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Unformatted text preview: Jose: and thats my answer? Megan Goldman: Yup. You'll need to poke it into the calculator to put it in nice decimal form, of course. Jose: yeah Jose: so Jose: The sum of the number of heads in all the tosses of the coin, plus the total number of times the die lands with an even number of spots showing on top (Q5) Megan Goldman: What do you think? Jose: (.5) change of heads Jose: 18 possibilities of even numbers Megan Goldman: Mmm, you're overthinking that part just a tad. =) Megan Goldman: The three die rolls are independent, so P(even) = .5 for each die. Megan Goldman: Melinda: What was your ch 11 problem? Megan Goldman: Jose: Welcome back. =) melinda: it was about q68 about the genetic problem in the populations Megan Goldman: Jose: Consider all the events have probability .5. See what you can sort out from there. Jose 2: my internet reloaded, sorry Megan Goldman: Melinda: Okay, how far have you gotten? melinda: i am having trouble starting the answer...but i think that it is binomial distribution Megan Goldman: Why do you think binomial is right? melinda: well because you are choosing ppl with the genetic marker so that could be like 1 adn 0 would be those without it right? Megan Goldman: Yeah. There are a few requirements for binomial: A fixed number of independent trials (We're told there's a certain sample size, and each person is independent from the others). Megan Goldman: You also need the same probability of success on every trial. Megan Goldman: We're all good on that so yes, the binomial disitribution is appropriate. melinda: right so would you do 50 C2 if it asks The chance that the sample would contain exactly two people with the genetic marker if the new theory is correct is Megan Goldman: Well, that would be looking for exactly 2 successes out of 50 trials. Megan Goldman: The problem has two probabilities: one "old" and one "new"... make sure you're using the right one in the problem. (You'll use each of them once, the first two questions of that section.) Megan Goldman: So, you'll use the binomial formula with n = 50, k = 2, and p = whatever your problem says....
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This note was uploaded on 02/12/2010 for the course STAT 21 taught by Professor Anderes during the Summer '08 term at University of California, Berkeley.
 Summer '08
 anderes
 Statistics, Probability

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