Answer:
The probability that these four will contain at least two more high-risk drivers than low-risk drivers is 0.0488.
Step-by-step explanation:
Denote the different kinds of drivers as follows:
L = low-risk drivers
M = moderate-risk drivers
H = high-risk drivers
The information provided is:
P (L) = 0.50
P (M) = 0.30
P (H) = 0.20
Now, it given that the insurance company writes four new policies for adults earning their first driver’s license.
The combination of 4 new drivers that satisfy the condition that there are at least two more high-risk drivers than low-risk drivers is:
S = {HHHH, HHHL, HHHM, HHMM}
Compute the probability of the combination {HHHH} as follows:
P (HHHH) = [P (H)]⁴
                 = [0.20]⁴
                 = 0.0016
Compute the probability of the combination {HHHL} as follows:
P (HHHL) =  × [P (H)]³ × P (L)
 × [P (H)]³ × P (L)
                = 4 × (0.20)³ × 0.50
                = 0.016
Compute the probability of the combination {HHHM} as follows:
P (HHHL) =  × [P (H)]³ × P (M)
 × [P (H)]³ × P (M)
                = 4 × (0.20)³ × 0.30
                = 0.0096
Compute the probability of the combination {HHMM} as follows:
P (HHMM) =  × [P (H)]² × [P (M)]²
 × [P (H)]² × [P (M)]²
                  = 6 × (0.20)² × (0.30)²
                  = 0.0216
Then the probability that these four will contain at least two more high-risk drivers than low-risk drivers is:
P (at least two more H than L) = P (HHHH) + P (HHHL) + P (HHHM)
                                                             + P (HHMM)
                                                   = 0.0016 + 0.016 + 0.0096 + 0.0216
                                                   = 0.0488
Thus, the probability that these four will contain at least two more high-risk drivers than low-risk drivers is 0.0488.