Question

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?

Answer 1

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

Answer 2

sample space is now
HHH
HHT
HT

Answer 3

Nice @ satellite 73

Answer 4

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

Answer 5

ty

Answer 6

given answer is actually 1/2

Answer 7

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

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

Answer 8

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

Answer 9

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

Answer 10

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

Answer 11

shouldn't we write complete sample T, HT, HHT, HHH

and given answer is 1/2 not 2/3 but it might be incorrect

Answer 12

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

Answer 13

exactly what you wrote yes

Answer 14

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

Answer 15

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

Answer 16

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

Answer 17

P(A) is correct, how to get P(A∩B)?

Answer 18

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 :/

Answer 19

ok i see the problem

Answer 20

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

Answer 21

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

Answer 22

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

Answer 23

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

Answer 24

ah, yeah..

Answer 25

ok now i understood lol

Answer 26

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

Answer 27

ok so now let us finish the problem.

Answer 28

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