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Question Number 973 by Yugi last updated on 10/May/15
The number of hits per minute on a website is 0.8. It is shown that the time T between two successive hits follows a negative exponential distribution f(t)=0.8e^(−0.8t) ( t ≥ 0 ). It is required to determine the probability that the time between the 1st hit and the 51st exceeds one hour. One approach to the calculation stated that the required probability is P(no. of hits in 60 mins < 50) ( ∗ ) Can anyone explain to me why it is that P(time between 1st hit and 51st hit > 60mins)= (∗) ? I sadly don′t understand the link made between the two wordings of the seemingly equal events contained in the probability expressions. Could a sufficiently detailed explanation of calculating ( ∗ ) be given also? Some sort of approximation using the normal distribition is needed but I′m unsure of the necessary procedure. Perhaps someone could suggest what book or exam papers I may use to improve my competence in noticing the equivalence of events in questions of this type.
$${The}\:{number}\:{of}\:{hits}\:{per}\:{minute}\:{on}\:{a}\:{website}\:{is}\:\mathrm{0}.\mathrm{8}.\:{It}\:{is}\:{shown}\:{that}\:{the}\:{time}\:{T} \\ $$$${between}\:{two}\:{successive}\:{hits}\:{follows}\:{a}\:{negative}\:{exponential}\:{distribution}\: \\ $$$${f}\left({t}\right)=\mathrm{0}.\mathrm{8}{e}^{−\mathrm{0}.\mathrm{8}{t}} \:\left(\:{t}\:\geqslant\:\mathrm{0}\:\right).\:{It}\:{is}\:{required}\:{to}\:{determine}\:{the}\:{probability}\:{that}\:{the}\:{time}\: \\ $$$${between}\:{the}\:\mathrm{1}{st}\:{hit}\:{and}\:{the}\:\mathrm{51}{st}\:{exceeds}\:{one}\:{hour}.\:{One}\:{approach}\:{to}\:{the}\: \\ $$$${calculation}\:{stated}\:{that}\:{the}\:{required}\:{probability}\:{is}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{P}\left({no}.\:{of}\:{hits}\:{in}\:\mathrm{60}\:{mins}\:<\:\mathrm{50}\right)\:\:\:\:\:\:\:\:\:\:\:\left(\:\ast\:\right) \\ $$$${Can}\:{anyone}\:{explain}\:{to}\:{me}\:{why}\:{it}\:{is}\:{that} \\ $$$${P}\left({time}\:{between}\:\mathrm{1}{st}\:{hit}\:{and}\:\mathrm{51}{st}\:{hit}\:>\:\mathrm{60}{mins}\right)=\:\left(\ast\right)\:? \\ $$$${I}\:{sadly}\:{don}'{t}\:{understand}\:{the}\:{link}\:{made}\:{between}\:{the}\:{two}\:{wordings}\:{of}\:{the} \\ $$$${seemingly}\:{equal}\:{events}\:{contained}\:{in}\:{the}\:{probability}\:{expressions}.\: \\ $$$${Could}\:{a}\:{sufficiently}\:{detailed}\:{explanation}\:{of}\:{calculating}\:\left(\:\ast\:\right)\:{be}\:{given}\:{also}? \\ $$$${Some}\:{sort}\:{of}\:{approximation}\:{using}\:{the}\:{normal}\:{distribition}\:{is}\:{needed}\:{but}\:{I}'{m} \\ $$$${unsure}\:{of}\:{the}\:{necessary}\:{procedure}. \\ $$$${Perhaps}\:{someone}\:{could}\:{suggest}\:{what}\:{book}\:{or}\:{exam}\:{papers}\:{I}\:{may}\:{use}\:{to}\:{improve} \\ $$$${my}\:{competence}\:{in}\:{noticing}\:{the}\:{equivalence}\:{of}\:{events}\:{in}\:{questions}\:{of}\:{this}\:{type}. \\ $$$$ \\ $$
Answered by prakash jain last updated on 11/May/15
Answer Part I: Equality of events Let us say 1st hit occurs at t_1 and 51st hit occurs att_(51) = t_1 +x, for time between if t_(51) −t_1 >60 then it is clear that only upto 50 hit can occur between t_1 and t_1 +60. The following 2 are equal 1. time between 1st hit and 51st hit >60 2. Number of event from t_1 to t_1 +60≤50 Now consider the following event: No of hit in 60 minutes ≤50. This event does not mention that the interval starting from t_1 is considered. So this is not the same statement. However, the given exponential distribution and poisson probability mass function depend only on the interval. So the event ′No of hits in 60 mins≤50′ is equal to the event ′time betwen 1st and 51st hit′ only for the given exponential distribution function.
$$\mathrm{Answer}\:\mathrm{Part}\:\mathrm{I}:\:\mathrm{Equality}\:\mathrm{of}\:\mathrm{events} \\ $$$$\mathrm{Let}\:\mathrm{us}\:\mathrm{say}\:\mathrm{1st}\:\mathrm{hit}\:\mathrm{occurs}\:\mathrm{at}\:{t}_{\mathrm{1}} \:\mathrm{and}\:\mathrm{51st}\:\mathrm{hit}\:\mathrm{occurs} \\ $$$$\mathrm{at}{t}_{\mathrm{51}} =\:{t}_{\mathrm{1}} +{x},\:\mathrm{for}\:\mathrm{time}\:\mathrm{between}\:\mathrm{if}\:{t}_{\mathrm{51}} −{t}_{\mathrm{1}} >\mathrm{60} \\ $$$$\mathrm{then}\:\mathrm{it}\:\mathrm{is}\:\mathrm{clear}\:\mathrm{that}\:\mathrm{only}\:\mathrm{upto}\:\mathrm{50}\:\mathrm{hit}\:\mathrm{can}\:\mathrm{occur} \\ $$$$\mathrm{between}\:{t}_{\mathrm{1}} \:\mathrm{and}\:{t}_{\mathrm{1}} +\mathrm{60}. \\ $$$$\mathrm{The}\:\mathrm{following}\:\mathrm{2}\:\mathrm{are}\:\mathrm{equal} \\ $$$$\mathrm{1}.\:\mathrm{time}\:\mathrm{between}\:\mathrm{1st}\:\mathrm{hit}\:\mathrm{and}\:\mathrm{51st}\:\mathrm{hit}\:>\mathrm{60} \\ $$$$\mathrm{2}.\:\mathrm{Number}\:\mathrm{of}\:\mathrm{event}\:\mathrm{from}\:{t}_{\mathrm{1}} \:\mathrm{to}\:{t}_{\mathrm{1}} +\mathrm{60}\leqslant\mathrm{50} \\ $$$$\mathrm{Now}\:\mathrm{consider}\:\mathrm{the}\:\mathrm{following}\:\mathrm{event}: \\ $$$$\mathrm{No}\:\mathrm{of}\:\mathrm{hit}\:\mathrm{in}\:\mathrm{60}\:\mathrm{minutes}\:\leqslant\mathrm{50}. \\ $$$$\mathrm{This}\:\mathrm{event}\:\mathrm{does}\:\mathrm{not}\:\mathrm{mention}\:\mathrm{that}\:\mathrm{the}\:\mathrm{interval} \\ $$$$\mathrm{starting}\:\mathrm{from}\:{t}_{\mathrm{1}} \:\mathrm{is}\:\mathrm{considered}.\:\mathrm{So}\:\mathrm{this}\:\mathrm{is}\:\mathrm{not} \\ $$$$\mathrm{the}\:\mathrm{same}\:\mathrm{statement}. \\ $$$$\mathrm{However},\:\mathrm{the}\:\mathrm{given}\:\mathrm{exponential}\:\mathrm{distribution}\: \\ $$$$\mathrm{and}\:\mathrm{poisson}\:\mathrm{probability}\:\mathrm{mass}\:\mathrm{function}\:\mathrm{depend} \\ $$$$\mathrm{only}\:\mathrm{on}\:\mathrm{the}\:\mathrm{interval}.\:\mathrm{So}\:\mathrm{the}\:\mathrm{event} \\ $$$$'\mathrm{No}\:\mathrm{of}\:\mathrm{hits}\:\mathrm{in}\:\mathrm{60}\:\mathrm{mins}\leqslant\mathrm{50}'\:\mathrm{is}\:\mathrm{equal}\:\mathrm{to}\:\mathrm{the}\:\mathrm{event} \\ $$$$'\mathrm{time}\:\mathrm{betwen}\:\mathrm{1st}\:\mathrm{and}\:\mathrm{51st}\:\mathrm{hit}'\:\mathrm{only}\:\mathrm{for}\:\mathrm{the} \\ $$$$\mathrm{given}\:\mathrm{exponential}\:\mathrm{distribution}\:\mathrm{function}.\: \\ $$$$ \\ $$
Commented by Yugi last updated on 11/May/15
I see now. Thanks for the explanation !
$${I}\:{see}\:{now}.\:{Thanks}\:{for}\:{the}\:{explanation}\:! \\ $$
Answered by prakash jain last updated on 11/May/15
Answer Part II: Probability of k events in interval T If the time between successive event is given by probability distibution function=λe^(−λt) then the probability function for k events in interval T is (((λT)^k e^(−λT) )/(k!)) In the question you are required to find probability of upto 50 events. You can calculate that using the above formula.
$$\mathrm{Answer}\:\mathrm{Part}\:\mathrm{II}:\:\mathrm{Probability}\:\mathrm{of}\:{k}\:\mathrm{events}\:\mathrm{in} \\ $$$$\mathrm{interval}\:{T} \\ $$$$\mathrm{If}\:\mathrm{the}\:\mathrm{time}\:\mathrm{between}\:\mathrm{successive}\:\mathrm{event}\:\mathrm{is}\:\mathrm{given} \\ $$$$\mathrm{by}\:\mathrm{probability}\:\mathrm{distibution}\:\mathrm{function}=\lambda{e}^{−\lambda{t}} \\ $$$$\mathrm{then}\:\mathrm{the}\:\mathrm{probability}\:\mathrm{function}\:\mathrm{for}\:{k}\:\mathrm{events} \\ $$$$\mathrm{in}\:\mathrm{interval}\:{T}\:\mathrm{is}\:\frac{\left(\lambda{T}\right)^{{k}} {e}^{−\lambda{T}} }{{k}!} \\ $$$$\mathrm{In}\:\mathrm{the}\:\mathrm{question}\:\mathrm{you}\:\mathrm{are}\:\mathrm{required}\:\mathrm{to} \\ $$$$\mathrm{find}\:\mathrm{probability}\:\mathrm{of}\:\mathrm{upto}\:\mathrm{50}\:\mathrm{events}.\:\mathrm{You} \\ $$$$\mathrm{can}\:\mathrm{calculate}\:\mathrm{that}\:\mathrm{using}\:\mathrm{the}\:\mathrm{above}\:\mathrm{formula}. \\ $$

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