Hi all. I'm working on a problem from my stats/probabilities class, and I think I know the answer. I just want to double check and make sure I understand what the problem's asking. The problem text is:
I assigned a random variable Y to be the number of flips until the first head. My reasoning is that we're looking for the probability that we need 12 or more flips to get head, but no upper bound is specified. They only say "at least 12 flips" so we could be looking at 12, 13, 14, ... 1000, ... one million, etc. Hence, the calculation I used was:
\(\displaystyle \displaystyle P(Y \ge 12 | \text{"first ten flips tails"}) = \sum_{y=2}^{\infty}(0.5 \cdot 0.5^{y-1}) = \sum_{y=2}^{\infty}(0.5^{y}) = 0.5\)
Mechanically, this calculation makes sense, but for some reason, I'm second guessing myself.
3.72)Given that we have already flipped a balanced coin ten times and obtained zero heads, what is the probability that we must toss the coin at least two more times to obtain the first head?
I assigned a random variable Y to be the number of flips until the first head. My reasoning is that we're looking for the probability that we need 12 or more flips to get head, but no upper bound is specified. They only say "at least 12 flips" so we could be looking at 12, 13, 14, ... 1000, ... one million, etc. Hence, the calculation I used was:
\(\displaystyle \displaystyle P(Y \ge 12 | \text{"first ten flips tails"}) = \sum_{y=2}^{\infty}(0.5 \cdot 0.5^{y-1}) = \sum_{y=2}^{\infty}(0.5^{y}) = 0.5\)
Mechanically, this calculation makes sense, but for some reason, I'm second guessing myself.