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Problem 6 Easy Difficulty

Defective Resistors A package contains 12 resistors, 3 of which are defective. If 4 are selected, find the probability of getting
$$
\begin{array}{l}{\text { a. } 0 \text { defective resistors }} \\ {\text { b. } 1 \text { defective resistor }} \\ {\text { c. } 3 \text { defective resistors }}\end{array}
$$

Answer

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Chapter 4

Probability and Counting Rules

Section 5

Probability and Counting Rules

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Video Transcript

Now, suppose we have 12 resistors and three of them are defective on DNA. Now we want to select four out of 12 and we want to know the probability for the following three situations. Um, so let's look at part A when there are no defective resistor out out of the four selected. And the probability of this should be the number of choices we have to select for resisters out of the nun the night non effective ones. So this should be nine choose for divided by the total number of choices we have. So this should be 12 choose for and this equals 126. Divided by 495 equals 0.255 Yeah, And that in part B, we wanna select one defective resistor, and the rest three are non defective ones. So this is actually a two step process. The first step. We want to select the defective ones, which should be three. Choose one, because we choose one out of three defective ones at times. Um, we want to choose 39 defective ones out of night, so this should be nine shoes three still divided by 12 Choose four, and this equals 84 times. We divided by 495 equals 0.5 or nine. And next in part C, we wanna select three defective, um, resistors out of out of three and select another one from the 99 defective ones. So first we choose three out of three. And then times, yeah, nine. Choose One divided by 12 chose for And this should be one times and I divided by 495 and this goes to point 01 night.

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