A factory produces fuses, which are packaged in boxes of 16. Three fuses are selected at random from each box for inspection. The box is rejected if at least one of the three fuses is defective. What is the probability that a box containing six defective fuses will be rejected?

p = probability defective = 6/16 = .375

1 - p = 10/16 = .625 = probability good

we do three trials on the box. They all must be good if it is to pass
probability of first trial fuse good = 10/16
probability of second choice being good fuse = 9/15
third = 8/14
so probability of passing = (10*9*8)/(16*15*14) = .0762
so probability of rejection = 1-.0762 = .924

arithmetic!

(10*9*8)/(16*15*14) = .2143
1 - .2143 = .7857

the probability of at least one being defective equals

1 minus the probability that all three are good

1 - (5/8)^3 = .756 (approx)

To calculate the probability that a box containing six defective fuses will be rejected, we can use the concept of the complement rule. The complement rule states that the probability of an event occurring is equal to one minus the probability of the event not occurring.

First, let's calculate the probability that a box with six defective fuses is not rejected. For the box to not be rejected, all three fuses selected for inspection must not be defective.

The probability of selecting a non-defective fuse from a box with six defective fuses is (10/16) since there are 10 non-defective fuses out of 16 total fuses in the box. We need to multiply this probability three times for three non-defective fuses:

(10/16) * (10/16) * (10/16) = (10/16)^3

Now, we can calculate the probability of the box being rejected by subtracting the probability of the box not being rejected from 1:

1 - (10/16)^3 ≈ 0.5781

Therefore, the probability that a box containing six defective fuses will be rejected is approximately 0.5781 or about 57.81%.