Answer: a) 0.25, b) 0.78, c) 0.71
Step-by-step explanation:
Since we have given that
Number of employees in marketing = 4
Number of employees in management = 7
We need to hire committee of 3 people.
a) Find the probability that the committee has exactly 2 employees from marketing.
So, Probability becomes
[tex]\dfrac{^4C_2\times ^7C_1}{^{11}C_3}\\\\=\dfrac{42}{165}\\\\=0.25[/tex]
b) Find the probability that the committee has at least one employee from marketing.
[tex]P(x\geq 1)=1-P(x=0)\\\\P(x\geq 1)=1-\dfrac{^7C_3}{^{11}C_3}=1-0.212=0.78[/tex]
c) Find the probability that the committee has at most one employee from management.
[tex]P(x\leq 1)=P(x=0)+P(x=1)\\\\P(x\leq 1)=\dfrac{^7C_3}{^{11}C_3}+\dfrac{^7C_2\times ^4C_1}{^{11}C_3}\\\\P(x\leq 1)=0.21+0.50=0.71[/tex]
Hence, a) 0.25, b) 0.78, c) 0.71
