Small probability problem

JCB

Hey all, I am doing this small question but I have two answers and I can't figure out which one is right:
The question is: The birthdays of three boys fall on the same week. Find the probability that only one of the birthdays falls on a Sunday:
The answer is either:
(a) 1/7 x 6/7 x 6/7 = 36/343 or
(b) [ 1/7 x 6/7 x 6/7] x 3 = 108/343..... x 3 because either of the 3 people could have their birthday on the Sunday.

Any help you could offer would be brilliant! Thanks.

smiles

JCB wrote:

(b) [ 1/7 x 6/7 x 6/7] x 3 = 108/343.....

I'd go with this one, but not sure.

The way I'd look at it is as follows. Name the boys A,B,C.

A's likelihood of having a b-day on a Sunday.. is (a), so is B, so is C, so P(Abday) + P(B) + P(C) = 3 * (a) = (b)

I'll have a think.

Yakuza

This is a binomially distributed type event, with the probablity of "success" is 1/7

With 3 "trials" (kids), the probability for one success is:

3!

---

* (1/7) * (6/7)²
1!(2!)
or, as a fraction 108 / 343. (Option b!)

For the record, the other probabilities are:
P(0 birthdays on Sunday) = 216/343
P(2 birthdays on Sunday) = 18/343
P(3 birthdays on Sunday) = 1/343

And the lot adds up to 343/343 (phew!)

JCB

Thanks very much for your thoughts. I never realised that it could be tackled from a binomial point of view!!!!
Cheers!

Yakuza

You're welcome Hope it helped :cool: It actually helped me to get back into thinking about probabilities etc - I'm going to be going back to the books again soon hopefully to finish off my Actuarial exams. I'll be up to my armpits in p's, q's and chi-squareds again