anonymous
  • anonymous
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?
Probability
schrodinger
  • schrodinger
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anonymous
  • anonymous
Binomial distribution. Do you know it?
anonymous
  • anonymous
No, can you help me solve this?
anonymous
  • anonymous
You've never even heard of binomial distribution?

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anonymous
  • anonymous
Yes I've heard of it but It's not clear to me.
anonymous
  • anonymous
Well. To get \(k\) success with \(n\) tries and each try has probability \(p\), we say the probability is: \[ {n\choose k}p^k(1-p)^{n-k} \]
anonymous
  • anonymous
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.
anonymous
  • anonymous
Can you try to figure out which number goes where?
anonymous
  • anonymous
not really, I really don't get it..
anonymous
  • anonymous
0.42(1-0.42)^4-^2 This doesn't even look correct. Is there a format that I should know?
anonymous
  • anonymous
In this case, there are 4 tries, so \(n=4\).
anonymous
  • anonymous
So we have, so far \[ {4\choose k}p^k(1-p)^{4-k} \]
anonymous
  • anonymous
The probability of success is given by batting average, so \(p=0.420\) and \(1-p=0.580\). Now we have: \[ {4\choose k}(0.420)^k(0.580)^{4-k} \]
anonymous
  • anonymous
At least two hits... that mean 2 hits, 3 hits, or 4 hits.
anonymous
  • anonymous
So then k is the # of hits then correct?
anonymous
  • anonymous
yes
anonymous
  • anonymous
Then I would replace the k with 2 to get the correct format?
anonymous
  • anonymous
no, you need to find probabilities for 2, 3, and 4. Then combine them.
anonymous
  • anonymous
okay

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