Grewal

Question

Suppose avg waiting time of a process to get chance in a queue is 5 min. What will the probability that process get chance at first minute is ________________

Answer 1

For the detailed solution , please see the pic below

Comments

@srestha mam, please see the pic below for the answer.

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Here the lifetime of the product is 0<T<3

T>=2 states that the product will work fine till 2 or more years i.e it includes 0<T<=2 and 2<T<3 and T>=3. 

But as the product breaks down in 3rd year, we have to remove T>=3 from above. 

Thus the required probability is P(T>=2) - P(T>=3).

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Probability will be $\frac{e^{-\frac{2}{4}}-e^{-\frac{3}{4}}}{e^{-\frac{2}{4}}}$

right??

i.e. $P\left ( A/B \right )=\frac{P\left ( A\cap B \right )}{P\left ( B \right )}$

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No.

Required probability= P(0<T<3) = P(T>=2)-P(T>=3)

I had given the explanation.

Answer 2

Avg waiting time $\lambda$ = $5$ min.

average rate is $1$ process in $5$ minutes i.e. $1/5$

According to exponential distribution,

$P(X<=1) = \int_{0}^{1}\frac{1}{5}e^{-\frac{t}{5}}dt = 1- e^{-\frac{1}{5}} = 1- 0.81=0.18$

Comments

@srestha mam,

pdf can take only continuous values,

pmf can take only discrete values,

but cdf can take both discrete and continuous values.

For continuous probability distribution, c.d.f take continuous values i.e. a range of values. 

For discrete probability distribution, c.d.f take discrete values.

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@SuvasishDutta

yes, right.

By the way, from where u read for  this probability portion, like exponential, pmf, pdf, cdf portion?

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From grewal book and made easy engg maths book

Answer 3

By Using the poisson distribution and substituting the average rate to 3,

Probality is 0.149

Comments

not clear