Respuesta :
A.
The number of different ways the computers can be chosen is given by a combination of 10 choose 4.
A combination of n choose p is given by the formula below:
[tex]C(n,p)=\frac{n!}{p!(n-p)!}[/tex]So we have:
[tex]C(10,4)=\frac{10!}{4!(10-4)!}=\frac{10\cdot9\cdot8\cdot7\cdot6!}{4\cdot3\cdot2\cdot6!}=210[/tex]B.
If the first computer chosen is the one defective, the probability of the first PC being defective is 4/10, the probability of the second one not being defective will be 6/9, for the third not being defective is 5/8 and for the fourth not being defective is 4/7.
Since the defective PC can be any of the 4 bought, we need to multiply the probability above by 4. So the final probability is:
[tex]P=4\cdot\frac{4}{10}\cdot\frac{6}{9}\cdot\frac{5}{8}\cdot\frac{4}{7}=0.3809[/tex]