- #1
Yagoda
- 46
- 0
Homework Statement
The following experiment involves a single coin with probability p of heads on anyone flip, where
0 < p < 1.
Step 1: Flip the coin. Let X = 1 if heads, 0 otherwise.
Step 2: Flip the coin (X + 1) times. Let Y = the number of heads obtained in this step.
Step 3: Flip the coin (X + Y + 1) times. Let Z = the number of heads obtained in this step.
Let T denote the total number of heads across all three steps.
What is P(X = 1|Z = 0)?
Homework Equations
[itex]P(A|B) = \frac{P(A \cap B)}{P(B)}[/itex]
The Attempt at a Solution
I think I have been thinking about this too long and am just confusing myself. My first gut reaction was to say that no matter what the outcome of Z, since the coin isn't changing, the probability of it coming up heads on any given flip (ie P(X = 1)) will be p.
But since you are flipping a variable number of times to get Z, it seems your chance of getting Z = 0 would be greater with a smaller number of flips, which would be more likely if you begin with X=0 than X=1. Does this even matter?
I tried using the above conditional probability formula as well, but it got ugly quickly in trying to calculate the numerator. Is there a less thorny method that I'm missing?