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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 :

1. $6.9 \times 10^6 \times e^{-20}$
2. $1.02 \times 10^6 \times e^{-20}$
3. $6.9 \times 10^3 \times e^{-20}$
4. $1.02 \times 10^3 \times e^{-20}$

20 request in 1 hour.. so we can expect 15 request in 45 minutes...

So, lemda = 15.. (expected value)

poission distribution formula: f(x, lemda) = p(X = x) = (lemda ^ x * e ^ - lemda)  / x!

Therefore p(one request) + p(3 request) + p(5 request)

= p(1; 15) + p(3; 15) + p(5; 15)

= 6.9 * 10^3 * e ^ -15..

= 6.9*103*e-15 = 6.9*103*e5*e-20 = 1.02*106*e-20..  Ans is (B)

selected

6.9*103*e-15 = 6.9*103*e5*e-20 = 1.02*106*e-20

omg.. thanku very much Danish..

answer coming ( 6.9*103 ) *e^5= ( 6.9*103 )*148.41 =1.02*106

yes B

why not can we take λ = 20 and then compute P(1)+P(2)+P(3) , which will give the probability of getting 1 or 3 or 5 request per hour and then take 3/4th of it.

Answer b is right only if we take  λ = 15 and then compute P(1)+P(2)+P(3), which will be in requests per 45 mins.