Question

To get to school you can travel by car, bus or bicycle. If you travel by car, there is a 50% chance you will be late because the roads are very busy. If you travel by bus, which uses special reserved lanes and the busway, the probability of being late is only 20%. If you travel by bicycle you are only late 1% of the time.
(a) Suppose that you are late one day to class. Since your teacher does not know which mode of transportation you usually use, he assumes each of the three possibilities are equally likely. If you are late, find the probability that you travelled to school that day by car?

(b) Suppose that a friend tells your teacher that you almost always ride your bicycle to school, never take the bus, but 10% of the time travel by car. If you are late, what is the new probability that you travelled to school that day by car?

Answer 1

Answer:

(a) 0.704

(b) 0.8475

Step-by-step explanation:

(a) Let 'A' be the event that you travel by car and late

Let 'B' be the event that you travel by bus and late

Let 'C' be the event that you travel by Bicycle and late

Then, P (A)  = 50% = $\frac{50}{100}$ = $\frac{1}{2}$

         P (B)   = 20% = $\frac{20}{100}$ = $\frac{1}{5}$

         P (C)   = 1%  = $\frac{1}{100}$  = $\frac{1}{100}$

A₁ = Student travels by car

B₁ = Student travels by bus

C₁ = Student travels by bicycle

Then according to teacher P(A₁) = $\frac{1}{3}$, P(B₁) = $\frac{1}{3}$, P(C₁) = $\frac{1}{3}$

Now we have to find "Student is already late and traveled to school that day by car." which will be given as $P(\frac{A}{L})$

where L : student is late

By using Bay's Theorem :

$P(\frac{A}{L})$  = $\frac{P(A)\times P(A_1)}{P(A)\times P(A_1)+P(B)\times P(B_1)+P(C)\times P(C_1)}$

= $\frac{\frac{1}{2}\times \frac{1}{3}}{\frac{1}{2}\times \frac{1}{3}+\frac{1}{5}\times \frac{1}{3}+\frac{1}{100}\times \frac{1}{3}}$

= $\frac{\frac{1}{6}}{\frac{1}{6}+\frac{1}{15}+\frac{1}{300}}$

= $\frac{\frac{1}{6}}{\frac{50+20+1}{300}}$

= $\frac{1}{6}\times \frac{300}{71}$

= $(\frac{50}{71})$

= 0.704

(b) Here P(A₁) = $(\frac{10}{100})$

              P(C₁) = $(\frac{90}{100})$

            $P(\frac{A}{L})$  = We have to find and known student is late and traveled by car.

$P(\frac{A}{L})$ = $\frac{P(A)\times P(A_1)}{P(A)\times P(A_1)+(P(C)\times P(C_1)}$

= $\frac{\frac{1}{2}\times \frac{1}{10}}{\frac{1}{2}\times \frac{1}{10}+\frac{1}{100}\times \frac{9}{10}}$

= $\frac{\frac{1}{20} }{\frac{1}{20}+\frac{9}{1000}}$

= $\frac{\frac{1}{20}}{\frac{50+9}{1000}}$

= $\frac{1}{20}\times \frac{1000}{59}$

= $(\frac{50}{59})$

= 0.8475