Question 1095502
Probability that one or zero are essay and subtract from 1.
Denominator is 64C14
Numerator is 10C0*54C14
That is 0.0678
Do for 1 which has 14 different ways of occurring
14*10C1*54C13 over the same denominator=0.3242
That sum is 0.3920
The complement is 0.6080, and that is the probability that two or more are essay.
This makes intuitive sense, since the expected value is just under 1/6 probability that the first one will be essay and with 14 problems, one would expect about 2 and a little more to be the most common value.