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order Group Title

A coin is tossed 8 times. Calculate the probability that the first 4 tosses and the last 4 tosses result in the same number of heads.

  • 2 years ago
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  1. lilg132 Group Title
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    i hate these coin questions :(

    • 2 years ago
  2. order Group Title
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    The answer is 0.273... so not 15/64... but I'm not sure how to arrive at the answer.

    • 2 years ago
  3. henkjan Group Title
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    0 heads: 0.5^8 1 head in each: 0.5^8*16 2 heads in each: 0.5^8*36 3 heads in each: 0.5^8*16 4 heads: 0.5 ^8

    • 2 years ago
  4. henkjan Group Title
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    which totals to 0.273, indeed

    • 2 years ago
  5. nitishanand Group Title
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    okay, what i punched in was the number of ways, you can get heads in the first four tosses, my bad

    • 2 years ago
  6. nitishanand Group Title
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    and i got the sample space wrong too, pellete

    • 2 years ago
  7. amistre64 Group Title
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    0 1 2 3 4 5 6 7 8 1 8 28 56 70 56 28 8 1 if my pascal is right ... 70 * .5^4 * .5^4 = 35/128 = .2734

    • 2 years ago
  8. order Group Title
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    henkjan, That doesn't add up to the answer... But for all those who want to know how to get it, I finally solved it on my own. First, by getting the binomial distributions and then timesing each distribution by 2, and then adding them together :) Also, Amistre is right.

    • 2 years ago
  9. nitishanand Group Title
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    thanks :)

    • 2 years ago
  10. henkjan Group Title
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    it does add up to the answer... I think you pressed the wrong button on your calculator.. total = 0.5^8 *(1+16+36+16+1) = 0.5^8*70 = 0.273

    • 2 years ago
  11. amistre64 Group Title
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    just to practice the latex :) \[{n \choose r}\ p^r\ q^r\] \[{8 \choose 4}\ .5^4\ .5^4\]

    • 2 years ago
  12. order Group Title
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    Hmmm. I'll have to double check that! Sorry..

    • 2 years ago
  13. order Group Title
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    It still doesn't add up...

    • 2 years ago
  14. henkjan Group Title
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    ok, that's strange, it works here... I don't know how to help you further than..

    • 2 years ago
  15. order Group Title
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    It's Ok, I already have the answer :) Amistre has the correct answer, and I eventually found it myself. But thanks for trying anyway...

    • 2 years ago
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