Question

Win-Win Consult ltd. bids for 10 jobs, believing that its chances of getting each of the jobs is 0.1.
i) What is the probability that it gets at least two of the jobs?
ii) What is the expected value of the number of jobs it will get?
iii) What is the standard deviation of the number of jobs it will get?

Answer 1

$n=10,p=0.1,(1-p)=0.9\\ i)\ p(x\geq 2)=1-p(x<2)\\ p(x\geq 2)=1-[p(x=0)+p(x=1)]\\ 1-[\binom{10}{0}(0.1)^{0}(0.9)^{10}+\binom{10}{1}(0.1)^{0}(0.9)^{9}]\\ 1-[0.3487+0.3874]\\ 1-0.7361\\ 0.2639\\ ii)\ E(x)=np=10*0.1=1\\ Expected\ value\ is\ 1\\ iii)\ Var(x)=npq\\ standard\ dev. =\sqrt{npq}=\sqrt{10 \times 0.1 \times0.9}\\ 0.9487\\ standard\ deviation\ is\ 0.9487$