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anonymous

  • 5 years ago

What is the probability of obtaining exactly 4 successes in 5 trials when the probability of success is 0.5. Round your answer to 4 decimal places.

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  1. anonymous
    • 5 years ago
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    i'm not so sure but i think the probability of success is 0.5, then the probability of not success is 1-0.5=0.5 obtaining 4 successes in 5 trials = 4 successes and 1 not success so, the probability of it is 0.5^4 x 0.5 = 0.5^5 = 1/32

  2. anonymous
    • 5 years ago
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    1/32 = 0.0312

  3. GG
    • 5 years ago
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    I think you are right. Those happenings are independent, so we just multiply probabilities of each.

  4. anonymous
    • 5 years ago
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    thats actually not the answer its (5)(0.5)^4(0.5)^5-4=0.1563

  5. GG
    • 5 years ago
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    5 times (0.5^4) times (0.5^5) -4 ?

  6. anonymous
    • 5 years ago
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    hmm maybe because there are 5 trials, the 1/32 that i got before should be multiplied by 5, therefore the answer is the same as yours ^_^

  7. anonymous
    • 5 years ago
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    1/32 x 5 = 0.03125 x 5 = 0.15625

  8. anonymous
    • 5 years ago
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    yes

  9. GG
    • 5 years ago
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    but why to multiply it by 5?

  10. anonymous
    • 5 years ago
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    i'm also confused, but i think it's because there are 5 trials

  11. GG
    • 5 years ago
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    but if we have 100 trials, than the possibility would be 3.1, and that is impossible, right? so, that is not the reason :)

  12. anonymous
    • 5 years ago
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    i figured it out by the book saying this Let T denote the outcome “Tail shows” (success) and let H denote the outcome “Head shows” (failure). Using Formula (2), in which k = 1 (one success), n = 6 (the number of trials), and (the probability of success using a fair coin), we obtain

  13. GG
    • 5 years ago
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    P= (n!/[k!(n-k)!] ) p^k * q^(n-k) use this formula. n is number of trials, k is number of successes, p is possibility of success, q=1-p :) is it ok now? :)

  14. anonymous
    • 5 years ago
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    oh yes, combination. 5C4 0.5^4 * 0.5. thank you

  15. GG
    • 5 years ago
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    problem solved :)

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