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Home  >>  CBSE XII  >>  Math  >>  Model Papers
+1 vote

A doctor is to visit a patient. From the past experience, it is known that the probabilities that he will come by train, bus, scooter or by any other means of transport are respectively \(\frac{3}{10}, \frac{1}{5}, \frac{1}{10} \) and \( \frac{2}{5} \) . The probabilities that he will be late are \( \frac{1}{4}, \frac{1}{3} \) and \( \frac{1}{12} \) if he comes by train, bus and scooter respectively, but if he comes by other means of transport, then he will not be late. When he arrives, he is late. What is the probability that he comes by train.

Can you answer this question?
 
 

1 Answer

0 votes

 

Solution:

 

$\Large\frac{1}{2} \Large $

 


Tool Box:
Recall Bayes Theorem - $P(A_{i } / B)= \Large\frac{P(A_{i})P( B/ A_{i })}{P(A_{1})P( B/ A_{1})+P(A_{2})P( B/ A_{2 })+\ldots +P(A_{n})P( B/ A_{n })} \Large $
 
Here, $A_{1}, A_{2 }, A_{3 ,}and A_{4}$ are the events that the doctor takes a train, scooter, train or other means respectively and event B is  the doctor is late.
 
Step 1: Identify what needs to be found and write it out symbolically. You have to find the probability that the doctor takes the train given that he is late. 
i.e. $P(A_{1 } / B)$
 
Step 2: Apply Bayes Theorem for the given problem.
$P(A_{1 } / B)= \Large \frac{P(A_{1})P( B/ A_{1})}{P(A_{1})P( B/ A_{1})+P(A_{2})P( B/ A_{2 })+P(A_{3})P( B/ A_{3 })+P(A_{4})P( B/ A_{4 })}\Large$
= $\Large \frac{P(A_{1}\cap B)}{P(A_{1}\cap B)+P(A_{2}\cap B)+P(A_{3}\cap B)+P(A_{4}\cap B)} \Large$
 
Note that $P(A_{1} / B)= \Large \frac{P(A_{1}\cap B)}{P(B)} \Large $.
 
Step 3:
$P(A_{1}\cap B)$ = the probability that the doctor takes the  train and he is late.
 = $ \Large \frac{3}{10} \Large. \Large \frac{1}{4} \Large = \Large \frac{3}{40}\Large$
 
$P(A_{2}\cap B)= $the probability that the doctor takes the bus and he  is late.
 = $ \Large \frac{2}{10} \Large. \Large \frac{1}{3} \Large = \Large \frac{1}{15} \Large $
 
$P(A_{3}\cap B)= $the probability that the doctor takes the scooter  and he is late.
= $|Large \frac{1}{10} \Large . \Large \frac{1}{12} \Large= \Large \frac{1}{120} \Large $
 
$P(A_{4}\cap B)=$ the probability that the doctor takes other means of  transport and he is late
= 0, because in this case the doctor is not late
 
Step 4: Substitute the values in the formula to arrive at $P(A_{1 } / B)$ 
 
Numerator =  $\Large \frac{3}{40} \Large $
 
Denominator = $ \Large \frac{3}{40}\Large + \Large\frac{1}{15}\Large +\Large\frac{1}{120} \Large +0 = \Large \frac{9+8+1}{120} \Large $
 
Therefore, $P(A_{1 } / B)$ = $ \Large \frac{3}{40} \Large . \Large \frac{120}{18}\Large $= $\Large \frac{9}{18 \Large }$ = $\Large \frac{1}{2} \Large$

 

answered Dec 21, 2012 by balaji.thirumalai
edited Dec 21, 2012 by balaji.thirumalai
 

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