1) You have to show that if for
and ,
then .
Hint: You may find the Cauchy-Schwartz Inequality useful.
2) You are correct.
I hope these aren't too hard for the basic statistics forum, but some friends and I are disagreeing on the answers to these two questions:
1) Show that if two dice are loaded with the same probability distribution, then the probability of doubles is at least 1/6.
2) If P2 is the probability of a 2 showing and we are told that at least one 2 has shown, what is the probability that upon looking at the pair of dice the sum will be 4?
Now for each question:
1) My initial solution to this was to show that the probability of doubles on die with even distributions are 1/6, so any variation in the distribution leads to a smaller number. But I can't quite prove this algebraically and I'm stuck. Is my approach wrong?
2) The simple answer (which some are sticking to) is that it's simple: The answer is P2. But I don't believe it could be that easy, and here's the answer I came up with:
Let A = the event where the sum is 4
B = The event where at least one 2 shows up
A is equal to the probability of getting a (1,3), (3,1), or (2,2). Since we know one die is a two, the probability of 1,3 or 3,1 is zero and P[A] = P[(2,2)].
As for B:
Now, we want to know P[A|B]:
I can't explain it...it just feels wrong. Like I did too much. Is the simple answer really correct or is this one right?
Yes. I am trying to figure out why some of these solutions don't work algebraically. I already have the correct answer, but I'm trying to figure out why some of these other approaches I might have used do not work.
I realize now that my assertion that P(D1 = 2 OR D2 = 2) = P(D1=2) + P(D2=2) doesn't work because the two events are not mutually exclusive. For anyone wondering, the formula should then be:
Which is correct. But now I have a different question. If this were seen as a bernoulli trial, we would then use the formula for a binomial distribution:
So does the binomial distribution not apply here?