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hwyl

  • one year ago

challenge! ready?

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  1. hwyl
    • one year ago
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  2. hwyl
    • one year ago
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    solve without using complement, that is \(\sf P(x) = 1 - P(x') \)

  3. anonymous
    • one year ago
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    1/5?

  4. hwyl
    • one year ago
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    @UsukiDoll

  5. anonymous
    • one year ago
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    Probability of at least 1 girl \[P(X \ge 1)=P(X=1)+P(X=2)+P(X=3)+P(X=4)+P(X=5)\]

  6. UsukiDoll
    • one year ago
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    I don't do stats problems.

  7. anonymous
    • one year ago
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    probability of 1 girl + probability of 2 girls + probability of 3 girls + probability of 4 girls + probability of 5 girls

  8. hwyl
    • one year ago
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    continue

  9. anonymous
    • one year ago
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    can i do it as well

  10. hwyl
    • one year ago
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    are you using \(\large \cup_{x=1} ^{n} X_i = P(x_1 \cup x_2 \cup x_3\cup x_4 \cup x_5) \)

  11. anonymous
    • one year ago
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    \[P(X \ge 1)=\frac{1}{32}+\frac{1}{16}+\frac{1}{8}+\frac{1}{4}+\frac{1}{2}=\frac{1+2+4+8+16}{32}=\frac{31}{32}\] another method \[P(X=0)=\frac{1}{32}\]\[P(X \ge 1)=1-P(X=0)=1-\frac{1}{32}=\frac{32-1}{32}=\frac{31}{32}\]

  12. hwyl
    • one year ago
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    haha wow

  13. hwyl
    • one year ago
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    i said without using complement

  14. anonymous
    • one year ago
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    I've done both with and without compliment, the compliment method is much faster anyway

  15. Michele_Laino
    • one year ago
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    another equivalent reasoning, can be this: the even space contains 32 possible events, nevertheless only one event, of them, is like this: BBBBB, where B stands for boy, so the total number of favorable events is 31

  16. anonymous
    • one year ago
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    I think you can also do this using binomial distribution where n=5 p=q=0.5

  17. anonymous
    • one year ago
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    why do u people make ur life so hard

  18. Michele_Laino
    • one year ago
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    oops..event space

  19. hwyl
    • one year ago
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    |dw:1438078796913:dw|

  20. hwyl
    • one year ago
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    |dw:1438079317682:dw|

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