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Sagot :
Solution:
Given:
[tex]\begin{gathered} P(Car)=0.6 \\ P(Air)=0.4 \end{gathered}[/tex][tex]\begin{gathered} P(Car\text{ and on time\rparen}=0.3 \\ P(air\text{ and on time\rparen}=0.65 \end{gathered}[/tex]i) The probability that he arrives in Lagos early for the appointment is;
[tex]\begin{gathered} P(early)=(0.3\times0.6)+(0.65\times0.4) \\ P(early)=0.18+0.26 \\ P(early)=0.44 \end{gathered}[/tex]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.
[tex]\begin{gathered} P(Car|Late)=\frac{P(car\text{ and late\rparen}}{P(Late)} \\ P(Car|Late)=\frac{P(car)\times 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}[/tex]We now get the probability of late;
[tex]\begin{gathered} P(late)=(0.6\times0.7)+(0.4\times0.35) \\ P(late)=0.42+0.14 \\ P(late)=0.56 \end{gathered}[/tex]Therefore, the probability that he went by car if he arrived late for his appointment is;
[tex]\begin{gathered} P(Car|Late)=\frac{P(car\text{ and late\rparen}}{P(Late)} \\ P(Car|Late)=\frac{P(car)\times 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)=\frac{0.6\times0.7}{0.56} \\ P(Car|Late)=\frac{0.42}{0.56} \\ P(Car|Late)=0.75 \end{gathered}[/tex]
Therefore, the probability that he went by car if he arrived late for his appointment is 0.75
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