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Probability: unfair coin toss, probably pretty easy

  1. Oct 1, 2009 #1
    1. The problem statement, all variables and given/known data

    A biased coin lands heads with probabilty 2/3. The coin is tossed 3 times

    a) Given that there was at least one head in the three tosses, what is the probability that there were at least two heads?

    b) use your answer in a) to find the probability that there was exactly one head, give that there was at least one head in the three tosses.


    3. The attempt at a solution

    a) So i figured the best way to do this would be to find the probability of the complement events and subtract one.

    So i calculated P(No heads)+ P(exactly one head)= 1/27 + 6/27 = 7/27. So my answer would be 20/27. Which isnt the answer in the back of the book.

    I assume i am wrong and i think it has to do with the following clause: "Given that there was at least one head in the three tosses"

    Dunno how to factor that into my work... any help appreciated
     
  2. jcsd
  3. Oct 1, 2009 #2
    For part A as your given that there was at least 1. would you than have to just work out the probability of at least 1 head in the other two tosses?

    If you use the binomial distribution

    [itex]_{n}C_{r}\times p^{r}\times q^{n-r}[/itex]

    where p = probability of success = 2/3 and q = the compliment (1-p) = 1/3

    n = trials r = number of events ie you need at least r from n tosses

    [itex]_{2}C_{1}\times p^{1}\times q^{2-1}[/itex]

    give that a go
     
  4. Oct 1, 2009 #3

    Dick

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    Yes, the problem is "Given that there was at least one head in the three tosses". For a) you want (P(2 heads)+P(3 heads))/(P(1 head)+P(2 heads)+P(3 heads)). P(no heads) doesn't enter into the sample space.
     
  5. Oct 1, 2009 #4
    Ok i calculated this out and got 20/26 which is the correct answer, but can you explain how you came to this formula?
     
  6. Oct 1, 2009 #5

    Dick

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    If you wanted the probability without the restriction "Given that there was at least one head in the three tosses" the answer would be (P(2 heads)+P(3 heads))/(P(0 heads)+P(1 head)+P(2 heads)+P(3 heads)), right? Where the denominator is clearly 1. I just threw P(0 heads) out of the problem. Since your restriction doesn't allow it. It's P(of everything I want)/P(everything that satisfies the restriction).
     
  7. Oct 1, 2009 #6
    Thanks dick, sorry you wasted your 12k post on me :/
     
  8. Oct 1, 2009 #7

    Dick

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    It's not a waste, if you understood what I said.
     
  9. Oct 27, 2009 #8
    can someone please solve this using sequences of events
     
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