If the 1500-lblb boom ABAB, the 220-lblb cage BCDBCD, and the 169-lblb man have centers of gravity located at points G1G1, G2G2
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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)
= 4 × (0.20)³ × 0.50
= 0.016
Compute the probability of the combination {HHHM} as follows:
P (HHHL) = × [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
= 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.
The required value of x is 4.
