## 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

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

sample space is now HHH HHT HT

3. Sry

Nice @ satellite 73

4. satellite73

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

5. satellite73

ty

6. Tomas.A

7. satellite73

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

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

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

10. satellite73

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

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

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

13. satellite73

exactly what you wrote yes

14. satellite73

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

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

16. satellite73

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

17. Tomas.A

P(A) is correct, how to get P(A∩B)?

18. Tomas.A

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

ok i see the problem

20. satellite73

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

21. satellite73

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

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

23. satellite73

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

24. Tomas.A

ah, yeah..

25. Tomas.A

ok now i understood lol

26. satellite73

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

27. satellite73

ok so now let us finish the problem.

28. Tomas.A

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

29. phi

|dw:1326555438102:dw|

30. satellite73

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

31. satellite73

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

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

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