A factory produces fuses, which are packaged in boxes of 17. Three fuses are selected at random from each box for inspection. The box is rejected if at least one of these three fuses is defective. What is the probability that a box containing two defective fuses will be rejected?
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To calculate the probability that a box containing two defective fuses will be rejected, we can use the concept of conditional probability.
Step 1: Determine the total number of possible outcomes.
Since there are three fuses selected from each box, and each fuse can either be defective (D) or non-defective (ND), there are 2^3 = 8 possible outcomes for the selection of three fuses.
Step 2: Determine the number of favorable outcomes.
In this case, the box will be rejected if at least one of the three selected fuses is defective. So, let's count the number of outcomes where at least one fuse is defective.
Let's consider the situation where two out of three fuses are defective:
There are three possible positions for the two defective fuses (indicated by D), and one position for a non-defective fuse (indicated by ND).
DDND, DNDN, NDDD
Therefore, there are 3 favorable outcomes when exactly two out of three fuses are defective.
Step 3: Calculate the probability.
The probability that a box containing two defective fuses will be rejected is the ratio of the number of favorable outcomes (3) to the total number of possible outcomes (8):
P(rejected box | 2 defective fuses) = 3/8 = 0.375
So, the probability is 0.375 or 37.5%.