Quantcast

Got Homework?

Connect with other students for help. It's a free community.

  • across
    MIT Grad Student
    Online now
  • laura*
    Helped 1,000 students
    Online now
  • Hero
    College Math Guru
    Online now

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

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

  • This Question is Closed
  1. satellite73 Group Title
    Best Response
    You've already chosen the best response.
    Medals 3

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

    • 2 years ago
  2. satellite73 Group Title
    Best Response
    You've already chosen the best response.
    Medals 3

    sample space is now HHH HHT HT

    • 2 years ago
  3. Sry Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    Nice @ satellite 73

    • 2 years ago
  4. satellite73 Group Title
    Best Response
    You've already chosen the best response.
    Medals 3

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

    • 2 years ago
  5. satellite73 Group Title
    Best Response
    You've already chosen the best response.
    Medals 3

    ty

    • 2 years ago
  6. Tomas.A Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    given answer is actually 1/2

    • 2 years ago
  7. satellite73 Group Title
    Best Response
    You've already chosen the best response.
    Medals 3

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

    • 2 years ago
  8. satellite73 Group Title
    Best Response
    You've already chosen the best response.
    Medals 3

    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

    • 2 years ago
  9. satellite73 Group Title
    Best Response
    You've already chosen the best response.
    Medals 3

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

    • 2 years ago
  10. satellite73 Group Title
    Best Response
    You've already chosen the best response.
    Medals 3

    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

    • 2 years ago
  11. Tomas.A Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

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

    • 2 years ago
  12. satellite73 Group Title
    Best Response
    You've already chosen the best response.
    Medals 3

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

    • 2 years ago
  13. satellite73 Group Title
    Best Response
    You've already chosen the best response.
    Medals 3

    exactly what you wrote yes

    • 2 years ago
  14. satellite73 Group Title
    Best Response
    You've already chosen the best response.
    Medals 3

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

    • 2 years ago
  15. Tomas.A Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

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

    • 2 years ago
  16. satellite73 Group Title
    Best Response
    You've already chosen the best response.
    Medals 3

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

    • 2 years ago
  17. Tomas.A Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    P(A) is correct, how to get P(A∩B)?

    • 2 years ago
  18. Tomas.A Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    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 :/

    • 2 years ago
  19. satellite73 Group Title
    Best Response
    You've already chosen the best response.
    Medals 3

    ok i see the problem

    • 2 years ago
  20. satellite73 Group Title
    Best Response
    You've already chosen the best response.
    Medals 3

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

    • 2 years ago
  21. satellite73 Group Title
    Best Response
    You've already chosen the best response.
    Medals 3

    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

    • 2 years ago
  22. satellite73 Group Title
    Best Response
    You've already chosen the best response.
    Medals 3

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

    • 2 years ago
  23. satellite73 Group Title
    Best Response
    You've already chosen the best response.
    Medals 3

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

    • 2 years ago
  24. Tomas.A Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    ah, yeah..

    • 2 years ago
  25. Tomas.A Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    ok now i understood lol

    • 2 years ago
  26. satellite73 Group Title
    Best Response
    You've already chosen the best response.
    Medals 3

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

    • 2 years ago
  27. satellite73 Group Title
    Best Response
    You've already chosen the best response.
    Medals 3

    ok so now let us finish the problem.

    • 2 years ago
  28. Tomas.A Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

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

    • 2 years ago
  29. phi Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    |dw:1326555438102:dw|

    • 2 years ago
  30. satellite73 Group Title
    Best Response
    You've already chosen the best response.
    Medals 3

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

    • 2 years ago
  31. satellite73 Group Title
    Best Response
    You've already chosen the best response.
    Medals 3

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

    • 2 years ago
  32. satellite73 Group Title
    Best Response
    You've already chosen the best response.
    Medals 3

    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!

    • 2 years ago
  33. satellite73 Group Title
    Best Response
    You've already chosen the best response.
    Medals 3

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

    • 2 years ago
    • Attachments:

See more questions >>>

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.