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?

- anonymous

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- anonymous

you already know that you didn't have a tail on the first toss, so change the words "3 times" to "two times"

- anonymous

sample space is now
HHH
HHT
HT

- anonymous

Nice @ satellite 73

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## More answers

- anonymous

in two out of the 3 cases it is thrown three times, unless i misinterpreted the question

- anonymous

ty

- anonymous

given answer is actually 1/2

- anonymous

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
\[A\cap B=\{HHH, HHT\}\]

- anonymous

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

- anonymous

no
\[P(A\cap B)\neq \frac{1}{2}\]

- anonymous

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

- anonymous

shouldn't we write complete sample T, HT, HHT, HHH
and given answer is 1/2 not 2/3 but it might be incorrect

- anonymous

{T, HT, HHT, HHH} only 4 possiblities

- anonymous

exactly what you wrote yes

- anonymous

so if we put the event A = first toss is heads, we know
\[P(A)=\frac{1}{2}\]easily, but what we need is
\[P(B|A)=\frac{P(A\cap B)}{P(A)}\]

- anonymous

\[\frac{3}{4}\div \frac{1}{2}>1\]

- anonymous

well i made a mistake of course, (not in the answer, in the computation)

- anonymous

P(A) is correct, how to get P(A∩B)?

- anonymous

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 :/

- anonymous

ok i see the problem

- anonymous

i am sorry i am an idiot this morning, so lets go slow

- anonymous

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

- anonymous

that is, each does not occur with probability 1/4

- anonymous

\[P(T)=\frac{1}{2}, P(HT)=\frac{1}{4},P(HHH)=\frac{1}{8},P(HHH)=\frac{1}{8}\]

- anonymous

ah, yeah..

- anonymous

ok now i understood lol

- anonymous

so i was completely wrong, and i apologize. should have a nice cup of coffee

- anonymous

ok so now let us finish the problem.

- anonymous

so \[ P(A \cap B)=\frac{1}{8}+\frac{1}{8}=\frac{1}{4}\]

- phi

|dw:1326555438102:dw|

- anonymous

we know
\[A\cap B = \{HHT, HHH\}\] and so what you said,
\[P(A\cap B)=\frac{1}{4}\]

- anonymous

now
\[P(A)=\frac{1}{2}\] still (at least we knew that much!) and so your answer is
\[\frac{1}{4}\div\frac{1}{2}=\frac{1}{2}\]

- anonymous

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!

- anonymous

sorry again for the error and taking up so much time being wrong

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