Question

(iii) Let X be the number of the cars being repaired at a repair shop. We have the following information: At any time, there are at most 3 cars repaired. The probability of having 2 cars at a shop is same as the probability of having 1 car. The probability of having no car is the same as the probability of having 3 cars. The probability of having 1 or 2 cars is double of the probability of having 0 or 3 cars. Find the p.m.f. of X.

Answer 1

Answer:

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Answer 2

Given : At any time, there are at most 3 cars repaired.

The probability of having 2 cars at a shop is same as the probability of having 1 car.

The probability of having no car is the same as the probability of having 3 cars.

The probability of having 1 or 2 cars is double of the probability of having 0 or 3 cars.

To Find :  the p.m.f. of X.

Solution:

p.m.f  is probability mass function ,

probability of having 0 cars =  3 cars = y

=> probability of having 1 car =  2 cars = 2y

P(0) + P(1) + P(2) + P(3) = 1

=> y + 2y + 2y + y = 1

=> 6y = 1

=> y = 1/6

  probability of having 0 cars =  3 cars = 1/6

 probability of having 1 car =  2 cars = 2(1/6)  = 1/3

p.m.f. of X is

$p_{X}(x)=\left \{ {{ \frac{1}{6} \quad x ~\in ~\{0,3\} } \atop { \frac{1}{3} \quad x ~\in ~\{1,2\}}} \right.$

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