Like

Report

A batch of 50 spare parts contains 7 defective spare parts.

a. If two spare parts are drawn randomly by one at a time without replacement, what is the probability that both spare parts are defective?

b. If this experiment is repeated, with replacement, what is the probability that both spare parts are defective?

No Related Subtopics

You must be signed in to discuss.

So we have ah, box of 50 parts, seven of which are defective. And we're calculating two different probabilities. Um, well, they're both the probability of two defective parts strong to defective parts one at a time. However, it's slightly different because the first experiment, we're not replacing the parts. And the second time, we are so doing this experiment twice. So we started the first probability when we're doing it without replacement. So we want to defective parts basically in a row, we pick one defective part, and then we pick another defective part. What is the probability of that happening? Well, on our first draw, um, the boxes full, we haven't taken anything out, So we still have seven defective parts out of 50 little parts will have seven out of 50. Now for our second draw. We are not replacing. So we took a part out, so we now have one less part. So that's how we get 49 in our denominator. And we're assuming we drew a defective part because we want the probability of two defective parts. So we drew already drew a defective parts of those will actually one last effective part as well, so we'll have six out of 40. Nice. There's one less in each, the numerator and the denominator, and then we'll multiply those together. And so our probability of two defective parts without replacement is 42 over 2450. Now, this probability is going to change. Would be are replacing. So now the probability of drawing to defective parts while our first draws still seven out of 50 because we haven't drawn anything yet. So everything is still in there. So we stopped the draw a defective part. And now we put that effective part back, right? We're replacing it. So when we draw it again, everything is still the same. We still have seven parts out of a total of 50. So now our probability becomes 49 over 25. That was Excuse me, 2500. And so we'll see. If we convert these two percentages, we can see how they differ. So this first percentages 0.17 and the 2nd 1 is 10196

Northern Arizona University

No Related Subtopics