Answer:
Step-by-step explanation:
We have 15 diesel engines, from which 13 are okay, and 2 are defected. Since there are 15 engines and 13 will be chosen two ship in the continer, the possibe number of outcomes is:
[tex]C(n,r) = \frac{n!}{r!(n-r)!}[/tex]
Where n = 15 and r=11.
Substituting, we find that:
[tex]C(15,11) = \frac{15!}{13!(15-13)!} = \frac{15!}{13!2!} = 105 [/tex]
Now we need to find the probability of choosing 13 effective engines and 0 defective engines, which is given by:
[tex]C(13,13) C(2,0) = \frac{13!}{13!0!} \frac{2!}{0!2!} = 1[/tex]
The probability of shipping the container is: [tex]\frac{1}{105}[/tex]
