Question 1094890
<br>This is a straightforward application of the binomial formula.  If the probability is .2 that each customer is willing to switch and .8 that they are not, then the probability that n of the 20 customers will be willing to switch is
{{{C(20,n)*(.8^n)*(.2^(20-n))}}}
where "C(20,n)" is "20 choose n".<br>
Since you want the probability that no more than 3 of the 20 customers will be willing to change plans, the calculations you need to perform are...<br>
{{{C(20,0)*(.8^20)*(.2^0)}}}  [0 willing to switch];
{{{C(20,1)*(.8^19)*(.2^1)}}}  [1 willing to switch];
{{{C(20,2)*(.8^18)*(.2^2)}}}  [2 willing to switch]; and
{{{C(20,3)*(.8^17)*(.2^3)}}}  [3 willing to switch]<br>
Then of course you need to add the probabilities for those 4 cases.<br>
The calculations are tedious with pencil and paper, and even with a scientific calculator; a spread sheet works nicely.<br>
And .411 is the right answer....