Question

A container has 50 electronic components, of which 10 are defective. If 8 components are drawn at random from the container, what is the probability that exactly 3 of them are defective?
A. 0.147
B. 0.203
C. 0.300
D. 0.375
E. 0.750

Answer 1

noted

Answer 2

I edited the question with multiple choice.

Answer 3

that was a mistake, sorry
you are choosing 8 out of 50 and the number of ways to do that is \(\binom{50}{8}\) that is your denominator for the probability

Answer 4

of those 8, 3 are defective and so evidently 5 are not
the number of ways to get 3 defective out of 10 is \(\binom{10}{3}\) and the number of ways to get 5 out of the 40 that are not defective is \(\binom{40}{2}\)

Answer 5

I am confused, I did not see a denominator is that what I'm suppose to solve?

Answer 6

final answer is
\[\huge \frac{\binom{10}{3}\binom{40}{5}}{\binom{50}{8}}\]

Answer 7

are you familiar with this notation for "50 choose 8" or do you write it as \(_{50}C_8\)?

Answer 8

No I am not familiar with that notation

Answer 9

What exactly is the answer then?????