Suppose a fair coin is tossed n times. Find simple formulae in terms of n and k for
a) P(k-1 heads |k-1 or k heads)
b) P(k heads | k-1 or k heads)
P(k heads in n fair tosses)=binomial(n,k)/2^n (0<=k<=n)
The Attempt at a Solution
You roll a die, and I roll a die. You win if the number showing on your die is strictly greater than the one on mine. If we play this game five times, what is the chance that you win at least four times? The answer is .1005
[b]2. Related Equations[\b]
There's two ways of doing it. Short way: given that you rolled a number, the probability of getting that number on the second roll is 1/6. But since we didn't get that number, we have 5/6. Now, the probability of getting one of those two numbers is 2/6. So we multiply 5/6*2/6=10/36.
Actually I just figured it out. It's similar to the birthday problem. Just create trees for multiplication rule for n events and you will get it.
Basic Probability Question—Rolling Die!!!
Suppose you roll a fair six-sided die repeatedly until the first time you roll a number that you have rolled before. A) for each r=1,2,... calculate the probability pr that you roll exactly r times.