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Tomas.A

  • 2 years ago

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?

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  1. satellite73
    • 2 years ago
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    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
    • 2 years ago
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    sample space is now HHH HHT HT

  3. Sry
    • 2 years ago
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    Nice @ satellite 73

  4. satellite73
    • 2 years ago
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    in two out of the 3 cases it is thrown three times, unless i misinterpreted the question

  5. satellite73
    • 2 years ago
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    ty

  6. Tomas.A
    • 2 years ago
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    given answer is actually 1/2

  7. satellite73
    • 2 years ago
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    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
    • 2 years ago
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    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
    • 2 years ago
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    no \[P(A\cap B)\neq \frac{1}{2}\]

  10. satellite73
    • 2 years ago
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    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
    • 2 years ago
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    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
    • 2 years ago
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    {T, HT, HHT, HHH} only 4 possiblities

  13. satellite73
    • 2 years ago
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    exactly what you wrote yes

  14. satellite73
    • 2 years ago
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    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
    • 2 years ago
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    \[\frac{3}{4}\div \frac{1}{2}>1\]

  16. satellite73
    • 2 years ago
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    well i made a mistake of course, (not in the answer, in the computation)

  17. Tomas.A
    • 2 years ago
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    P(A) is correct, how to get P(A∩B)?

  18. Tomas.A
    • 2 years ago
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    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
    • 2 years ago
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    ok i see the problem

  20. satellite73
    • 2 years ago
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    i am sorry i am an idiot this morning, so lets go slow

  21. satellite73
    • 2 years ago
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    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
    • 2 years ago
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    that is, each does not occur with probability 1/4

  23. satellite73
    • 2 years ago
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    \[P(T)=\frac{1}{2}, P(HT)=\frac{1}{4},P(HHH)=\frac{1}{8},P(HHH)=\frac{1}{8}\]

  24. Tomas.A
    • 2 years ago
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    ah, yeah..

  25. Tomas.A
    • 2 years ago
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    ok now i understood lol

  26. satellite73
    • 2 years ago
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    so i was completely wrong, and i apologize. should have a nice cup of coffee

  27. satellite73
    • 2 years ago
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    ok so now let us finish the problem.

  28. Tomas.A
    • 2 years ago
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    so \[ P(A \cap B)=\frac{1}{8}+\frac{1}{8}=\frac{1}{4}\]

  29. phi
    • 2 years ago
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    |dw:1326555438102:dw|

  30. satellite73
    • 2 years ago
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    we know \[A\cap B = \{HHT, HHH\}\] and so what you said, \[P(A\cap B)=\frac{1}{4}\]

  31. satellite73
    • 2 years ago
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    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
    • 2 years ago
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    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
    • 2 years ago
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    sorry again for the error and taking up so much time being wrong

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