In a multi-user operating system on an average, 20 requests are made to use a…
2007
In a multi-user operating system on an average, 20 requests are made to use a particular resource per hour. The arrival of requests follows a Poisson distribution. The probability that either one, three or five requests are made in 45 minutes is given by :
- A.
6.9 × 10⁶ × e⁻²⁰
- B.
1.02 × 10⁶ × e⁻²⁰
- C.
6.9 × 10³ × e⁻²⁰
- D.
1.02 × 10³ × e⁻²⁰
Attempted by 2 students.
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Correct answer: B
Let X be the number of requests in 45 minutes. Since requests follow a Poisson distribution and the average rate is 20 per hour, the mean for 45 minutes is:
λ = 20 × 45/60 = 15.
Therefore,
P(X = 1 or 3 or 5)
= e⁻¹⁵(15¹/1! + 15³/3! + 15⁵/5!)
= e⁻¹⁵(15 + 3375/6 + 759375/120)
= e⁻¹⁵(15 + 562.5 + 6328.125)
= 6905.625e⁻¹⁵
≈ 6.9 × 10³e⁻¹⁵.
Since the options are written in terms of e⁻²⁰:
e⁻¹⁵ = e⁵e⁻²⁰, and e⁵ ≈ 148.4.
So,
6.9 × 10³e⁻¹⁵ ≈ 6.9 × 10³ × 148.4 × e⁻²⁰
≈ 1.02 × 10⁶e⁻²⁰.