Answer :
The probability that 6 or 7 of the employees are right handers are 0.50436,the probability that at most two of the employees are left handers is 0.9 and the expected number of employees who are left handers is 3.
Given that there are 75% right hander employees in a company.
We are required to find the probability that 6 or 7 of the employees are right handers, the probability that at most two of the employees are left handers.
Probability is the calculation of likeliness of happening an event among all the events possible. It lies between 0 and 1.
Probability=Number of items/ Total number of items
Probability that the employee will be selected if 1 employee is selected from all the employees=75/100=3/4
a)Probability that 6 or 7 of the employees are right handers=
=9[tex]C_{6}(0.75)^{6}(0.25)^{3}[/tex]+9[tex]C_{7}(0.75)^{7} (0.25)^{2}[/tex]
=84*0.17*0.015+36*0.13*0.062
=0.2142+0.29016
=0.50436
b) Probability that at most two of the employees are left handers=
=[tex]9C_{0}(0.25)^{0} (0.75)^{9}+9C_{1}(0.25)^{1} (0.75)^{8}+9C_{2}(0.25)^{2} (0.75)^{7}[/tex]
=0.075+0.225+0.6
=0.9
c) The expected number of employees who are left handers=9*0.25=2.25
Approximately 3 employees.
Hence the probability that 6 or 7 of the employees are right handers are 0.50436,the probability that at most two of the employees are left handers is 0.9 and the expected number of employees who are left handers is 3.
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