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\[p(y) = p*(1-p)^{y-1}\]
p = .5
y = k

by the way, the answer is supposed to be 2^-k

^is there a way to think* sorry, typo haha

kind of a strange way to approach this however

\(k=1\) you get \(\frac{1}{2}\)
\(k=2\) it is \(\left(\frac{1}{2}\right)^2\) etc

that's true. hmm. it would only be
.5*[(.5^k)/(.5)]
if we start at 0, right @satellite73.

you can't get a head for the first time on the zeroth trial

i am not saying the answer is wrong, just saying it is somewhat complicated way to think about it

yeah, you have to start from 1.

\[(0.5)^{k-1}=\frac{0.5^{k}}{0.5}\]

sorry, i had a brain block
(.5)^(k-1) = (.5^k)*(.5^-1)

\[p(k)=0.5\times 0.5^{k-1}=0.5\times \frac{0.5^{k}}{0.5}=0.5^{k}=2^{-k}\]

.5^k = 1/(2^k) = 2^(-k)

@windsylph Are you there?

You're welcome :)