1) Suppose we roll a fair die. Whatever number comes up, we toss a coin that many times. What is the probability mass function of the number of heads? Let X=number of heads Then I know that X can take on the values 0,1,2,3,4,5,6. How to proceed from here I have no clue... Could someone please explain? Any help would be appreciated!
There's probably (P=1) a more elegant way, but I'd do a separate table for each die result, then multiply each item in each table by 1/6 to get a combined table... if that makes sense.
How about the coin??? By the way, is this a "birvariate" question? How can I define the second random variable?
I meant that for each roll of the die, draw a separate probability table for the coin flips. For example: Code: 4 on die: Heads | P 0 | 0.0625 1 | 0.25 2 | 0.375 3 | 0.25 4 | 0.0625 Then combine the tables, dividing each of the listed probabilities by six and adding matching items from each table. For example, to get the combined probability for 2 heads, you'd divide the "2 head" probability from each table by six and add them together: Code: 2 on die: Heads | P 2 | 0.25 3 on die: Heads | P 2 | 0.375 4 on die: Heads | P 2 | 0.375 5 on die: Heads | P 2 | 0.3125 6 on die: Heads | P 2 | 0.234375 Total: --------------- Heads | P 2 | 0.2578125 (I think that's right... but you'd better check!) It does appear to be a bivariate problem, with the two variables being the die roll and the coin flip. But I'm not sure of what that means for the "correct" way to approach the problem. I'm approaching in a brute force intuitive kind of way.
This is the binomial probability distribution because the coin toss satisfies the conditions for a Bernoulli Trial. 1/ Binomial outcome ; H or T 2/ Each toss is independent of the other events P(r) = C(n,r) p^n-r (1 - p)^r Plot P(r) verses r to get the distribution n = the result of the die roll If the die comes up 4, n = 4 and r is 0, 1,2,3,4, the possible outcomes for which the probability is determined by the distribution above.