Twenty percent of drivers driving between 10 pm and 3 am are drunken drivers. In a random sample of 12 drivers driving between 10pm and 3 am, find the probability that: a) Exactly two will be drunken drivers. b) Three or four will be drunken drivers. c) At least 7 will be drunken drivers. d) At most 5 will be drunken drivers.

Answer:(a) 0.28347(b) 0.36909(c) 0.0039(d) 0.9806Step-by-step explanation:Given information:n=12p = 20% = 0.2q = 1-p = 1-0.2 = 0.8Binomial formula:[tex]P(x=r)=^nC_rp^rq^{n-r}[/tex](a) Exactly two will be drunken drivers.[tex]P(x=2)=^{12}C_{2}(0.2)^{2}(0.8)^{12-2}[/tex][tex]P(x=2)=66(0.2)^{2}(0.8)^{10}[/tex][tex]P(x=2)=\approx 0.28347[/tex]Therefore, the probability that exactly two will be drunken drivers is 0.28347.(b)Three or four will be drunken drivers.[tex]P(x=3\text{ or }x=4)=P(x=3)\cup P(x=4)[/tex][tex]P(x=3\text{ or }x=4)=P(x=3)+P(x=4)[/tex]Using binomial we get[tex]P(x=3\text{ or }x=4)=^{12}C_{3}(0.2)^{3}(0.8)^{12-3}+^{12}C_{4}(0.2)^{4}(0.8)^{12-4}[/tex][tex]P(x=3\text{ or }x=4)=0.236223+0.132876[/tex][tex]P(x=3\text{ or }x=4)\approx 0.369099[/tex]Therefore, the probability that three or four will be drunken drivers is 0.3691.(c)At least 7 will be drunken drivers.[tex]P(x\geq 7)=1-P(x<7)[/tex][tex]P(x\leq 7)=1-[P(x=0)+P(x=1)+P(x=2)+P(x=3)+P(x=4)+P(x=5)+P(x=6)][/tex][tex]P(x\leq 7)=1-[0.06872+0.20616+0.28347+0.23622+0.13288+0.05315+0.0155][/tex][tex]P(x\leq 7)=1-[0.9961][/tex][tex]P(x\leq 7)=0.0039[/tex]Therefore, the probability of at least 7 will be drunken drivers is 0.0039.(d) At most 5 will be drunken drivers.[tex]P(x\leq 5)=P(x=0)+P(x=1)+P(x=2)+P(x=3)+P(x=4)+P(x=5)[/tex][tex]P(x\leq 5)=0.06872+0.20616+0.28347+0.23622+0.13288+0.05315[/tex][tex]P(x\leq 5)=0.9806[/tex]Therefore, the probability of at most 5 will be drunken drivers is 0.9806.