Arrivals at a telephone booth are considered to be Poisson, with an average time of 10 minutes between successive arrivals. The length of a phone call is distributed exponentially with mean 3 minutes. The probability that an arrival does not have to wait before service is

0.3

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0.7

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0.5

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0.9

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Solution

The correct option is B 0.7
$Given data:$

$λ = 0.1 per minute,$

$Service rate, μ = 0.33 per minute$

$ρ = \frac{0.1}{0.33} = 0.3$

Probability that an arrival does not have to wait before service or probability that a system is idle,

$P_{0} = 1 - ρ = 1 - 0.3 = 0.7$

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Arrivals at a telephone booth are considered to be Poisson, with an average time of 10 minutes between successive arrivals. The length of a phone call is distributed exponentially with mean 3 minutes. The probability that an arrival does not have to wait before service is

The correct option is B 0.7Given data: λ=0.1 per minute, Service rate, μ=0.33 per minute ρ=0.10.33=0.3 Probability that an arrival does not have to wait before service or probability that a system is idle, P0=1−ρ=1−0.3=0.7

The correct option is B 0.7
$Given data:$

$λ = 0.1 per minute,$

$Service rate, μ = 0.33 per minute$

$ρ = \frac{0.1}{0.33} = 0.3$

Probability that an arrival does not have to wait before service or probability that a system is idle,

$P_{0} = 1 - ρ = 1 - 0.3 = 0.7$