A person goes to office either by car, scooter, bus or train probability of which being 17,37,27 and 17 respectively. Probability that he reaches office late, if he takes car, scooter, bus or train is 29,19,49 and 19 respectively. Given that he reached office in time, then the probability that he travelled by a car is given by , then the value of 7 is equal to …..

P(C)=17 (by car) P(B)=27 (by bus) P(S)=37 (by scooter) P(T)=17 (person goes by train) P(LC)=29, P(LS)=19, P(LB)=49 & P(LT)=19We have to find P(CL¯)(since reaches in time ≡ not late) By bayes theorem= 79×1779×17+89×37+59×27+89×17=17=P

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A person goes to office either by car, scooter, bus or train probability of which being $\frac{1}{7}, \frac{3}{7}, \frac{2}{7}$ and $\frac{1}{7}$ respectively. Probability that he reaches office late, if he takes car, scooter, bus or train is $\frac{2}{9}, \frac{1}{9}, \frac{4}{9}$ and $\frac{1}{9}$ respectively. Given that he reached office in time, then the probability that he travelled by a car is given by 𝑃, then the value of 7 𝑃 is equal to …..

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Detailed Solution

$P (C) = \frac{1}{7} (by car) P (B) = \frac{2}{7} (by bus) P (S) = \frac{3}{7} (by scooter) P (T) = \frac{1}{7} (person goes by train) P (\frac{L}{C}) = \frac{2}{9}, P (\frac{L}{S}) = \frac{1}{9}, P (\frac{L}{B}) = \frac{4}{9} & P (\frac{L}{T}) = \frac{1}{9}$

We have to find $P (\frac{C}{\bar{L}})$
(since reaches in time $\equiv$ not late)
By bayes theorem
$= \frac{\frac{7}{9} \times \frac{1}{7}}{\frac{7}{9} \times \frac{1}{7} + \frac{8}{9} \times \frac{3}{7} + \frac{5}{9} \times \frac{2}{7} + \frac{8}{9} \times \frac{1}{7}} = \frac{1}{7} = P$