A lamp holds 3 bulbs. There are two kinds of bulbs, short-lived red on

grasaladae5

grasaladae5

Answered question

2021-11-26

A lamp holds 3 bulbs. There are two kinds of bulbs, short-lived red ones with mean lifetime 1 day, and long-lived green ones with mean lifetime 3 days. All bulb lifetimes have exponential distributions independent of each other. Whenever a green light bulb burns out, it is replaced by a red one. Whenever a red light bulb burns out, it is replaced by a green one. Define variables and write equations that, if solved, would yield the probability distribution of the number of green light bulbs at a random moment when a bulb has just been replaced.

Answer & Explanation

barcelodurazo0q

barcelodurazo0q

Beginner2021-11-27Added 13 answers

Step 1
Given information:
Given that there are 3 lamp bulbs.
The mean life-time of red bulbs is 1 day and the mean life-time of green bulbs is 3 days.
The life-time of both bulbs is exponentially distributed and independent of each other.
Step 2
Find the probability distribution of the number of green bulbs:
The total number of bulbs in the lamp is 3.
Suppose that the number of green blubs is X.
The number of red bulbs is 3X.
The number of green bulbs at a random moment will be {0,1,2,3}.
The probability distribution is as given below:
P(X=x)=3Cxpx(1p)3x.
Here, p is the probability of green bulbs when a bulb is just replaced.

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