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) = 0.6526

Step-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^r

x = number of drivers

p = 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 = 0.2

q = 1 - p = 1 - 0.2 = 0.8

n = 12

We 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 = 0.2835

p (x=3) = 12C3 * 0.8^9 * 0.2^3 = 0.2362

p (x=4) = 12C4 * 0.8^8 * 0.2^4 = 0.1329

p (2 lesser than or equal to x lesser than or equal to 4) = 0.2835 + 0.2362 + 0.1329 = 0.6526