Answer: 0.357
Step-by-step explanation:
Given : The number of Hatfield = 6
The number of McCoys = 2
The number of companies = 8
The number of construction jobs -3
Now, the required probability is given by :-
[tex]\dfrac{^6C_3\times^2C_0}{^8C_3}\\\\=\dfrac{\dfrac{6!}{3!(6-3)!}}{\dfrac{8!}{3!(8-3)!}}=0.357142857143\approx0.357[/tex]
Hence, the probability that all 3 jobs go to Hatfields =0.357
