Answer:
The reuired probability is 0.756
Step-by-step explanation:
Let the number of trucks be 'N'
1) Trucks on interstate highway N'= 76% of N =0.76N
2) Truck on intra-state highway N''= 24% of N = 0.24N
i) Number of trucks flagged on intrastate highway = 3.4% of N'' = [tex]\frac{3.4}{100}\times 0.24N=0.00816N[/tex]
ii) Number of trucks flagged on interstate highway = 0.7% of N' = [tex]\frac{0.7}{100}\times 0.76N=0.00532N[/tex]
Part a)
The probability that the truck is an interstate truck and is not flagged for safety is [tex]P(E)=P_{1}\times (1-P_{2})[/tex]
where
[tex]P_{1}[/tex] is the probability that the truck chosen is on interstate
[tex]P_{2}[/tex] is the probability that the truck chosen on interstate is flagged
[tex]\therefore P(E)=0.76\times (1-0.00532)=0.756[/tex]
