Vehicles arriving at an intersection from one of the approach roads follow the Poisson distribution. The mean rate of arrival is $900$ vehicles per hour. If a gap is defined as the time difference between two successive vehicle arrivals (with vehicles assumed to be points), the probability (up to four decimal places) that the gap is greater than 8 seconds is....

0.1353

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Solution

The correct option is A 0.1353
Probability of time headway being greater than $8 s e c$
$P (h \geq 8) = e^{- λ t} = e^{- \frac{900}{3600} \times 8}$
$= e^{- 2} = 0.1353$

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The correct option is A 0.1353Probability of time headway being greater than 8sec P(h≥8)=e−λt=e−9003600×8 =e−2=0.1353

The correct option is A 0.1353
Probability of time headway being greater than $8 s e c$
$P (h \geq 8) = e^{- λ t} = e^{- \frac{900}{3600} \times 8}$
$= e^{- 2} = 0.1353$