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ssky · 07-21-2008, 06:31 PM

I also got 1/5 being (1,3 ; 3,6 ; 4,5) as 3 fav sets. If SaGo is correct then there is more mystery to it as there will be many more total sets than 15 since 1,2 will be different than 2,1.

btit · 07-21-2008, 08:25 PM

I dont get any of the ans.

Total selection is 6C2 = 15.

Total choices are

1, 3 (or 3,1)
2, 2
3, 6 (or 6, 3)
4, 5 ( or 5,4)

Total choices are either 4 or 7

So, probability is either 4/15 or 7/15..

Am I missarabally wrong ?

ssky · 07-22-2008, 02:54 PM

btit, you forgot to read the question. The question explicitly says that two diffrent numbers to be selected so (2,2) is not an fav. set.

Fiver · 07-22-2008, 04:09 PM

Has to be 1/5 without replacement and 1/6 with replacement.

khushi · 07-22-2008, 04:48 PM

even i got 1/5

gmat731 · 07-22-2008, 06:33 PM

1st Pick Comment No of ways
1 only possible with 3. 1
2 None 0
3 Possible with 1 and 6 2
4 Only with 5 1
5 Only with 4 1
6 Only with 3 1

6/6c2 = 6/15 = 2/5

Agree with Sago.

gmat731 · 07-22-2008, 06:42 PM

ssky, can you please clarify what you meant by there will more sets than 15?

ssky · 07-23-2008, 08:22 PM

I meant that the favourable events will be 1,3 ; 3,6 ; 4,5 ; 3,1 ; 6,3 ; 5,4

saketbansal · 07-24-2008, 01:17 AM

We have identified total 15 pairs , and i think the order was not consideration when we identified the 15 pairs , so how can we take the desired pairs by considering the order.

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07-21-2008, 06:31 PM	#11 (permalink)
ssky Confucius is thinking !! Join Date: Jul 2008 Location: US Posts: 109	I also got 1/5 being (1,3 ; 3,6 ; 4,5) as 3 fav sets. If SaGo is correct then there is more mystery to it as there will be many more total sets than 15 since 1,2 will be different than 2,1.

07-21-2008, 08:25 PM	#12 (permalink)
btit Eager! Join Date: May 2008 Posts: 92	I dont get any of the ans. Total selection is 6C2 = 15. Total choices are 1, 3 (or 3,1) 2, 2 3, 6 (or 6, 3) 4, 5 ( or 5,4) Total choices are either 4 or 7 So, probability is either 4/15 or 7/15.. Am I missarabally wrong ?

07-22-2008, 02:54 PM	#13 (permalink)
ssky Confucius is thinking !! Join Date: Jul 2008 Location: US Posts: 109	btit, you forgot to read the question. The question explicitly says that two diffrent numbers to be selected so (2,2) is not an fav. set.

07-22-2008, 04:09 PM	#14 (permalink)
Fiver Go-May be only a sec Away Join Date: Jul 2008 Posts: 506	Has to be 1/5 without replacement and 1/6 with replacement.

07-22-2008, 04:48 PM	#15 (permalink)
khushi Mission GMAT Join Date: Jun 2007 Location: India Posts: 219	even i got 1/5

07-22-2008, 06:33 PM	#16 (permalink)
gmat731 get'in it within grasp Join Date: Jun 2008 Location: Farmington, MI Posts: 84	1st Pick Comment No of ways 1 only possible with 3. 1 2 None 0 3 Possible with 1 and 6 2 4 Only with 5 1 5 Only with 4 1 6 Only with 3 1 6/6c2 = 6/15 = 2/5 Agree with Sago.

07-22-2008, 06:42 PM	#17 (permalink)
gmat731 get'in it within grasp Join Date: Jun 2008 Location: Farmington, MI Posts: 84	ssky, can you please clarify what you meant by there will more sets than 15?

07-23-2008, 08:22 PM	#18 (permalink)
ssky Confucius is thinking !! Join Date: Jul 2008 Location: US Posts: 109	I meant that the favourable events will be 1,3 ; 3,6 ; 4,5 ; 3,1 ; 6,3 ; 5,4

07-24-2008, 01:17 AM	#19 (permalink)
saketbansal Eager! Join Date: Dec 2007 Posts: 60	We have identified total 15 pairs , and i think the order was not consideration when we identified the 15 pairs , so how can we take the desired pairs by considering the order.

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