Question

# A. A bank receives on average 1.038 customers every 10 minutes. What is the probability that...

A. A bank receives on average 1.038 customers every 10 minutes. What is the probability that the bank receives 1 customers in the next hour?

B. A company is running adds on the internet. The probability of a reader of the add clicking on it is 0.81 and the probability that the clicking reader buys the product is 0.45. What is the probability of selling 6 or more products after displaying 10 ads?

We know that the average customers per 10 minutes is 1.038. Thus, the average per hour is 1.038*6= 6.228

Thus, lambda= 6.228

We know that the P(X=x)= (e-lambda * lambdax ) / x!

Substituting the value of lambda as 6.228 and x as 1, we get

P(X=1)= 0.01229

On displaying an ad, we know that the probability of selling the product is 0.81*0.45= 0.3645

Thus, probability of success,p= 0.3645

n=10

We need to find the probability that X>=6.

This is a binomial distribution.

We know

P(X=r)= nCr * pr * q n-r

Thus, P(X>=6)= P(X=6)+P(X=7)+P(X=8)+P(X=9)+P(X=10)

= 0.0803279549+0.0263275617+0.00566270431+0.0007217609+0.00004139762

= 0.11308137943

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