Question 1087907
<font color="black" face="times" size="3">We have n = 11 as the sample size, p = 0.9 as the probability of success, and k = 8 as the number of successes (I'm using k instead of x)


--------------------------------------


Plug in n = 11 and k = 8 into the combination formula below


n C k = (n!)/(k!*(n-k)!)


11 C 8 = (11!)/(8!*(11-8)!)


11 C 8 = (11!)/(8!*3!)


11 C 8 = (11*10*9*8!)/(8!*3!)


11 C 8 = (11*10*9)/(3!)


11 C 8 = (11*10*9)/(3*2*1)


11 C 8 = 990/6


11 C 8 = 165


--------------------------------------


Using that combination value, we can say


P(X = k) = (n C k)*(p)^(k)*(1-p)^(n-k)


P(X = 8) = (11 C 8)*(0.9)^(8)*(1-0.9)^(11-8)


P(X = 8) = (11 C 8)*(0.9)^(8)*(0.1)^(3)


P(X = 8) = (165)*(0.9)^(8)*(0.1)^3


P(X = 8) = (165)*(0.43046721)*(0.001)


P(X = 8) = 0.07102708965


P(X = 8) = 0.07


P(X = 8) = <font color=red size=4>7%</font> (move the decimal 2 spots to the right to go from 0.07 to 7%)


This means that the answer is <font color=red size=4>Choice B</font></font>