g Twenty percent of drivers driving between 11 PM and 3 AM are drunken drivers. Using the binomial probability formula, find the probability that in a random sample of 12 drivers driving between ll PM and 3 AM, two to four will be drunken drivers. (Round to 4 digits, ex. 0.1234)

Answer: p(2 lesser than or equal to x lesser than or equal to 4) = 0.6526Step-by-step explanation:20% of drivers driving between 11 PM and 3 AM are drunken drivers. We want to use the binomial distribution to determine the probability that in a random sample of 12 drivers driving between 11 PM and 3 AM, two to four will be drunken drivers. The formula for binomial distribution is P( x = r) = nCr × q^n-r × p^rx = number of driversp = probability that the drivers that are drunken.q= 1-p = probability that the drivers are not drunken.n = number of sampled drivers.From the information given,p = 20/100 = 0.2q = 1 - p = 1 - 0.2 = 0.8n = 12We want to determine p(2 lesser than or equal to x lesser than or equal to 4)It is equal to p(x=2) + p(x= 3) + p(x=4)p(x=2) = 12C2 × 0.8^10 × 0.2^2 = 0.2835p(x=3) = 12C3 × 0.8^9 × 0.2^3 = 0.2362p(x=4) = 12C4 × 0.8^8 × 0.2^4 = 0.1329p(2 lesser than or equal to x lesser than or equal to 4) = 0.2835 + 0.2362 + 0.1329 = 0.6526