Answer: 0.357
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 :-
![(^6C_3*^2C_0)/(^8C_3)\\\\=((6!)/(3!(6-3)!))/((8!)/(3!(8-3)!))=0.357142857143\approx0.357](https://img.qammunity.org/2020/formulas/mathematics/college/rn19x38hybf51t55s2wtuxuuyuuzkwcjsn.png)
Hence, the probability that all 3 jobs go to Hatfields =0.357