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anonymous
 2 years 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?
anonymous
 2 years 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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anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0Binomial distribution. Do you know it?

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0No, can you help me solve this?

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0You've never even heard of binomial distribution?

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0Yes I've heard of it but It's not clear to me.

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0Well. To get \(k\) success with \(n\) tries and each try has probability \(p\), we say the probability is: \[ {n\choose k}p^k(1p)^{nk} \]

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0Will 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
 2 years ago
Best ResponseYou've already chosen the best response.0Can you try to figure out which number goes where?

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0not really, I really don't get it..

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.00.42(10.42)^4^2 This doesn't even look correct. Is there a format that I should know?

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0In this case, there are 4 tries, so \(n=4\).

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0So we have, so far \[ {4\choose k}p^k(1p)^{4k} \]

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0The probability of success is given by batting average, so \(p=0.420\) and \(1p=0.580\). Now we have: \[ {4\choose k}(0.420)^k(0.580)^{4k} \]

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0At least two hits... that mean 2 hits, 3 hits, or 4 hits.

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0So then k is the # of hits then correct?

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0Then I would replace the k with 2 to get the correct format?

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0no, you need to find probabilities for 2, 3, and 4. Then combine them.
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