## Tomas.A Group Title 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? 2 years ago 2 years ago

1. satellite73 Group Title

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

2. satellite73 Group Title

sample space is now HHH HHT HT

3. Sry Group Title

Nice @ satellite 73

4. satellite73 Group Title

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

5. satellite73 Group Title

ty

6. Tomas.A Group Title

7. satellite73 Group Title

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 $A\cap B=\{HHH, HHT\}$

8. satellite73 Group Title

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

9. satellite73 Group Title

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

10. satellite73 Group Title

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

11. Tomas.A Group Title

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

12. satellite73 Group Title

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

13. satellite73 Group Title

exactly what you wrote yes

14. satellite73 Group Title

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

15. Tomas.A Group Title

$\frac{3}{4}\div \frac{1}{2}>1$

16. satellite73 Group Title

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

17. Tomas.A Group Title

P(A) is correct, how to get P(A∩B)?

18. Tomas.A Group Title

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 :/

19. satellite73 Group Title

ok i see the problem

20. satellite73 Group Title

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

21. satellite73 Group Title

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

22. satellite73 Group Title

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

23. satellite73 Group Title

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

24. Tomas.A Group Title

ah, yeah..

25. Tomas.A Group Title

ok now i understood lol

26. satellite73 Group Title

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

27. satellite73 Group Title

ok so now let us finish the problem.

28. Tomas.A Group Title

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

29. phi Group Title

|dw:1326555438102:dw|

30. satellite73 Group Title

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

31. satellite73 Group Title

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

32. satellite73 Group Title

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!

33. satellite73 Group Title

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