Question 416795
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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 = .5 and q = .5 
nCx = {{{n!/(x!(n-x)!)}}}
Assume sales calls are independent and the employee is doing 2 sales calls
P(at least one of the two sales calls is successful) = 1-P(none)= 1 - (.5)^2
P(at least one of the two sales calls is successful) = .75