Full solution

Q. A box contains

16

transistors,

4

of which are defective. If

4

are selected at random, find the probability of the statements below.

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a. All are defective

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b. None are defective

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a. The probability is

\square

\newline

(Type a fraction. Simplify your answer.)

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b. The probability is

\square

\newline

(Type a fraction. Simplify your answer.)

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 $\frac{4}{16}$ .
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 $\frac{3}{15}$ .
Third transistor defective: For the third transistor, there are now $2$ defective transistors left out of $14$ total transistors. The probability is therefore $\frac{2}{14}$ .
Fourth transistor defective: Finally, for the fourth transistor, there is $1$ defective transistor left out of $13$ total transistors. The probability is rac{1}{13}.
Product of probabilities: The probability of all $4$ transistors being defective is the product of the individual probabilities: $(\frac{4}{16}) \times (\frac{3}{15}) \times (\frac{2}{14}) \times (\frac{1}{13})$ .
Calculate probabilities: Calculating this product gives us $(\frac{4}{16}) \times (\frac{3}{15}) \times (\frac{2}{14}) \times (\frac{1}{13}) = \frac{1}{560}$ .
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 $\frac{12}{16}$ , 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 $\frac{11}{15}$ .
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 $\frac{10}{14}.$
Product of probabilities: Finally, for the fourth transistor, there are $9$ non-defective transistors left out of $13$ total transistors. The probability is $\frac{9}{13}$ .
Product of probabilities: Finally, for the fourth transistor, there are $9$ non-defective transistors left out of $13$ total transistors. The probability is $\frac{9}{13}$ .The probability of none of the transistors being defective is the product of the individual probabilities: $\left(\frac{12}{16}\right) \times \left(\frac{11}{15}\right) \times \left(\frac{10}{14}\right) \times \left(\frac{9}{13}\right)$ .
Product of probabilities: Finally, for the fourth transistor, there are $9$ non-defective transistors left out of $13$ total transistors. The probability is $\frac{9}{13}$ .The probability of none of the transistors being defective is the product of the individual probabilities: $(\frac{12}{16}) \times (\frac{11}{15}) \times (\frac{10}{14}) \times (\frac{9}{13})$ .Calculating this product gives us $(\frac{12}{16}) \times (\frac{11}{15}) \times (\frac{10}{14}) \times (\frac{9}{13}) = \frac{33}{560}$ .