elemprob-fall2010-page15

elemprob-fall2010-page15 - E X = np . An easier way is to...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
5 Binomial distributions A random variable X has a binomial distribution with parameters n and p if P ( X = k ) = ± n k ² p k (1 - p ) n - k , k = 0 , 1 ,...,n. The number of successes in n trials is a binomial. Suppose one performs an experiment (trial) n times, and there is probability p of success and probabil- ity 1 - p of failure. The probability of any particular sequence of 4 successes out of 7 trials, such as SSFSSFF , would be p 4 (1 - p ) 3 . This would be the same as the probability of FFFSSSS and SSSFFFS . So the total probbility of 4 successes would be this number times the number of ways we can choose 4 trials out of 7 in which there were successes. This is ± 7 4 ² . In general, the probability of k successes in n trials is given by the formula above. After some cumbersome calculations one can derive
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: E X = np . An easier way is to realize that if X is binomial, then X = Y 1 + + Y n , where the Y i are independent random variables with P ( Y i = 1) = p and P ( Y i = 0) = 1-p , so E X = E Y 1 + + E Y n = np . The cumbersome way is as follows. E X = n X k =0 k n k p k (1-p ) n-k = n X k =1 k n k p k (1-p ) n-k = n X k =1 k n ! k !( n-k )! p k (1-p ) n-k = np n X k =1 ( n-1)! ( k-1)!(( n-1)-( k-1))! p k-1 (1-p ) ( n-1)-( k-1) = np n-1 X k =0 ( n-1)! k !(( n-1)-k )! p k (1-p ) ( n-1)-k = np n-1 X k =0 n-1 k p k (1-p ) ( n-1)-k = np. 15...
View Full Document

Ask a homework question - tutors are online