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Manager
Joined: 28 Jun 2009
Posts: 51
Location: Moscow Russia

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27 Jul 2009, 02:32
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1. If the probability of rain on any givven day in City X is 50 percent (1/2), what is the probability that it rains on exactly 3 days out of 5 day period?
A 8/125
B 2/25
C 5/16
D 8/25
E 3/4

2. What is the probability of bearing two boys from four born babies
A 1/5
B 1/8
C 1/2
D 3/8
E 6/11

3. The host of dinner party must determin how seat himself and five guests around a cidrcular table. How many different seating arrangements are possible if the host always chooses the same seat for himself?

A 6
B 15
C 21
D 120
E 720

(I think that the answer whould be (5-1)! as there are five guest left and they sit around the table) and the host's seat does not change, so I do not include him in the could!

pic for Q4
Attachment:

Untitled.jpg [ 7.69 KiB | Viewed 1416 times ]

4. In how many ways it is possible to go from point A to point B if only one up and one right step from cell to a cell are permited
A 10
B 15
C 30
D 60
E 120

Answ
1.c
2.d
3.d
4.b
Senior Manager
Joined: 25 Jun 2009
Posts: 293

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27 Jul 2009, 04:42
1
KUDOS
1. If the probability of rain on any givven day in City X is 50 percent (1/2), what is the probability that it rains on exactly 3 days out of 5 day period?
A 8/125
B 2/25
C 5/16
D 8/25
E 3/4

2. What is the probability of bearing two boys from four born babies
A 1/5
B 1/8
C 1/2
D 3/8
E 6/11

3. The host of dinner party must determin how seat himself and five guests around a cidrcular table. How many different seating arrangements are possible if the host always chooses the same seat for himself?

A 6
B 15
C 21
D 120
E 720

(I think that the answer whould be (5-1)! as there are five guest left and they sit around the table) and the host's seat does not change, so I do not include him in the could!

pic for Q4
Attachment:
Untitled.jpg

4. In how many ways it is possible to go from point A to point B if only one up and one right step from cell to a cell are permited
A 10
B 15
C 30
D 60
E 120

Answ
1.c
2.d
3.d
4.b

1. As it can rain on any of the 3 days so that means 5C3 X ( probability of of raining on 3 days X probability of not raining on 3 days ) = 5C3 ( 1/2 x 1/2 x 1/2 x 1/2 x 1/2 ) .. [ 1/2 is the probablity of raining as well of not raining )
so it becomes 5 !/2!x3! x 1/32 = 10/32 = 5/16

2. Total no. of options/permutations can 2x2x2x2 = 16 ( As for every baby it can be a girl or a boy)
4C2 = 2 boys bearing out of 4 babies = 6
Probability = 6/16 = 3/8

3. I think you are right here,

4. I am not sure about this one as the questions says only one right and one up is possible. I think either question is not properly written or I am not able to understand the question
Director
Joined: 05 Jun 2009
Posts: 735
WE 1: 7years (Financial Services - Consultant, BA)

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27 Jul 2009, 05:17
1
KUDOS
agree with Nitish on Q1,2

1) 5C2 * (1/2)^5
2) 4C2 * (1/2)^4
3) There are Host + five guests
Host is fixed. So, in this case it should be 5! not 5-1!
In circular table we do (n-1)! because the first and last position are same, in this case they are not same.

4) for fourth this is what I understood.
(*Not sure if interpreted the question 4 correctly)
x x x x B
x x x x x
A x x x x

There are 2 possible upticks.

second uptick will occur is same column of first uptick or the columns after it.

total 5 columns

if first uptick is in :
=> first column then second possible upticks are 5 (1st to 5th column)
=> second column = 4
=> 3
=> 2
=> 1
Total = 5+4+3+2+1 =15
_________________

Consider kudos for the good post ...
My debrief : http://gmatclub.com/forum/journey-670-to-720-q50-v36-long-85083.html

Manager
Joined: 15 Apr 2008
Posts: 50
Location: Moscow

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27 Jul 2009, 05:32
1
KUDOS
Another approach for 4):
We have to make a total of 6 moves: 4 left's and 2 up's
Total number of ways: 6! (number of ways of selecting 6 elements), divided by 4! (as among these elements there is a set of 4 equal elements "left's") and by 2! (as among these elements there is another set of 2 equal elements "up's"): $$\frac{6!}{4!*2!}=15$$
Senior Manager
Joined: 25 Jun 2009
Posts: 293

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27 Jul 2009, 05:39
sudeep wrote:
agree with Nitish on Q1,2

1) 5C2 * (1/2)^5
2) 4C2 * (1/2)^4
3) There are Host + five guests
Host is fixed. So, in this case it should be 5! not 5-1!
In circular table we do (n-1)! because the first and last position are same, in this case they are not same.

4) for fourth this is what I understood.
(*Not sure if interpreted the question 4 correctly)
x x x x B
x x x x x
A x x x x

There are 2 possible upticks.

second uptick will occur is same column of first uptick or the columns after it.

total 5 columns

if first uptick is in :
=> first column then second possible upticks are 5 (1st to 5th column)
=> second column = 4
=> 3
=> 2
=> 1
Total = 5+4+3+2+1 =15

Sudeep, Can you explain the logic for the 4th question, I think I misinterpreted the question. So, basically as u said there are 5 columns and in any question like this one ( for e.g OG 11 Q 195 ) can we simply calculate by counting the no. of columns and then just adding the no.s ? ( Because in OG 11 Q 195 they have manually calculated the ways)
Director
Joined: 05 Jun 2009
Posts: 735
WE 1: 7years (Financial Services - Consultant, BA)

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27 Jul 2009, 05:50
1
KUDOS
Sunchaser20 wrote:
Another approach for 4):
We have to make a total of 6 moves: 4 left's and 2 up's
Total number of ways: 6! (number of ways of selecting 6 elements), divided by 4! (as among these elements there is a set of 4 equal elements "left's") and by 2! (as among these elements there is another set of 2 equal elements "up's"): $$\frac{6!}{4!*2!}=15$$

Thanks, nice explanation!
=> You have explained the reasoning of nCr

Another View:

Form A to B there are 6 moves (only moves possible are left and right)

how many combination of moves will have 4 lefts => 6C4
or
how many combination of moves will have 2 upticks ==> 6C2 = 6C4
_________________

Consider kudos for the good post ...
My debrief : http://gmatclub.com/forum/journey-670-to-720-q50-v36-long-85083.html

Senior Manager
Joined: 25 Jun 2009
Posts: 293

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27 Jul 2009, 05:56
Sunchaser20 wrote:
Another approach for 4):
We have to make a total of 6 moves: 4 left's and 2 up's
Total number of ways: 6! (number of ways of selecting 6 elements), divided by 4! (as among these elements there is a set of 4 equal elements "left's") and by 2! (as among these elements there is another set of 2 equal elements "up's"): $$\frac{6!}{4!*2!}=15$$

Nice Approach, clears the all clouds of confusion. +1 for this !
Cheers
Manager
Joined: 28 Jun 2009
Posts: 51
Location: Moscow Russia

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27 Jul 2009, 11:16
sudeep wrote:
3) There are Host + five guests
Host is fixed. So, in this case it should be 5! not 5-1!
In circular table we do (n-1)! because the first and last position are same, in this case they are not same.

How do you know that you do not need to do the -1 thing!!
Director
Joined: 05 Jun 2009
Posts: 735
WE 1: 7years (Financial Services - Consultant, BA)

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27 Jul 2009, 11:55
1
KUDOS
sudeep wrote:
3) There are Host + five guests
Host is fixed. So, in this case it should be 5! not 5-1!
In circular table we do (n-1)! because the first and last position are same, in this case they are not same.

How do you know that you do not need to do the -1 thing!!

I'll try to explain in brief

Lets's take example of four person A,B,C and D

in linear sitting.
ABCD and BCDA are different

in circular table it is not. Why?

circular table will like for ABCD

A - B
| - |
D - C

So if you see the same group can represent ABCD, BCDA, CDAB, DABC ( they all are same, which is not the case in linear sitting)

To solve the issue what we do is we fix one person and arrange the rest, so everyone else get arranged w.r.t fixed person
=(n-1)!

In this case as you see, Host is already fixed, so we just need to arrange the rest 5 guests.

HTH!!
_________________

Consider kudos for the good post ...
My debrief : http://gmatclub.com/forum/journey-670-to-720-q50-v36-long-85083.html

Manager
Joined: 28 Jun 2009
Posts: 51
Location: Moscow Russia

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27 Jul 2009, 12:15
sudeep wrote:
sudeep wrote:

So if you see the same group can represent ABCD, BCDA, CDAB, DABC ( they all are same, which is not the case in linear sitting)
HTH!!

So the ABCD, BCDA, CDAB, DABC WILL be diifferent as the E will be inserted!

Now I get it !!!

Thank you for the contribution!

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