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Consider X to be an exponentially distributed random variable with parameter $\lambda =0.5$ that represents the time (in hour) required to repair a machine.What is the conditional probability that it will take atleast 6 hour to repair, given that time exceeds 5 hours?

$e^{-0.5} = 0.6065$ ??

@nitish don't you think it should be

1 - e-0.5 = 0.3935 ?

@Ashwin,

$P(Atleast(6)\cap >5)=\int_{6}^{\infty }0.5e^{-0.5x}dx$

$P(>5)=\int_{5}^{\infty}0.5e^{-0.5x}dx$

$P(Atleast(6)/>5)=\frac{P(Atleast(6)\cap >5)}{P(>5)}$$=e^{-0.5} =0.6065$

1-(1 - e-0.5 }=0.6065

$\lambda = 0.5$

Now question mentions that time exceeds 5 hours. So we need to calculate probability that repairing machine will take atleast 1 hour more.

$P(X \leq x) = 1 - e^{- \lambda x}$

$P(X \leq 1) = 1 - e^{-0.5}$

$P(X \geq 1) = 1 - P(X \leq 1)$

$P(X \geq 1) = 1 - 1 + e^{-0.5} = 0.6065$

@Mk Utkarsh

The formula u r using here is Poisson or something else??

1 vote