Probability No Member Gets >1 Medal: Math Club

[resurgance2001]

Homework Statement

There are 15 members of a maths club. There are 4 different medals to be randomly given to the members of the club. What is the probability that no member will receive more than one of the medals.

Homework Equations

Try to find the number of combinations where no member receives more than 1 medal and divide by the total number of ways the medals can be distributed.

The total number of ways the medals can be distributed is 15^4 = 50625. Use nCr. Or nPr but really not sure

The Attempt at a Solution

I have the answer from the mark scheme is (15 x 14 x 14 x 12)/15^4

I just don't understand though how this answer was derived. I have tried experimenting with smaller numbers on paper but can't for the life of me figure it out and have not been able to find any similar questions in the textbook or online.

 

[Charles Link]

If you give out the medals one a a time, what is the probability you can do it successfully with the first medal, without having one person with two medals? Then, once you have given out one medal, what is the fraction (probability ) of these cases, that you can give out the second medal, and still have no one with two medals? Then continue the process two more times. $\\$ And I think you have a typo in the answer that you gave.

 

[Ray Vickson]

Homework Statement

There are 15 members of a maths club. There are 4 different medals to be randomly given to the members of the club. What is the probability that no member will receive more than one of the medals.

Homework Equations

Try to find the number of combinations where no member receives more than 1 medal and divide by the total number of ways the medals can be distributed.

The total number of ways the medals can be distributed is 15^4 = 50625. Use nCr. Or nPr but really not sure

The Attempt at a Solution

I have the answer from the mark scheme is (15 x 14 x 14 x 12)/15^4

I just don't understand though how this answer was derived. I have tried experimenting with smaller numbers on paper but can't for the life of me figure it out and have not been able to find any similar questions in the textbook or online.

The probability that the second medal goes to a different member than the first is 14/15, because each member has probability 1/15 of getting the second medal, and 14 of those members are different from the first recipient. The probability that the third recipient is different from both of the first two is 13/15, etc. (If I had been writing the answer I would not have bothered with the first factor 15/15 = 1.)

 

[resurgance2001]

Thank you - I think you have explained it very well and in the simplest terms.

 

[resurgance2001]

So the first factor of 15/15 which you said is not really necessary is just saying that there is a 15/15 or probability of 1 that one of the 15 will get the first meddle - which is trivial. Thanks