
A person goes to the office by either by car, scooter, bus, or train probability of which being $ \dfrac{1}{7},\dfrac{3}{7},\dfrac{2}{7} $ and $ \dfrac{1}{7} $ respectively. The probability that he reaches office late, if he takes a car, scooter, bus, or train is $ \dfrac{2}{9},\dfrac{1}{9},\dfrac{4}{9} $ and $ \dfrac{1}{9} $ respectively. Given that he reached office in time, then if the probability that he travelled by car is $ \dfrac{1}{p} $ . Find p.
Answer
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Hint: In order to solve this problem, we need to find the probability of the person reaches on time by taking the complement of each probability. And the probability that the person travelled by car and reached in time is $ P\left( \dfrac{C}{\overline{L}} \right)=\dfrac{P\left( \dfrac{\overline{L}}{C} \right).P\left( C \right)}{P\left( \overline{L} \right)}. $ , where $ P\left( C \right) $ is the probability that the person travel by car. $ P\left( \overline{L} \right) $ is the probability that the person reached on time.
Complete step-by-step answer:
Let C be the event of a person going by car.
Let S be the event of a person going by scooter.
Let B be the event of a person going by bus.
Let T be the event of the person going by train.
According to the question,
Probability of going by car is $ P\left( C \right)=\dfrac{1}{7} $ .
Probability of going by scooter is $ P\left( S \right)=\dfrac{3}{7} $ .
Probability of going by bus is $ P\left( B \right)=\dfrac{2}{7} $ .
Probability of going by train is $ P\left( T \right)=\dfrac{1}{7} $ .
Let L be the event of reaching office late.
Therefore, the event of being in time is $ \overline{L} $ .
Now, we have given the probabilities of car, bus, scooter and train by giving the condition that the person reaches late.
Therefore, $ P\left( \dfrac{L}{C} \right)=\dfrac{2}{9} $ , $ P\left( \dfrac{L}{S} \right)=\dfrac{1}{9} $ , $ P\left( \dfrac{L}{B} \right)=\dfrac{4}{9} $ , $ P\left( \dfrac{L}{T} \right)=\dfrac{1}{9} $
Taking the inverse probabilities, we get,
$ P\left( \dfrac{\overline{L}}{C} \right)=\dfrac{7}{9} $ , $ P\left( \dfrac{\overline{L}}{S} \right)=\dfrac{8}{9} $ , $ P\left( \dfrac{\overline{L}}{B} \right)=\dfrac{5}{9} $ , $ P\left( \dfrac{\overline{L}}{T} \right)=\dfrac{8}{9} $
We need to find the probability that the person reaches office in time.
Therefore the probability that we need to find is $ P\left( \dfrac{C}{\overline{L}} \right) $ .
The formula is given by $ P\left( \dfrac{C}{\overline{L}} \right)=\dfrac{P\left( \dfrac{\overline{L}}{C} \right).P\left( C \right)}{P\left( \overline{L} \right)}......................(i) $ ,
We need to find the value of $ P\left( \overline{L} \right) $ separately.
To find the probability of reaching in time we need to add all the probabilities of all modes of transport.
Therefore,
The probability of reaching in time by car = $ P\left( C \right)\times P\left( \dfrac{\overline{L}}{C} \right) $ = $ \dfrac{1}{7}\times \dfrac{7}{9}=\dfrac{1}{9} $ .
The probability of reaching in time by scooter = $ P\left( S \right)\times P\left( \dfrac{\overline{L}}{S} \right) $ = $ \dfrac{3}{7}\times \dfrac{8}{9}=\dfrac{24}{63} $ .
The probability of reaching in time by bus = $ P\left( B \right)\times P\left( \dfrac{\overline{L}}{B} \right) $ = $ \dfrac{2}{7}\times \dfrac{5}{9}=\dfrac{10}{63} $ .
The probability of reaching in time by train = $ P\left( T \right)\times P\left( \dfrac{\overline{L}}{T} \right) $ = $ \dfrac{8}{9}\times \dfrac{1}{7}=\dfrac{8}{63} $ .
Substituting all the values we get,
$ P\left( \overline{L} \right)=\dfrac{1}{9}+\dfrac{24}{63}+\dfrac{10}{63}+\dfrac{8}{63}=\dfrac{49}{63}=\dfrac{7}{9} $ .
Substituting the values in equation (i), we get,
$ P\left( \dfrac{C}{\overline{L}} \right)=\dfrac{P\left( \dfrac{\overline{L}}{C} \right).P\left( C \right)}{P\left( \overline{L} \right)}=\dfrac{\dfrac{7}{9}\times \dfrac{1}{7}}{\dfrac{7}{9}}=\dfrac{1}{7} $
Comparing $ \dfrac{1}{p} $ the value of p is 7.
Note: We need to understand to write the probability that includes the condition with it. When any condition is applied we need to multiply the probability. Also, in the question, we are given the probabilities of reaching late and we are asked the probability of reaching in time. Therefore, we need to take the complement of all the events to work the problem.
Complete step-by-step answer:
Let C be the event of a person going by car.
Let S be the event of a person going by scooter.
Let B be the event of a person going by bus.
Let T be the event of the person going by train.
According to the question,
Probability of going by car is $ P\left( C \right)=\dfrac{1}{7} $ .
Probability of going by scooter is $ P\left( S \right)=\dfrac{3}{7} $ .
Probability of going by bus is $ P\left( B \right)=\dfrac{2}{7} $ .
Probability of going by train is $ P\left( T \right)=\dfrac{1}{7} $ .
Let L be the event of reaching office late.
Therefore, the event of being in time is $ \overline{L} $ .
Now, we have given the probabilities of car, bus, scooter and train by giving the condition that the person reaches late.
Therefore, $ P\left( \dfrac{L}{C} \right)=\dfrac{2}{9} $ , $ P\left( \dfrac{L}{S} \right)=\dfrac{1}{9} $ , $ P\left( \dfrac{L}{B} \right)=\dfrac{4}{9} $ , $ P\left( \dfrac{L}{T} \right)=\dfrac{1}{9} $
Taking the inverse probabilities, we get,
$ P\left( \dfrac{\overline{L}}{C} \right)=\dfrac{7}{9} $ , $ P\left( \dfrac{\overline{L}}{S} \right)=\dfrac{8}{9} $ , $ P\left( \dfrac{\overline{L}}{B} \right)=\dfrac{5}{9} $ , $ P\left( \dfrac{\overline{L}}{T} \right)=\dfrac{8}{9} $
We need to find the probability that the person reaches office in time.
Therefore the probability that we need to find is $ P\left( \dfrac{C}{\overline{L}} \right) $ .
The formula is given by $ P\left( \dfrac{C}{\overline{L}} \right)=\dfrac{P\left( \dfrac{\overline{L}}{C} \right).P\left( C \right)}{P\left( \overline{L} \right)}......................(i) $ ,
We need to find the value of $ P\left( \overline{L} \right) $ separately.
To find the probability of reaching in time we need to add all the probabilities of all modes of transport.
Therefore,
The probability of reaching in time by car = $ P\left( C \right)\times P\left( \dfrac{\overline{L}}{C} \right) $ = $ \dfrac{1}{7}\times \dfrac{7}{9}=\dfrac{1}{9} $ .
The probability of reaching in time by scooter = $ P\left( S \right)\times P\left( \dfrac{\overline{L}}{S} \right) $ = $ \dfrac{3}{7}\times \dfrac{8}{9}=\dfrac{24}{63} $ .
The probability of reaching in time by bus = $ P\left( B \right)\times P\left( \dfrac{\overline{L}}{B} \right) $ = $ \dfrac{2}{7}\times \dfrac{5}{9}=\dfrac{10}{63} $ .
The probability of reaching in time by train = $ P\left( T \right)\times P\left( \dfrac{\overline{L}}{T} \right) $ = $ \dfrac{8}{9}\times \dfrac{1}{7}=\dfrac{8}{63} $ .
Substituting all the values we get,
$ P\left( \overline{L} \right)=\dfrac{1}{9}+\dfrac{24}{63}+\dfrac{10}{63}+\dfrac{8}{63}=\dfrac{49}{63}=\dfrac{7}{9} $ .
Substituting the values in equation (i), we get,
$ P\left( \dfrac{C}{\overline{L}} \right)=\dfrac{P\left( \dfrac{\overline{L}}{C} \right).P\left( C \right)}{P\left( \overline{L} \right)}=\dfrac{\dfrac{7}{9}\times \dfrac{1}{7}}{\dfrac{7}{9}}=\dfrac{1}{7} $
Comparing $ \dfrac{1}{p} $ the value of p is 7.
Note: We need to understand to write the probability that includes the condition with it. When any condition is applied we need to multiply the probability. Also, in the question, we are given the probabilities of reaching late and we are asked the probability of reaching in time. Therefore, we need to take the complement of all the events to work the problem.
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