Got Homework?
Connect with other students for help. It's a free community.
Here's the question you clicked on:
 0 viewing
windsylph
Group Title
Probability: What is the probability that a fair coin lands as heads for the first time on the kth toss?
 one year ago
 one year ago
windsylph Group Title
Probability: What is the probability that a fair coin lands as heads for the first time on the kth toss?
 one year ago
 one year ago

This Question is Closed

kropot72 Group TitleBest ResponseYou've already chosen the best response.2
This can be found from the binomial distribution: \[\left(\begin{matrix}k \\ 1\end{matrix}\right)0.5(0.5)^{k1}\]
 one year ago

tomo Group TitleBest ResponseYou've already chosen the best response.0
i think that this might be a geometric distribution, rather than a binomial distribution. @kropot72
 one year ago

windsylph Group TitleBest ResponseYou've already chosen the best response.0
is there my way to think of the answer more intuitively? the book asks this question even before those topics are discussed
 one year ago

tomo Group TitleBest ResponseYou've already chosen the best response.0
\[p(y) = p*(1p)^{y1}\] p = .5 y = k
 one year ago

windsylph Group TitleBest ResponseYou've already chosen the best response.0
by the way, the answer is supposed to be 2^k
 one year ago

windsylph Group TitleBest ResponseYou've already chosen the best response.0
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?
 one year ago

windsylph Group TitleBest ResponseYou've already chosen the best response.0
^is there a way to think* sorry, typo haha
 one year ago

tomo Group TitleBest ResponseYou've already chosen the best response.0
.5 for the one trial that you get correct. .5^(k1) for the trials that you get incorrect. its k1, because you get the kth trial correct.
 one year ago

windsylph Group TitleBest ResponseYou've already chosen the best response.0
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
 one year ago

windsylph Group TitleBest ResponseYou've already chosen the best response.0
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?
 one year ago

kropot72 Group TitleBest ResponseYou've already chosen the best response.2
You are right @tomo. The geometric distribution applies: \[p(k)=p(1p)^{k1}\] p = 0.5 \[p(k)=0.5(0.5)^{k1}=0.5\times \frac{0.5^{k}}{0.5}=0.5^{k}=\frac{1^{k}}{2^{k}}=2^{k}\]
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
kind of a strange way to approach this however
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
\(k=1\) you get \(\frac{1}{2}\) \(k=2\) it is \(\left(\frac{1}{2}\right)^2\) etc
 one year ago

tomo Group TitleBest ResponseYou've already chosen the best response.0
that's true. hmm. it would only be .5*[(.5^k)/(.5)] if we start at 0, right @satellite73.
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
you can't get a head for the first time on the zeroth trial
 one year ago

kropot72 Group TitleBest ResponseYou've already chosen the best response.2
@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\[0.5^{k}=2^{k}\]
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
i am not saying the answer is wrong, just saying it is somewhat complicated way to think about it
 one year ago

tomo Group TitleBest ResponseYou've already chosen the best response.0
yeah, you have to start from 1.
 one year ago

tomo Group TitleBest ResponseYou've already chosen the best response.0
i am just a bit confused @kropot72 on your second to last comment. wouldn't p(k) = .5*.5*(.5)^k? wouldn't (.5)^(k1) = .5*.5^(k)
 one year ago

kropot72 Group TitleBest ResponseYou've already chosen the best response.2
\[(0.5)^{k1}=\frac{0.5^{k}}{0.5}\]
 one year ago

tomo Group TitleBest ResponseYou've already chosen the best response.0
sorry, i had a brain block (.5)^(k1) = (.5^k)*(.5^1)
 one year ago

kropot72 Group TitleBest ResponseYou've already chosen the best response.2
\[p(k)=0.5\times 0.5^{k1}=0.5\times \frac{0.5^{k}}{0.5}=0.5^{k}=2^{k}\]
 one year ago

tomo Group TitleBest ResponseYou've already chosen the best response.0
.5^k = 1/(2^k) = 2^(k)
 one year ago

windsylph Group TitleBest ResponseYou've already chosen the best response.0
@satellite73, how is it that k=2 is (1/2)^2 ? your response is more of what I was looking for, thanks
 one year ago

kropot72 Group TitleBest ResponseYou've already chosen the best response.2
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 subevent.
 one year ago

kropot72 Group TitleBest ResponseYou've already chosen the best response.2
@windsylph Are you there?
 one year ago

kropot72 Group TitleBest ResponseYou've already chosen the best response.2
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)}^(k1) * P(H) = (0.5)^(k1) * 0.5^1 = (0.5)^k = 2^k
 one year ago

windsylph Group TitleBest ResponseYou've already chosen the best response.0
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.
 one year ago

kropot72 Group TitleBest ResponseYou've already chosen the best response.2
You're welcome :)
 one year ago
See more questions >>>
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.