Question

Your desktop, on average, crashes 6 times in every 24 hours.

1. You are working from home and you have work to complete in 2 hours. What is the probability that your desktop will not crush in these 2 hour period?

2. What is the probability that your desktop will crush exactly twice in the next 5 hours?

3. What is the probability that 2 crushes happen in the next 5 hours?

Please give clear explanation of all the steps involved. Thanks

Answer 1

on an average. a 20 9 o 6 Given that, desktop crashes 6 times sin every 24 hours of times the desktop crashes in a particular time Lemon X be a mandom variable denoting, the number interval. Then, x takes values: x=0,102. Here, we use a Poisson distribution to model the probability distribution of X, with parameterea Croate ). The pim.ß of X is given by PCX = m). 2=0,1,2,3,4,..., botherwise. 1. Foro a 2 hours time -interval, we need to find value of a birst. Pero hour, rate of desktop crashing , i8= 24 : Foro 2 hours, the rate is: 1= 2x + 4 The p.m.f of x with parameter, az 1 is & P(X=36) x=0, 1, 2, 3, ...... 20. o gotherwise. To Bind: probability, that, desktop will not crash in 'these 2 hours period, ie, X will take value=0. Required probability, id = P(x=0) e © Hiv 2 Se () 9 . 0! 1 [():4,0!=1 112 e Ĉ0.6065-- Answer.

5 2. Given, time - period = 5 hours. Hence, the rate is - a = (5x5) - .: The p.m.f of x with parameter 2= gisa 5 4 a P(x+x)=fe @). & o olu 9 X=0,1,2,3, al botherwise. To find: Probability twice that desktop will crash exo exactly in these 5 hours périodsie, & will take valueza. -- Requined probability, is = P(X=2) =ē CG 21 25 [*: 2:=2] (0.2865 x 0.48125) ñ 0.2238 - Answer, 3. Given, time-period=5 hours. Again, we have, To find: p(x-2) = ~0.2238. (same as part 2). = e -X 2 / 2 16 2 5 n = 2 26

(P.S: part 3 is exactly same as part 2!)

*A note why Poisson distribution is used:

Poisson distribution is appropriate when we have non negative integered valued random variable such that, the observed variable describes a count data such that, the event we are counting occurs randomly and independently over time.