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?
Answer:
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: HHHH, HHHL, HHHM, HHMM
The HHHH 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
=12