A shipment with 25 computers contains 3 without all their accessories. What is the probability that at least one is found without all its attachments? A shipment with 25 computers contains 3 without all their accessories (cables or mouse). 5 without substitution are chosen for inspection.
What is the probability that at least one is found without all its attachments?
I think its $\frac{\binom51+ \binom52 + \binom53}{\binom{25}5}$
 A: Probability of all of the 5 are good is $\frac{C_5^{22}}{C_5^{25}}$, probability of at least 1 without all accessories is $1-\frac{C_5^{22}}{C_5^{25}} \approx 0.504$.
A: Alternative approach, consistent with OP's approach:
$$\frac{\left[\binom{3}{1} \times \binom{22}{4}\right]
+ \left[\binom{3}{2} \times \binom{22}{3}\right]
+ \left[\binom{3}{3} \times \binom{22}{2}\right]}
{\binom{25}{5}}$$
A: What you have done in your numerator is choose positions for the defective items out of the 5 draws, so let's build on that. The denominator will be the total number of ways to pick $5$ ordered items, which is $25*24*23*22*21$. For each of the numerator items, you will need to multiply by the number of defectives of the number of spots you have chosen for them, sequentially, as well as the same thing for the number of non-defectives.
$$\frac{{5\choose 1}3*22*21*20*19+{5\choose 2}*3*2*22*21*20+{5\choose 3}*3*2*1*22*21}{25*24*23*22*21}$$
user2661923's approach (+1) is the 2nd way of doing it, and probably the preferred way of the two. BStar's approach is clever too.
