Question 1153172
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Use the binomial probability distribution to solve this problem.



Probability  P(k successes in n trials) = {{{C[n]^k * p^k * (1-p)^(n-k)}}}, where {{{C[n]^k}}} = {{{n! / (k! * (n-k)!)}}}.


For this problem n = 4, k = 2, p = 0.52.


P ( 2 of 4 orders are "to go" ) = {{{C[4]^2 * 0.52^2 * (1 - 0.52)^(4-2)}}} = {{{((4*3)/(1*2)) * 0.52^2 * 0.48^2}}} = 0.3738 = 37.38%.    <U>ANSWER</U>
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