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<math> q=(n-k)/n</math>
 
<math> q=(n-k)/n</math>
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Correction:  What you did was find the pr[T} which was 6/10. So the pr[H] = 1 - pr[T] = 4/10
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Jonathan Morales

Revision as of 16:04, 3 November 2008

it probably will be obvious for most of you guys,but just in case.... for part a), after we take the drivitive we get something like:

$ 4*q^3*(1-q)^6-6*(1-q)^5*q^4=0 $

you may try to solve it this way: divide both side by q^3 and (1-q)^5, thus we get:

$ 4*(1-q)-6*q=0 $

then we can easily find q=6/10

for part two we can divide both side with:

$ (1-q)^{k-1}*q^{n-k-1} $ then we get:

$ q=(n-k)/n $

Correction: What you did was find the pr[T} which was 6/10. So the pr[H] = 1 - pr[T] = 4/10 Jonathan Morales

Alumni Liaison

Abstract algebra continues the conceptual developments of linear algebra, on an even grander scale.

Dr. Paul Garrett