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

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i think that this might be a geometric distribution, rather than a binomial distribution. @kropot72

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0is there my way to think of the answer more intuitively? the book asks this question even before those topics are discussed

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[p(y) = p*(1p)^{y1}\] p = .5 y = k

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0by the way, the answer is supposed to be 2^k

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so 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
 3 years ago
Best ResponseYou've already chosen the best response.0^is there a way to think* sorry, typo haha

anonymous
 3 years ago
Best 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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0even 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
 3 years ago
Best ResponseYou've already chosen the best response.0it 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
 3 years ago
Best ResponseYou've already chosen the best response.2You are right @tomo. The geometric distribution applies: \[p(k)=p(1p)^{k1}\] p = 0.5 \[p(k)=0.5(0.5)^{k1}=0.5\times \frac{0.5^{k}}{0.5}=0.5^{k}=\frac{1^{k}}{2^{k}}=2^{k}\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0kind of a strange way to approach this however

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\(k=1\) you get \(\frac{1}{2}\) \(k=2\) it is \(\left(\frac{1}{2}\right)^2\) etc

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0you can't get a head for the first time on the zeroth trial

kropot72
 3 years ago
Best 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\[0.5^{k}=2^{k}\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i am not saying the answer is wrong, just saying it is somewhat complicated way to think about it

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yeah, you have to start from 1.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i 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)

kropot72
 3 years ago
Best ResponseYou've already chosen the best response.2\[(0.5)^{k1}=\frac{0.5^{k}}{0.5}\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0sorry, i had a brain block (.5)^(k1) = (.5^k)*(.5^1)

kropot72
 3 years ago
Best ResponseYou've already chosen the best response.2\[p(k)=0.5\times 0.5^{k1}=0.5\times \frac{0.5^{k}}{0.5}=0.5^{k}=2^{k}\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0.5^k = 1/(2^k) = 2^(k)

anonymous
 3 years ago
Best 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

kropot72
 3 years ago
Best ResponseYou've already chosen the best response.2If 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.

kropot72
 3 years ago
Best ResponseYou've already chosen the best response.2@windsylph Are you there?

kropot72
 3 years ago
Best ResponseYou've already chosen the best response.2Perhaps 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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Thank 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.
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