A box contains 16 transistors, 4 of which are defective. If 4 are selected at random, find the probability of the statements below. a. All are defective b. None are defective a. The probability is ◻. (Type a fraction. Simplify your answer.) b. The probability is ◻. (Type a fraction. Simplify your answer.)

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Calculate probabilities: To find the probability of all 4 transistors being defective, we need to calculate the probability of choosing a defective transistor each time we select one, without replacement.First transistor defective: The probability of the first transistor being defective is 4 out of 16, or 4/16.Second transistor defective: After one defective transistor is chosen, there are 3 defective transistors left and 15 transistors in total. So, the probability of the second transistor being defective is 3/15.Third transistor defective: For the third transistor, there are now 2 defective transistors left out of 14 total transistors. The probability is therefore 2/14.Fourth transistor defective: Finally, for the fourth transistor, there is 1 defective transistor left out of 13 total transistors. The probability is 1/13.Product of probabilities: The probability of all 4 transistors being defective is the product of the individual probabilities: (4/16) * (3/15) * (2/14) * (1/13).Calculate probabilities: Calculating this product gives us (4/16) * (3/15) * (2/14) * (1/13) = 1/560.First transistor non-defective: Now, to find the probability of none of the transistors being defective, we need to calculate the probability of choosing a non-defective transistor each time we select one, without replacement.Second transistor non-defective: The probability of the first transistor being non-defective is 12 out of 16, or 12/16, since there are 12 non-defective transistors.Third transistor non-defective: After one non-defective transistor is chosen, there are 11 non-defective transistors left and 15 transistors in total. So, the probability of the second transistor being non-defective is 11/15.Fourth transistor non-defective: For the third transistor, there are now 10 non-defective transistors left out of 14 total transistors. The probability is therefore 10/14.Product of probabilities: Finally, for the fourth transistor, there are 9 non-defective transistors left out of 13 total transistors. The probability is 9/13.The probability of none of the transistors being defective is the product of the individual probabilities: (12/16) * (11/15) * (10/14) * (9/13).Calculating this product gives us (12/16) * (11/15) * (10/14) * (9/13) = 33/560.

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