7in 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

a.   since the profit made in 2012 is equal to the profit made in 2014, the profit is symmetric about the line y = 120,000.

step-by-step explanation: