The probability density function of the time customers arrive at a terminal (in minutes after 8:00 A.M) is(adsbygoogle = window.adsbygoogle || []).push({});

f(x)= (e^(-x/10))/10 for 0 < x

c) Determine the probability that:

two or more customers arrive before 8:40 A.M among five that arrive at the terminal. Assume arrivals are independent

my logic is the following:

Probability= 1-Probability 0 or 1 customers arrive before 8:40 A.M

the answer is the following:

P(X1>40)+ P(X1<40 and X2>40)= e-4+(1- e-4) e-4= 0.0363

from what is written above, it seems to be the probability that no one arrives before 8:40 P(X1>40) and the probability that one arrives before 8:40 (X1<40) and another arrives after 8:40 (X2>40).

i tihnk i just need some help on understanding why X2 is brought in.

Thanks!

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# Continuous random variable (stats)

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