Quick probability prob?

Nappy

Hey, there are 7 major accidents in a city every week, assuming accidents are equally likely what is the probability in there being 1 accident each day??

Would I be right in assuming it is

7!/(7)7

*(7)7 = 7 to the power of 7..

Thanks..

Nappy

Am studying for repeat by the way...

Any help much appreciated..

there are 6 pairs of shoes, a set of 6 shoes are chosen, whats the probability that there is at least 1 pair of shoes in the set? all shoes are different..

Am I right >> (12.10.8.6.4.2)/12C6

Cheers!

Fremen

Yes, I think you're right on both counts.

First problem is a calculation involving the Multinomial distribution. The example on the wiki page is similar to your problem.

Second problem is fairly straightforward. Number of ways to choose shoes without making pairs is 12*10.....*2, and clearly the total number is 12C6.

Good work .

P.s. Not sure I like the phrasing of the first question. Did it come from a course in DCU?

ray giraffe

The first question is ambiguous. 2 possible meanings:

(a)If we know that in a particular week, there were 7 accidents, what is the probability that there was 1 each day - this is 7!/(7^7)

(b)If we know that in an average week there are 7 accidents, what is the probability there is one each day in a particular week? - poisson distribution with expected value 1 for each day - gives e^(-7)

The answer to the second question would surely be 1 - ((2^6)/12C6). The number of ways of choosing shoes without pairs is 2^6 -simply choose left or right for each.

Fremen

You're right about the second problem Ray, my bad. The number of ways to choose the shoes without making a pair is
12*10*8*6*4*2/6! = 2^6

I misread the question and was looking for the probability of not making a pair. Bit careless of me.