Probability: What is the probability that a fair coin lands as heads for the first time on the kth toss?

- anonymous

- katieb

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- kropot72

This can be found from the binomial distribution:
\[\left(\begin{matrix}k \\ 1\end{matrix}\right)0.5(0.5)^{k-1}\]

- anonymous

i think that this might be a geometric distribution, rather than a binomial distribution. @kropot72

- anonymous

is there my way to think of the answer more intuitively? the book asks this question even before those topics are discussed

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## More answers

- anonymous

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

- anonymous

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

- anonymous

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?

- anonymous

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

- anonymous

.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.

- anonymous

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

- anonymous

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?

- kropot72

You are right @tomo. The geometric distribution applies:
\[p(k)=p(1-p)^{k-1}\]
p = 0.5
\[p(k)=0.5(0.5)^{k-1}=0.5\times \frac{0.5^{k}}{0.5}=0.5^{k}=\frac{1^{k}}{2^{k}}=2^{-k}\]

- anonymous

kind of a strange way to approach this however

- anonymous

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

- anonymous

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

- anonymous

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

- kropot72

@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\[0.5^{k}=2^{-k}\]

- anonymous

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

- anonymous

yeah, you have to start from 1.

- anonymous

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)

- kropot72

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

- anonymous

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

- kropot72

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

- anonymous

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

- anonymous

@satellite73, how is it that k=2 is (1/2)^2 ? your response is more of what I was looking for, thanks

- kropot72

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.

- kropot72

@windsylph Are you there?

- kropot72

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

- anonymous

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.

- kropot72

You're welcome :)

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