have 40 students and section II have 60 students
If me and my friend enter section I, then there are 38 more students to be selected from I. But since 2 gone already, we have 98 students to chose from. 98C38. If we went to section II, 58 students would have to be chosen from 98. Since either is possible

98C38+ 98C58. The total way to choose is either 100C40 or 100C60. But note ACB = ACD if B+D=A.

So P(a)= (98C38+ 98C58)/100C40

After tedious and boring calculations you end up with
17/33.
GMAT will never ask this question. Too tedious for the test.

Use similar logic for b but after the above calculation, I leave that one for now.

The question is not as boring and tedious as is your approach, its like catching your ear with farther hand when nearer one will do the job.

as for as relevance with gmat is concerned see the BIRTHDAY problem given in OG which is very much on similar line as these one..

as for as solving it within a minute is concerned, here goes the logic...

Let us say the groups are color so people are balls so there are 40 Red ball and 60 Green ball, one is choosing two balls, and needs to find out are they of similar or different color,

case 1: When they need to fall in one section, it can happen in following two ways: either the two selections are from Red or from Green i.e. probability will be:
(40C2+60C2)/100C2
= (40*39+60*59)/100*99
= 20(78+177)/100*99
= 20*255/100*99
= 20*5*17*3/20*5*3*33
= 17/33

Where is the calulation? just a few addition single digit multiplication or cancelling the common factor...

case 2: The selection is from both section, one from Red another from Green:
40C1*60C1/100C2 = 40*60*2/100*99
= 20*2*3*20*2/20*5*3*33
= 16/33

Where is the long tedious calculation, if one has idea can be solved in a minute or so, laughingly some people are self declared authority on what should be asked in gmat what not...huh..

Bring on the hostility. I relish it.