Question 627278
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Hi,
Note: The probability of x successes in n trials is: 
P = nCx* {{{p^x*q^(n-x)}}} where p and q are the probabilities of success and failure respectively. 
In this case p & q are = 1/2 or .5
nCx = {{{n!/(x!(n-x)!)}}}
 (a) Exactly five of the patients will return? 9C5(.5)^5(.5)^4 
 (b) All nine will return? 9C9(.5)^9
 (c) At least eight will return? 1 - 9C9(.5)^9
 (d) At least one will return? 9C0(.5)^0(.5^9) + 9C1(.5)^1(.5)^8
 (e) How many patients would be expected to return to the same dentist, i.e., what is the mean of the distribution? {{{.5*9 = 4.5}}}