The probability that a customer of a network operator has a problem about you needing technical staff's help in a month is 0.01. This operator installs internet for 500 households in a residential area a, Calculate the average number of households in this residential area having internet problems in a certain month b, Calculate the probability that in 6 consecutive months there is only one month that no customer in this area has a network problem that needs the help of technical staff​

(a) average calls = 5  

(b) probability that there is exactly one call in 6 consecutive monts = 0.038

Step-by-step explanation:

Let event of a customer requiring help in a particular month = H

and event of a customer not requiring help in a particular month = ~H

Given

p= 0.01,  therefore

Number of households, n = 500.

Binomial distribution:

x = number of households requiring help in a particular month

P(x,n,p) = C(x,n)*p^x*(1-p)^(n-x)

where

C(x,n) = n!/(x!(n-x)!) is the the number of combinations of x objects out of n

(a) Average number of households requiring help = np = 500*0.01 = 5

(b)

Probability that there are no calls requiring help in a particular month

P(0), q= C(0,n)*p^0(1-p)^(n-0)

= 1*1*0.99^500

= 0.006570483

Applying binomial probability over six months,

q = 0.006570483

n = 6

x = 1

P(x,n,q)

= C(x,n)*q^x*(1-q)^(n-x)

= 6!/(1!*5!) * 0.006570483^1 * (1-0.006570483)^5

= 0.038145

Therefore the probability that in 6 consecutive months there is exactly one month that no customer has called for help = 0.038

what is the information provided?

step-by-step explanation:

ok thats very interesting