Question 1087906
<font color="black" face="times" size="3">Given Information:


n = 5 (sample size), p = 0.65 (probability of success), k = 3 (number of successes; I'm using k in place of x)


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Use the combination formula with n = 5 and k = 3


n C k = (n!)/(k!*(n-k)!)
5 C 3 = (5!)/(3!*(5-3)!)
5 C 3 = (5!)/(3!*2!)
5 C 3 = (5*4*3!)/(3!*2!)
5 C 3 = (5*4)/(2!)
5 C 3 = (5*4)/(2*1)
5 C 3 = 20/2
5 C 3 = 10


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With that combination in mind, we can find the following binomial probability


P(X = k) = (n C k)*(p)^(k)*(1-p)^(n-k)
P(X = 3) = (5 C 3)*(0.65)^(3)*(1-0.65)^(5-3)
P(X = 3) = (5 C 3)*(0.65)^(3)*(0.35)^(2)
P(X = 3) = (10)*(0.65)^(3)*(0.35)^2
P(X = 3) = (10)*(0.274625)*(0.1225)
P(X = 3) = 0.336415625
P(X = 3) = 0.34 (round to the nearest hundredth)
P(X = 3) = <font color=red size=4>34%</font> which is the final answer. You move the decimal point 2 spots to the right to go from 0.34 to 34%</font>