n=5,p=0.5
(a) Find the probability that the flight will be overbooked.
"P(X=4)+P(X=5)=\n\\binom{5}{4}(0.5)^4(1-0.5)^{5-4}+\\binom{5}{5}(0.5)^5(1-0.5)^{5-5}=\n(5+1)(0.5)^5=0.1875"
(b) Find the probability that there will be empty seats.
"P(X=0)+P(X=1)+P(X=2)=\\binom{5}{0}(0.5)^0(1-0.5)^{5-0}+\\binom{5}{1}(0.5)^1(1-0.5)^{5-1}+\\binom{5}{2}(0.5)^2(1-0.5)^{5-2}=(1+5+10)(0.5)^5=0.5"
Comments
Leave a comment