Question

The probability of an employee receiving the Superstar of the Month Award for a large company is 22%. What is the probability of exactly 10 employees receiving the award out of one department with 30 employees

Answer 1

0.055

Explanation:

We have been given that the probability of an employee receiving the Superstar of the Month Award for a large company is 22%. We are asked to find the probability of exactly 10 employees receiving the award out of one department with 30 employees.

We will use Bernoulli's trails to solve our given problem.

`P(X=x)=^nC_x\cdot p^x(1-p)^(n-x)`

Upon substituting our given values, in above formula, we will get:

`P(X=10)=^(30)C_(10)\cdot 0.22^(10)(1-0.22)^(30-10)`

`P(X=10)=(30!)/(10!(30-10)!)\cdot 0.22^(10)(0.78)^(20)`

`P(X=10)=(30*29*28*27*26*25*24*23*22*21*20!)/(10*9*8*7*6*5*4*3*2*20!)\cdot 0.0000002655992279\cdot 0.0069485158708622`

`P(X=10)=29*3*13*5*23*11*21\cdot 0.00000000184552045035189`

`P(X=10)=30045015\cdot 0.00000000184552045035189`

`P(X=10)=0.055448689613`

`P(X=10)\approx 0.055`

Therefore, the probability that exactly 10 employees will receive the award out of one department with 30 employees, would be 0.055.