Probability of Defective Electric Toothbrushes

E

3/20*2/19 = 3/190

To solve this problem, we need to calculate the probability that the first two electric toothbrushes sold will be defective.

Let's break down the problem step by step:

Step 1: Determine the total number of toothbrushes The problem states that there are three defective toothbrushes and 17 nondefective toothbrushes, resulting in a total of 20 toothbrushes (3 + 17 = 20).

Step 2: Calculate the probability of selecting a defective toothbrush first Since there are three defective toothbrushes out of a total of 20, the probability of selecting a defective toothbrush first is 3/20.

Step 3: Calculate the probability of selecting a defective toothbrush second After selling the first toothbrush, there will be two defective toothbrushes remaining out of the 19 remaining toothbrushes. Therefore, the probability of selecting a defective toothbrush second is 2/19.

Step 4: Calculate the overall probability To find the probability of both events happening (selling two defective toothbrushes in a row), we multiply the probabilities of each event. Therefore, the overall probability is (3/20) * (2/19) = 6/380 = 3/190.

So, the correct answer is option E: 3/190 or approximately 0.01579.

Please note that this explanation assumes that the toothbrushes are selected without replacement, meaning that once a toothbrush is sold, it is not put back into the pool of available toothbrushes.