# Thread: joint and marginal p.m.f.s

1. ## joint and marginal p.m.f.s

In a smoking survey among boys between the ages of 12 and 17, 78% prefer to date
nonsmokers. 1% prefer to date smokers, and 21% don’t care. Suppose seven such boys are
selected randomly and X equals the number who prefer to date nonsmokers and Y equals the
number who prefer to date smokers.

(a) Determine the joint p.m.f. of X and Y. Be sure to include the support of the p.m.f.

$(\frac{7!}{x!7-x!})*(\frac{7-x!}{y!7-x-y!})*(0.78^{x})*(0.01^{y})*(0.21)^{7-x-y}$?

(b) Find the marginal p.m.f. of X. Again, include the support.

$(\frac{7!}{x!7-x!})*(0.78^{x})*(0.21)^{7-x-y}$?

Thank you!

2. (a) looks right, but you need to place () before the factorials and cancel that common term. This is a multinomial.
(b) does not make sense mathematically. You can't have y in a function that is only a function on x. This is just a binomial.

3. Here are my corrections.......

Originally Posted by redpack
In a smoking survey among boys between the ages of 12 and 17, 78% prefer to date
nonsmokers. 1% prefer to date smokers, and 21% don’t care. Suppose seven such boys are
selected randomly and X equals the number who prefer to date nonsmokers and Y equals the
number who prefer to date smokers.

(a) Determine the joint p.m.f. of X and Y. Be sure to include the support of the p.m.f.

$\frac{7!}{x!y!(7-x-y)!}(0.78)^x(0.01)^y(0.21)^{7-x-y}$

(b) Find the marginal p.m.f. of X. Again, include the support.

$\frac{7!}{x!(7-x)!}(0.78)^x(0.22)^{7-x}$ where x=0,...,7.

Thank you!
You combine the .21 and the .01 for the q in the 'failure' category.

4. In part b, you can combine the 0.01 and 0.21 and just forget the y? the exponent doesn't make a difference?

5. a function of just x can only have constants and x's, no other variables.

6. IN (b) a success is to prefer to date nonsmokers, so p=.78 and q=.22
The complement is {don't care or date smokers}.