Question 609927
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Hi
Note: The probability of x succesP(0)ses 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= .70 & q = .30 and n = 13
nCx = {{{n!/(x!(n-x)!)}}}
P( making >5) = 1 - [P(0)+P(1)+  P(2) + P(3) + P(4)+ P(5)]
P = 1-({{{ (.7)^0*(.3)^(13) }}} {{{13(.7)^1*(.3)^(12) }}}+{{{78(.7)^2*(.3)^(11)}}}+{{{286(.7)^3*(.3)^(10)}}}+{{{715(.7)^4*(.3)^(9)}}}+{{{1287(.7)^5*(.3)^(8)}}})= .9960
Recommend using TI or stattrek.com for statistical calculations.
Above is an explanatory presenation.  Hope this helps.