Answer to Question #166970 in Statistics and Probability for Ajay Singh

Question

1.     Samuel woke up late in the morning on the day that he has to go to school to take an important test. He can either take the shuttle bus which is usually running late 15% of the time or ride his unreliable motorcycle which breaks down 40% of the time. He decides to toss a fair coin to make his choice.

 

a.      If Samuel, in fact, gets to the test on time, what is the probability that he took the bus? (2)

 

(b)If Samuel is late to the test, what is the probability that he took the bus?

Answer 1

a. Let   T = the event that Samuel gets to the test on time

              C = the event that Samuel took his car

              B = the event that Samuel took the bus

The probability that Samuel gets to the test on time, given that he took the bus is $P(T|B)=1-0.15=0.85$ and the probability that Samuel gets to the test on time, given that he took his car is $P(T|C)=1-0.4=0.6.$

If he flips a “fair” coin, to choose between the bus and car, then $P(B)=P(C)=0.5.$

Then, Baye’s Rule states:

$$P(B|T)=\dfrac{P(T|B)P(B)}{P(T|B)P(B)+P(T|C)P(C)}$$

$$P(B|T)=\dfrac{0.85(0.5)}{0.85(0.5)+0.6(0.5)}\approx 0.5862$$

b.The probability that Samuel is late to the test, given that he took the bus is $P(T'|B)=0.15$ and the probability that Samuel is late to the test, given that he took his car is $P(T'|C)=0.4.$

If he flips a “fair” coin, to choose between the bus and car, then $P(B)=P(C)=0.5.$

Then, Baye’s Rule states:

$$P(B|T')=\dfrac{P(T'|B)P(B)}{P(T'|B)P(B)+P(T'|C)P(C)}$$

$$P(B|T')=\dfrac{0.15(0.5)}{0.15(0.5)+0.4(0.5)}\approx 0.2727$$