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wildmango

  • 9 months ago

If a baseball player has a batting average of 0.420, what is the probability that the player will get at least 2 hits in the next four times at bat?

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  1. wio
    • 9 months ago
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    Binomial distribution. Do you know it?

  2. wildmango
    • 9 months ago
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    No, can you help me solve this?

  3. wio
    • 9 months ago
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    You've never even heard of binomial distribution?

  4. wildmango
    • 9 months ago
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    Yes I've heard of it but It's not clear to me.

  5. wio
    • 9 months ago
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    Well. To get \(k\) success with \(n\) tries and each try has probability \(p\), we say the probability is: \[ {n\choose k}p^k(1-p)^{n-k} \]

  6. wio
    • 9 months ago
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    Will will bat 4 times. His probability of hitting it is 0.420. We want the probability he its it at least 2 times, meaning he hits it 2, 3, or 4 times.

  7. wio
    • 9 months ago
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    Can you try to figure out which number goes where?

  8. wildmango
    • 9 months ago
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    not really, I really don't get it..

  9. wildmango
    • 9 months ago
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    0.42(1-0.42)^4-^2 This doesn't even look correct. Is there a format that I should know?

  10. wio
    • 9 months ago
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    In this case, there are 4 tries, so \(n=4\).

  11. wio
    • 9 months ago
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    So we have, so far \[ {4\choose k}p^k(1-p)^{4-k} \]

  12. wio
    • 9 months ago
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    The probability of success is given by batting average, so \(p=0.420\) and \(1-p=0.580\). Now we have: \[ {4\choose k}(0.420)^k(0.580)^{4-k} \]

  13. wio
    • 9 months ago
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    At least two hits... that mean 2 hits, 3 hits, or 4 hits.

  14. wildmango
    • 9 months ago
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    So then k is the # of hits then correct?

  15. wio
    • 9 months ago
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    yes

  16. wildmango
    • 9 months ago
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    Then I would replace the k with 2 to get the correct format?

  17. wio
    • 9 months ago
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    no, you need to find probabilities for 2, 3, and 4. Then combine them.

  18. wildmango
    • 9 months ago
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    okay

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