Question

) Four items are taken at random from a box of 12 items and inspected. The box is rejected if more than 1 item is found to be faulty. If there are 3 faulty items in the box, find the probability that the box is accepted.

Answer 1

We can use combination ($C^n_k$ ) which shows how may ways to pick n items from a collection of k items, when the order of selection does not matter. $C^k_n=n!/(k!(n-k)!)$.

We can choose 4 items from 12 in $C^4_{12}=12!/(4!*8!)=495$ ways. This is a total number of different variants.

Now we should find how many ways to pick 4 good items or 3 good items and 1 faulty item and sum them:

We can choose 4 good items in $C^4_{9}=9!/(4!*5!)=126$ ways (because we have only 9 good items in the box and we pick 4 of them).

We can choose 3 good items and 1 faulty item in in $C^3_9*C^1_3=9!/(3!*6!)*3!/(1!*2!)=252$ ways (because we picks 3 good items out of 9 and 1 faulty item out of 3).

Thus, the probability that the box is accepted:

$(C^4_{9}+C^3_9*C^1_3)/C^4_{12}=(126+252)/495=0.7636$