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Coin can be tossed until a tail appears or until it has been tossed 3 times. Given that tail does not occur on the first toss, whats probability that the coin is tossed 3 times? I think I got answer by thinking but i don't know how to write it out So I need to use conditional probability and P(no tail on 1 try)=0.5 and how to get others?

Mathematics
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you already know that you didn't have a tail on the first toss, so change the words "3 times" to "two times"
sample space is now HHH HHT HT
Nice @ satellite 73

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Other answers:

in two out of the 3 cases it is thrown three times, unless i misinterpreted the question
ty
given answer is actually 1/2
if you want a formula, you can write A= tail is tossed on first tryt B= coin is tossed 3 times and then you want \[P(B|A)=\frac{P(A\cap B)}{P(A)}\] but this begs the question because you still have to compute \[P(A\cap B)\] which just requires the same work, computing \[A\cap B=\{HHH, HHT\}\]
i had a typo above, it should have been A = heads tossed on first try ok first of all the answer, and it is \[\frac{2}{3}\] because out of the three events HHH HHT HT two are favorable
no \[P(A\cap B)\neq \frac{1}{2}\]
experiment is "Coin can be tossed until a tail appears or until it has been tossed 3 times." so lets write out a complete sample space, as it is small
shouldn't we write complete sample T, HT, HHT, HHH and given answer is 1/2 not 2/3 but it might be incorrect
{T, HT, HHT, HHH} only 4 possiblities
exactly what you wrote yes
so if we put the event A = first toss is heads, we know \[P(A)=\frac{1}{2}\]easily, but what we need is \[P(B|A)=\frac{P(A\cap B)}{P(A)}\]
\[\frac{3}{4}\div \frac{1}{2}>1\]
well i made a mistake of course, (not in the answer, in the computation)
P(A) is correct, how to get P(A∩B)?
if we look at full sample T, HT, HHT, HHH then HHT and HHH suits A and B, no? but it means 0.5 and 0.5/0.5=1 and again it's incorrect :/
ok i see the problem
i am sorry i am an idiot this morning, so lets go slow
first of all there are 4 possible outcomes, T HT HHT HHH but they are not EQUALLY LIKELY so we cannot use the uniform distribution
that is, each does not occur with probability 1/4
\[P(T)=\frac{1}{2}, P(HT)=\frac{1}{4},P(HHH)=\frac{1}{8},P(HHH)=\frac{1}{8}\]
ah, yeah..
ok now i understood lol
so i was completely wrong, and i apologize. should have a nice cup of coffee
ok so now let us finish the problem.
so \[ P(A \cap B)=\frac{1}{8}+\frac{1}{8}=\frac{1}{4}\]
  • phi
|dw:1326555438102:dw|
we know \[A\cap B = \{HHT, HHH\}\] and so what you said, \[P(A\cap B)=\frac{1}{4}\]
now \[P(A)=\frac{1}{2}\] still (at least we knew that much!) and so your answer is \[\frac{1}{4}\div\frac{1}{2}=\frac{1}{2}\]
now let us make sense out of this our new sample space is HT HHT HHH that is we knew the first toss was an H we might as well have written T HT HH and now it is clear that on the second toss you get tails with probability 1/2 and if you don't get tails you will toss again and have three tosses!
sorry again for the error and taking up so much time being wrong

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