3. The duration in hours of a certain incandescent light bulb is a continuous ra

Question

3. The duration in hours of a certain incandescent light bulb is a continuous random variable X having the probability density function (PDF) 100 x > 100; , fx(x) = 0, otherwise. (a) [6 points What is the probability that a bulb will last less than 200 hours, given that it is still ously, what s the prola bility that exactly (c) 8 points What is the minimum number of bulbs that have to be connected in a parallel system working after 150 hours of operation? one has to be replaced after 150 hours of operation? so that the probability of having the system working after 150 hours of operation is 0.999?

Explanation / Answer

Given X = life (in hours) of a certain incandescent light bulb, with pdf,

f(x) = 100/x2, for x >100 and 0, otherwise.

Back-up Theory

P(X < t) = integral [100, t]{f(x)dx}for t >100…………………………………………(1)

If A and B are two events such that probability of B is influenced by occurrence or otherwise of A, then Conditional Probability of B given A, denoted by P(B/A) = P(B A)/P(A)..….(2)

If X ~ B(n, p). i.e., X has Binomial Distribution with parameters n and p, where n = number of trials and p = probability of one success, then probability mass function (pmf) of X is given by p(x) = P(X = x) = (nCx)(px)(1 - p)n – x, x = 0, 1, 2, ……. , n …………………..(2A)

[The above probability can also be directly obtained using Excel Function of Binomial Distribution]

Preparatory Work

Probability the bulb lasts at least 150 hours

= P(X > 150)

= 1 - P(X < 150)

= 1 - integral [100, 150]{( 100/x2)dx} [vide (1) under Back-up Theory]

= 1- [-{100/x}](100, 150)]

= 1 + (100/150) – (100/100)

= 2/3 = 0.6667……….……………………………………………………………………(3)

Part (a)

Let A represent the event that bulb is still working after 150 hours of operation and B represent the event that bulb will last less than 200 hours. Then, the required probability

= P(B/A)

= P(B A)/P(A) [vide (2) under Back-up Theory]………………………………………….(4)

Now, P(B A) = P(150 < X < 200)

= integral [150, 200]{( 100/x2)dx} [vide (1) under Back-up Theory]

= [-{100/x}](150, 200)]

= 1/6 …………………………………………………………………………………(5)

(3), (4) and (5) => required probability = (1/6)/(2/3)

= ¼ = 0.25 ANSWER

Part (b)

Let Y = number of bulbs to be replaced after150 hours of operation when 3 bulbs are installed simultaneously. Then, Y ~ B(3, p), where p = Probability the bulb lasts at least 150 hours = 2/3 [vide (3) above].

Required probability = P(Y = 1)

= (3C1)(2/3)(1/3)2 [vide (2A) under Back-up Theory]

= 0.22 ANSWER

Part (c)

Let n be the minimum number of bulbs to be connected in parallel so that probability of the system working after 150 hours.

Since the connection is in parallel, the system works if at least one bulb works and hence

probability of the system working after 150 hours

= probability of at least one bulb working after 150 hours

= 1 - probability of none of n bulbs working after 150 hours

= 1 - (nC0)(2/3)0(1/3)n [vide (2A) under Back-up Theory]

= 1 - (1/3)n

We want this probability to be 0.999 which in turn implies:

(1/3)n = 0.001

Taking log to base 10, - nlog3 = log0.001

Or n = 3/log3

= 3/0.477121

= 6.29

= 7 ANSWER

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