
Patrick B. answered 10/20/20
Math and computer tutor/teacher
Here is the binomial table with N=4 trials, probability of success p=0.13;
k N-k N choose K p^k q^(N-k) prob cumulative prob.
0 4 1 1 0.57289761 0.57289761 0.57289761
1 3 4 0.13 0.658503 0.34242155999999996 0.9153191699999998
2 2 6 0.0169 0.7569 0.07674966000000001 0.9920688299999999
3 1 4 0.002197 0.87 0.007645560000000001 0.99971439
4 0 1 0.00028561 1 0.00028561 1
So the probability of at least two is Prob(x>=2) = 1 -Prob(x<2) = 1 - (Prob(0) + Prob(1))
=1 - 0.9153197
= 0.0846803
Note that this answer agrees with Prob(x=2)+Prob(X=3)+Prob(X=4)
Final answer 8.46803+ %