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order

  • 3 years ago

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.

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  1. lilg132
    • 3 years ago
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    i hate these coin questions :(

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

  3. henkjan
    • 3 years ago
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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

  4. henkjan
    • 3 years ago
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    which totals to 0.273, indeed

  5. nitishanand
    • 3 years ago
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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

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

  7. amistre64
    • 3 years ago
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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

  8. order
    • 3 years ago
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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.

  9. nitishanand
    • 3 years ago
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    thanks :)

  10. henkjan
    • 3 years ago
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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

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

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

  13. order
    • 3 years ago
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    It still doesn't add up...

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

  15. order
    • 3 years ago
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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...

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