Open study

is now brainly

With Brainly you can:

  • Get homework help from millions of students and moderators
  • Learn how to solve problems with step-by-step explanations
  • Share your knowledge and earn points by helping other students
  • Learn anywhere, anytime with the Brainly app!

A community for students.

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

I got my questions answered at in under 10 minutes. Go to now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly


Get your free account and access expert answers to this and thousands of other questions

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

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

\[p(y) = p*(1-p)^{y-1}\] 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)
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)
@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.
You're welcome :)

Not the answer you are looking for?

Search for more explanations.

Ask your own question