91Ó°ÊÓ

First, we need to find the total number of combinations when picking 2 cartridges from the shelf. We are given that there are 80 cartridges in total. The formula for combinations is given by: \(C(n,r) = \frac{n!}{(n-r)!r!}\), where n represents the total number of items, r represents the number of items being selected, and ! denotes the factorial function. Here, we are given that \(n = 80\) and we are selecting 2 cartridges, so \(r = 2\). Plugging these values into the formula, we get: \(C(80, 2) = \frac{80!}{(80-2)!2!} = \frac{80!}{78!2!} = \frac{80 \times 79}{2} = 3160\) So, there are 3160 possible combinations when selecting 2 cartridges from the shelf.

Next, we need to find the number of combinations when both selected cartridges are defective. We are given there are 6 defective cartridges. Therefore, we need to find the combination of 2 from these 6 defective cartridges. Using the combination formula again, we get: \(C(6, 2) = \frac{6!}{(6-2)!2!} = \frac{6!}{4!2!} = \frac{6 \times 5}{2} = 15\) So, there are 15 combinations where both selected cartridges are defective. Now, using the total number of combinations (3160) calculated in step 1, we can find the probability that both selected cartridges are defective: \(P(\text{both defective}) = \frac{\text{number of combinations with both defective}}{\text{total number of combinations}} = \frac{15}{3160} = \frac{3}{632}\) Therefore, the probability that both selected cartridges are defective is \(\frac{3}{632}\).

To find the probability that at least 1 cartridge is defective, we can first calculate the probability that both selected cartridges are not defective (meaning both are non-defective) and then subtract that from 1 because the probability of all outcomes should always sum to 1. We have 80 total cartridges, 6 of which are defective, so there are 74 non-defective cartridges. To find the number of combinations where both selected cartridges are non-defective, we need to find the combination of 2 from these 74 non-defective cartridges using the combination formula: \(C(74, 2) = \frac{74!}{(74-2)!2!} = \frac{74!}{72!2!} = \frac{74 \times 73}{2} = 2701\) So, there are 2701 combinations where both selected cartridges are non-defective. Now, using the total number of combinations (3160) calculated in step 1, we can find the probability that both selected cartridges are non-defective: \(P(\text{both non-defective}) = \frac{\text{number of combinations with both non-defective}}{\text{total number of combinations}} = \frac{2701}{3160} \) Now, we can find the probability that at least one selected cartridge is defective by subtracting the probability that both are non-defective from 1: \(P(\text{at least 1 defective}) = 1 - P(\text{both non-defective}) = 1 - \frac{2701}{3160} = \frac{459}{3160}\) Therefore, the probability that at least 1 selected cartridge is defective is \(\frac{459}{3160}\). In conclusion, the probability that: a. Both selected cartridges are defective is \(\frac{3}{632}\). b. At least 1 selected cartridge is defective is \(\frac{459}{3160}\).