m (2b AJ Hartnett moved to 4.2b AJ Hartnett: Improperly named originally) |
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P(x = 3) = (n-1)/n * (n-2)/(n-1) * 1/n | P(x = 3) = (n-1)/n * (n-2)/(n-1) * 1/n | ||
| − | P(x = k) = (n-1)/n * ... * (n - (k-1))/(n-k) * 1/(n-(k-1)) | + | P(x = k) = (n-1)/n * ... * (n - (k-1))/(n-(k-2)) * 1/(n-(k-1)) |
= [(n-1)*...*(n-k-1)] / [n*...*(n-k-1)] | = [(n-1)*...*(n-k-1)] / [n*...*(n-k-1)] | ||
= [(n-1)!/(n-k)!]/[n!/(n-k)!] | = [(n-1)!/(n-k)!]/[n!/(n-k)!] | ||
Latest revision as of 07:47, 15 October 2008
An important part to figuring out part b is to understand why P(x=k) is 1/n:
P(x = 1) = 1/n
P(x = 2) = (n-1)/n * 1/(n-1)
P(x = 3) = (n-1)/n * (n-2)/(n-1) * 1/n
P(x = k) = (n-1)/n * ... * (n - (k-1))/(n-(k-2)) * 1/(n-(k-1))
= [(n-1)*...*(n-k-1)] / [n*...*(n-k-1)]
= [(n-1)!/(n-k)!]/[n!/(n-k)!]
= [ (n-1)!/ (n!)]
= 1/n
You can then use this probability while solving the rest of 2b.
