Answer to Question #108226 in Statistics and Probability for mazen

Question

There are 20 boys and 10 girls in the class. Teacher selects 4 students and gives one question to
each of them. Let X is the number of questions given to the boys.
a) Find a probability density function f(x) and a cumulative distributional function F(x);
b) What is the mean and the variance?
c) What is the probability that at least 2 boys were selected to answer to the questions?

Answer 1

$a) pmf\\ f(x)=P\{X=x\}=\\ \frac{C_{10}^4}{C_{30}^4}=0.0077, x=0\\ \frac{C_{10}^3 C_{20}^1}{C_{30}^4}=0.0876, x=1\\ \frac{C_{10}^2 C_{20}^2}{C_{30}^4}=0.312, x=2\\ \frac{C_{10}^1 C_{20}^3}{C_{30}^4}=0.416, x=3\\ \frac{C_{20}^4}{C_{30}^4}=0.1768, x=4.$

$CDF\\ F(x)=P\{X\leq x\}=\\ 0, x<0\\ 0+0.0077=0.0077, 0\leq x<1\\ 0.0077+0.0876=0.0953, 1\leq x<2\\ 0.0953+0.312=0.4073, 2\leq x<3\\ 0.4073+0.416=0.8233, 3\leq x<4\\ 0.8233+0.1768\approx 1, x\geq 4.$

$b) MX=2.6668$ (sum of products of probabilities and values of random variable).

$DX=MX^2-(MX)^2\\ MX^2=0^2\cdot (0.0077)+1^2\cdot (0.0876)+2^2\cdot (0.312)+3^2\cdot (0.416)+\\+ 4^2\cdot(0.1768)=7.9084.\\ DX=7.9084-(2.6668)^2=0.7966.\\ c) P\{X\geq 2\}=1-P\{X<2\}=1-0.0953=0.9047.$