7 in 10 auto accidents involve a single vehicle. Suppose 14 accidents are randomly selected. (Round your answers to five decimal places.) a. What is the probability that exactly four involve a single vehicle? b. What is the probability that at most four involve a single vehicle? c. What is the probability that exactly five involve multiple vehicles?

Step-by-step explanation:

Assuming a binomial distribution for the number of auto accidents. If 7 in 10 auto accidents involve a single vehicle, the probability, p = 7/10 = 0.7

Then the probability that the accident involved multiple vehicles is

q = 1 - p = 1 - 7/10 = 3/10 = 0.3

Since 14 accidents are randomly selected, n = 14

The formula for binomial distribution is expressed as

P (x = r) = nCr * q^ (n - r) * p^r

a) we want to determine P (x = 4) =

P (x = 4) = 14C4 * 0.3^ (14 - 4) * 0.7^4

P (x = 4) = 0.00142

b) we want to find P (x lesser than or equal to 4) = P (x = 0) + P (x = 1) + P (x = 2) + P (x = 3) + P (x = 4)

P (x = 0) = 14C0 * 0.3^ (14 - 0) * 0.7^0 = 0.00000004783

P (x = 1) = 14C1 * 0.3^ (14 - 1) * 0.7^1 = 0.00011

P (x = 2) = 14C2 * 0.3^ (14 - 2) * 0.7^2 = 0.00002

P (x = 3) = 14C3 * 0.3^ (14 - 3) * 0.7^3 = 0.00022

P (x = 4) = 0.00142

P (x lesser than or equal to 4) = 0.00000004783 + 0.00011 + 0.00002 + 0.00022 + 0.00142 = 0.00177

c) p = 0.7, q = 0.3

P (x = 5) = 14C5 * 0.7^ (14 - 5) * 0.3^5 = 0.19631