A certain Bookstore that sells A Million books wants to hire you, and you get the job if you can answer these problems correctly. Suppose that the number of books that the store sells each minute has a Poisson distribution with a known rate of 1.3 books per minute. a) What's the probability that the store will sell 2 books in the next minute? b) What's the probability that the store will sell less than 4 books in the next minute? c) What's the probability that the store will sell at least 3 books in the next minute given that it sells at least 2 books in the next minute?

a) P ( X = 2 ) = 0.23028

b) P ( X < 4 ) = 0.95689

c) P ( X ≥ 3 | X ≥ 2 ) = 0.38292

Step-by-step explanation:

Given:-

- The parameter for the poisson distribution is given, λ = 1.3.

- Declare a random variable (X) which is the number of books sold in the next minute:

                               X ~ Po (1.3)

Find:-

a) What's the probability that the store will sell 2 books in the next minute? b) What's the probability that the store will sell less than 4 books in the next minute? c) What's the probability that the store will sell at least 3 books in the next minute given that it sells at least 2 books in the next minute?

Solution:-

a) The required probability P ( X = 2 ). Can be computed by using the pmf for the poisson distribution:

Where, "k" is the number of books sold in next minute.

- Now compute P ( X = 2 ) :

b) The required probability P ( X < 4 ). Can be computed by using the pmf for the poisson distribution and summing individual terms from 0 - 3:

c) The required probability P ( X ≥ 3 | X ≥ 2 ). We have to consider the conditional probability as follows:

                   P ( X ≥ 3 | X ≥ 2 ) = P ( X ≥ 3 & X ≥ 2 ) / P (X ≥ 2 )

                                               = P ( X ≥ 3 ) / P (X ≥ 2 )

                                               = P ( X > 2 ) / P ( X > 1 )

                                               = [ 1 - P ( X ≤ 2 ) ] / [ 1 - P ( X ≤ 1 ) ]  

- Compute P ( X ≤ 2 ) & P ( X ≤ 1 ) using pmf:

                     P ( X ≤ 2 ) = 0.27253 + 0.35429 + 0.23028

                                      = 0.8571

                     P ( X ≤ 1 ) = 0.27253 + 0.35429

                                      = 0.62682

- Use the expression developed for conditional probability to evaluate the required probability:

                     P ( X ≥ 3 | X ≥ 2 ) = [ 1 - P ( X ≤ 2 ) ] / [ 1 - P ( X ≤ 1 ) ]

                                                  = [ 1 - 0.8571 ] / [ 1 - 0.62682 ]

                                                  = 0.38292

answer:

it is general formula that is used to write value to its excet position

a11 a12 a13

a21 a22 a23

it is the example of order 2 by 3 matrix in this

a11 means value occur in 1 st row 1st column

a12 means value occurs in 1st row 2nd column

a13 means value occur in 1st row and 3rd column

a21 means value occurs in 2nd row and 1st column

a22 means value occurs in 2nd row and in 2nd column

a23 means value occurs in 2nd row and in 3rd column