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pakinam

I throw darts repeatedly. Assume that on each single throw, my chance of hitting the bullseye is 10%, independently of all other throws. I decide to throw until I have hit the bullseye 3 times. What is the chance that I throw exactly 30 times?

  • 11 months ago
  • 11 months ago

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  1. reemii
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    you have this info: 30 throws, last throw is a BE (bullseye). P(BE)=0.1. But you don't know where the other two BE's are. So it's like that: (29 hits with 2BE's) and 1BE. Does this help?

    • 11 months ago
  2. pakinam
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    ahh ok

    • 11 months ago
  3. reemii
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    you have to use a combintation "C(n,k)" factor in the computation of the probability of (29 throws with 2 BE's), then multiply by P(1BE).

    • 11 months ago
  4. sarah10
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    is the ans 40.6

    • 11 months ago
  5. reemii
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    No, the answer must be a number of [0,1]. but it should be \(P(\text{2 BE's in 29 throws})\times P(\text{1 BE}) = \binom{29}{2}0.1^{2}\,0.9^{27} \times 0.1\). I got 0.02360879322323428... as anwer.

    • 11 months ago
  6. sarah10
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    is it the right ans ..... actually in frst attempt i got the the wrong ans

    • 11 months ago
  7. reemii
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    it's right.

    • 11 months ago
  8. sarah10
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    thank you soooo much 4 helping me

    • 11 months ago
  9. reemii
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    yw

    • 11 months ago
  10. pakinam
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    0.0236 is wrong

    • 11 months ago
  11. reemii
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    what's your answer ?

    • 11 months ago
  12. pakinam
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    i got wrong when i entered this answer

    • 11 months ago
  13. Khlara
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    @pakinam i gt it correct

    • 11 months ago
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