Solution:
Given:
![\begin{gathered} P(Car)=0.6 \\ P(Air)=0.4 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/vu71zgh31uiikg3k3cfrtps5ppmymrh4kf.png)
![\begin{gathered} P(Car\text{ and on time\rparen}=0.3 \\ P(air\text{ and on time\rparen}=0.65 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/d8kae0cdekafvoqnkc7vy2rf1usv35ll1d.png)
i) The probability that he arrives in Lagos early for the appointment is;
![\begin{gathered} P(early)=(0.3*0.6)+(0.65*0.4) \\ P(early)=0.18+0.26 \\ P(early)=0.44 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/wwvx823gxb2n844hypabnbhrh35l4iijlu.png)
Therefore, the probability that he arrives in Lagos early for the appointment is 0.44
ii) The probability that he went by car if he arrived late for his appointment is;
To get the probability, we use conditional probability.
![\begin{gathered} P(Car|Late)=\frac{P(car\text{ and late\rparen}}{P(Late)} \\ P(Car|Late)=\frac{P(car)* P(car\text{ and late\rparen}}{P(late)} \\ P(car\text{ and late\rparen}=1-0.3=0.7 \\ P(car)=0.6 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/3xbdiponp2qu0nt4qn3kkdm1blzwk2qv07.png)
We now get the probability of late;
![\begin{gathered} P(late)=(0.6*0.7)+(0.4*0.35) \\ P(late)=0.42+0.14 \\ P(late)=0.56 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/tsdprphzgjqkz70q4fm9b3rt4u6k7yoq0q.png)
Therefore, the probability that he went by car if he arrived late for his appointment is;
![\begin{gathered} P(Car|Late)=\frac{P(car\text{ and late\rparen}}{P(Late)} \\ P(Car|Late)=\frac{P(car)* P(car\text{ and late\rparen}}{P(late)} \\ P(car\text{ and late\rparen}=1-0.3=0.7 \\ P(car)=0.6 \\ P(Car|Late)=(0.6*0.7)/(0.56) \\ P(Car|Late)=(0.42)/(0.56) \\ P(Car|Late)=0.75 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/6onfl9gaadc0vf6zqqsa8k8yh6dbipgk4n.png)
Therefore, the probability that he went by car if he arrived late for his appointment is 0.75