Question

4. [5 marks] Suppose that Best Buy sells 4 TVs per day on average.

a) [1 marks] What is the probability that 8 TVs will be sold in a day?

b) [2 marks] What is the probability that fewer than 3 TVs (inclusive) will be sold in a day?

c) [2 marks] Suppose 5 customers enter Best Buy independently. The probability that a single customer will buy a TV is 0.1. What is the probability that at least one of these 5 customers buys a TV? Do not round your answer.

Answer 1

Poisson distribution :

Poisson random variable define number of event occurs in an interval of event per unit.

If x follows Poisson distribution with rate then probability distribution function is,

, x = 0 , 1 , 2 , 3 , ...

a)

Let , X be the number of TVs will be sold in a day.

Average sell of TVs in day = = 4

X follows Poisson distribution with rate   = 4

We have to find  probability that 8 TVs will be sold in a day

i.e P( X = 8 )

Using Formula of pdf of Poisson distribution,

{ = 0.018315639 }

The probability that 8 TVs will be sold in a day is 0.0298

b)

We have to find probability that fewer than 3 TVs (inclusive) will be sold in a day

i.e P( X <= 3 )

The probability that fewer than 3 TVs (inclusive) will be sold in a day is 0.4335

c)

Suppose 5 customers enter Best Buy independently. i.e n = 5

The probability that a single customer will buy a TV = p = 0.1

Let , X be the number of customers buys a TV

X follows Binomial distribution with n = 5 and p = 0.1

We have to find probability that at least one of these 5 customers buys a TV.

I.e P( x >= 1 )

P( x >= 1 ) = 1 - P( x = 0 )

Using Binomial distribution formula,

  { nCx = n! / x!(n-x)! }

So, P( x >= 1 ) = 1 - 0.59049 = 0.40951