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
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