CIS 2033
Alternative Solutions to Homework Problem 11, Chapter 4
TA’s note
Question 4.11. You decide to play monthly in two different lotteries, and you stop playing as soon as you win a prize in one (or both) lotteries of at least one million dollars.
Suppose that every time you participate in these lotteries, the probability to win at least
one million dollars is p1 for one of the lotteries and p2 for the other. Let M be the number
of times you participate in these lotteries until winning at least one prize. What kind of
distribution does M have, and what is its parameter?
Answer:
Let S be the event that you stop playing the lottery in a given month. According to
the problem, S occurs if you either
• Win lottery 1 - we’ll call this event A
• Win lottery 2 - we’ll call this event B
• Win both lotteries - this would be event A ∩ B
And there’s no other way to make you stop playing.
Now, there are at least two ways of looking at this.
1st way - the longer way:
One way to do this is to note that S can be decomposed into the following mutually
exclusive events:
{Stop Playing} = {Win both lotteries} OR {Win 1st , not 2nd } OR {Win 2nd , not 1st }.
That is,
S = (A ∩ B) ∪ (A ∩ B c ) ∪ (Ac ∩ B).
Therefore
P (S) = p1 p2 + p1 (1 − p2 ) + (1 − p1 )p2
1
(1)
Thus M would have a geometric distribution with parameter P (S).
2nd way - maybe a better way:
Another way to do this is to simply notice that you don’t stop playing as long as you
are loosing. That is, you don’t stop playing as long as you’ve won neither lootery 1 nor
lottery 2:
S c = Ac ∩ B c
And therefore
P (S c ) = (1 − p1 )(1 − p2 )
Since we want P (S), the desired answer is:
P (S) = 1 − P (S c ) = 1 − (1 − p1 )(1 − p2 )
Now, it’s an easy check to see that this works out to be the same as (1).
I think the second way is better. You?
2
(2)