Math Is Fun Forum

coolwind

Member

Registered: 2005-10-30

Posts: 30

Some Probability Problem~~

1.At tht Mary sorority the 15 sisters who are seniors line up in a random manner for a graduation picture.
Two of these sisters are Columba and Piret.What is the probability that this graduation picture will find
(a)Piret and Columba standing next to each other?(b)exact five sisters standing between Columba and
Piret?

2. Consider the 2^19 compositions of 20.(a)How many have each summand
    even?(b)How many have each summand a multiple of 4?

Thank you very much~~:cool:

luca-deltodesco

Member

Registered: 2006-05-05

Posts: 1,470

Re: Some Probability Problem~~

1.a)
you have 15 places, for one sister can be. for the other sister, there are 14 more places, and given that sister A isnt on the edge, 2 places out of the 14 she could be in, if A is on the edge, there is only 1 out of 14 places, so the answer is

P(0<A<15) = 13/15, P(B next to A | 0<A<15) = 2/14
P(A=0 or A=15) = 2/15, P(B next to A | A=0 or A=15) = 1/14
P(B next to A) = 13/15*2/14 + 2/15*1/14 = 28/210 = 2/15 = 0.1333...

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pi man

Member

Registered: 2006-07-06

Posts: 251

Re: Some Probability Problem~~

I come up with a different solution. 

A:   Consider all of the different pairings the 2 could be in.   There are (15 choose 2)possibilities.     That's 105 different pairings.   Now count how many pairings are next to each other:  (1,2), (2,1), (2,3), (3,2), ... (14,15) and (15, 14).   That's 28 pairings out of the possible 105.   That's .2666, which is double Luca's answer.   And that explains my mistake.    There are 15 * 14 possible pairings, not (15 choose 2).     So that makes it 28 possibilities out of 210 which is .133333.   

B:  You still have the 210 different pairings but the number of pairings that meet the requirements is smaller:  (1,7), (7,1), (2, 8), (8,2),..., (9,15) & (15,9).   There are 18 good pairings out of the 210, or .0857

2.   I have no idea what you're asking for here.

coolwind

Member

Registered: 2005-10-30

Posts: 30

Re: Some Probability Problem~~

I come up with a different solution. 

A:   Consider all of the different pairings the 2 could be in.   There are (15 choose 2)possibilities.     That's 105 different pairings.   Now count how many pairings are next to each other:  (1,2), (2,1), (2,3), (3,2), ... (14,15) and (15, 14).   That's 28 pairings out of the possible 105.   That's .2666, which is double Luca's answer.   And that explains my mistake.    There are 15 * 14 possible pairings, not (15 choose 2).     So that makes it 28 possibilities out of 210 which is .133333.   

B:  You still have the 210 different pairings but the number of pairings that meet the requirements is smaller:  (1,7), (7,1), (2, 8), (8,2),..., (9,15) & (15,9).   There are 18 good pairings out of the 210, or .0857

2.   I have no idea what you're asking for here.

Thank you,and the second problem is a combinations with repetition problem.

coolwind

Member

Registered: 2005-10-30

Posts: 30

Re: Some Probability Problem~~

1.a)
you have 15 places, for one sister can be. for the other sister, there are 14 more places, and given that sister A isnt on the edge, 2 places out of the 14 she could be in, if A is on the edge, there is only 1 out of 14 places, so the answer is

P(0<A<15) = 13/15, P(B next to A | 0<A<15) = 2/14
P(A=0 or A=15) = 2/15, P(B next to A | A=0 or A=15) = 1/14
P(B next to A) = 13/15*2/14 + 2/15*1/14 = 28/210 = 2/15 = 0.1333...

Good!