Answer:
a) P=0.3174
b) P=0.4232
c) P=0.2594
d) The shape of the hypergeometric, in this case, is like a binomial with mean np=1.
Step-by-step explanation:
The appropiate distribution to model this is the hypergeometric distribution:
[tex]P(X=x)=\frac{\binom{s}{x}\binom{N-s}{M-x}}{\binom{N}{M}}=\frac{\binom{6}{x}\binom{54}{10-x}}{\binom{60}{10}}[/tex]
a) What is the probability that none of the questions are essay?
[tex]P(X=0)=\frac{\binom{6}{0}\binom{54}{10-0}}{\binom{60}{10}}\\\\P(X=0)=\frac{1*(54!/(10!*44!)}{60!/(10!*50!)} =\frac{2.3931*10^{10}}{7.5394*10^{10}} = 0.3174[/tex]
b) What is the probability that at least one is essay?
[tex]P(X=1)=\frac{\binom{6}{1}\binom{54}{9}}{\binom{60}{10}}\\\\P(X=1)=\frac{6*(54!/(9!*43!)}{60!/(10!*50!)} =\frac{3.1908*10^{10}}{7.5394*10^{10}} =0.4232[/tex]
c) What is the probability that two or more are essay?
[tex]P(X\geq2)=1-(P(0)+P(1))=1-(0.3174+0.4232)=1-0.7406=0.2594[/tex]
