Here's the question you clicked on:
windsylph
Probability: What is the probability that a fair coin lands as heads for the first time on the kth toss?
This can be found from the binomial distribution: \[\left(\begin{matrix}k \\ 1\end{matrix}\right)0.5(0.5)^{k-1}\]
i think that this might be a geometric distribution, rather than a binomial distribution. @kropot72
is there my way to think of the answer more intuitively? the book asks this question even before those topics are discussed
\[p(y) = p*(1-p)^{y-1}\] p = .5 y = k
by the way, the answer is supposed to be 2^-k
so for the first trial, it's 0.5. because obviously heads and tails are the only possibilities. however, for the second trial, it's 0.25.. but why?
^is there a way to think* sorry, typo haha
.5 for the one trial that you get correct. .5^(k-1) for the trials that you get incorrect. its k-1, because you get the kth trial correct.
even independence is not discussed at this point (0.5*0.5*0.5... because the one trial should not affect the outcome of the next), so I'm wondering if we could reason this out in a more intuitive way
it makes sense that as k increases, the probability decreases, because obviously the P(1st trial) = 0.5 and P(32984th trial) = really small number.. if you only got a head after the 32984th trial, then yes the probability is indeed smaller.. but how exactly is it 2^-k?
You are right @tomo. The geometric distribution applies: \[p(k)=p(1-p)^{k-1}\] p = 0.5 \[p(k)=0.5(0.5)^{k-1}=0.5\times \frac{0.5^{k}}{0.5}=0.5^{k}=\frac{1^{k}}{2^{k}}=2^{-k}\]
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
@satellite73 Very true. The outcome of each toss is independent of the others. So the probability of a head for the first time is simply\[0.5^{k}=2^{-k}\]
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.
i am just a bit confused @kropot72 on your second to last comment. wouldn't p(k) = .5*.5*(.5)^k? wouldn't (.5)^(k-1) = .5*.5^(k)
\[(0.5)^{k-1}=\frac{0.5^{k}}{0.5}\]
sorry, i had a brain block (.5)^(k-1) = (.5^k)*(.5^-1)
\[p(k)=0.5\times 0.5^{k-1}=0.5\times \frac{0.5^{k}}{0.5}=0.5^{k}=2^{-k}\]
@satellite73, how is it that k=2 is (1/2)^2 ? your response is more of what I was looking for, thanks
If I can answer in the absence of satellite73: The probability of getting a head on the first toss (k = 1) is 1/2 = 0.5. If tails comes up on the first toss then the probability of getting heads for the first time on the second toss (k = 2) is 1/2 * 1/2 = (1/2)^2 This result makes use of the multiplication theorem of probability which says: If a compound event is made up of a number of separate and independent events, the probability of the compound event is the product of the separate probabilities of each sub-event.
@windsylph Are you there?
Perhaps this explanation will help you: P(H) = 0.5 P(T) = 0.5 The probability of heads coming up for the first time on the fourth toss is: P(T) * P(T) * P(T) * P(H) = 1/2 * 1/2 * 1/2 * 1/2 = (1/2)^4 = 2^-4 The probability of heads coming up on the k th toss is: {(P(T)}^(k-1) * P(H) = (0.5)^(k-1) * 0.5^1 = (0.5)^k = 2^-k
Thank you very much, @kropot72. I guess I'm wrong that there's a more basic explanation that they are independent events. I'm sorry, I already slept during the time you answered.