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anonymous
 3 years ago
tough probability problem
anonymous
 3 years ago
tough probability problem

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0why is it (1+p)^(mr) instead of (p)^(mr)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@zarkon, when you come on can you take a look?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0whew i can explain that one (and not embarrass myself) because it is algebra

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0the probability that you decide to buy a ticket and it loses is \(\frac{1p}{2}\) viewing this as bernoulli trials (independent repeated experiments with only two outcome) the the formula is \[P(x=k)=\dbinom{n}{k}p^k(1p)^{nk}\] but in this case \[p=\frac{1p}{2}\] and therefore \[1p=1\frac{1p}{2}=\frac{1+p}{2}\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i guess what i wrote is silly, i should have made the first \(p\) a \(p^*\) or some other variable

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I see, all those 'p' were confusing me
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