drivers have no prior driving record, an insurance company considers each driver to be randomly selected from the pool. This month, the insurance company writes 4 new policies for adults earning their first drivers license. What is the probability that these 4 will contain at least two more high-risk drivers than low-risk drivers

Complete question is:

A large pool of adults earning their first drivers license includes 50% low- risk drivers, 30% moderate-risk drivers, and 20% high-risk drivers. Because these drivers have no prior driving record, an insurance company considers each driver to be randomly selected from the pool. This month, the insurance company writes 4 new policies for adults earning their first drivers license. What is the probability that these 4 will contain at least two more high-risk drivers than low-risk drivers?

probability that these 4 will contain at least two more high-risk drivers than low-risk drivers = 0.0488

Step-by-step explanation:

Let H represent High risk

M represent moderate risk

L represent Low risk.

The following combinations will satisfy the condition that there are at least two more high-risk drivers than low-risk drivers: , HHHL, HHHM, HHMM

The case has probability 0.2 ⁴ = 0.0016

The HHHL case has probability 4 × 0.2³ × 0.3 = 0.0096 (This is because L can be in four different places)

Similarly, the HHHM case has probability 4 × 0.2 ³ × 0.5 = 0.016

Lastly, the HHMM case has probability 6 × 0.2 ² × 0.3 ² = 0.0216 (This is because the number of ways to choose places for two M letters in this way is 6)

Summing all these probabilities, we have;

0.0016 + 0.0096 + 0.016 + 0.0216 = 0.0488

the equation cos(35) =a/25 can be used to find the length of bc what is the length of bc? round to the nearest tenth

step-by-step explanation:

ou would have 12 of each