Question

There are 17 broken light bulbs in a box of 100 light bulbs. A random sample of 3 light bulbs is chosen without replacement.

How many light bulbs contain exactly exactly 1 broken light bulb?

Answer 1

Please expand: \((p+q)^{3}\) Trust me on this.

Answer 2

p^3+3qp^2+3pq^2+q^3

Answer 3

That's pretty good.

p = Probability that a light is working
q = Probability that a light is broken = 1-p

Can you now solve the problem?

Answer 4

No >.< . Besides, I was asked to use combinations/Permutations .

Answer 5

Why didn't you say so?

You need to:

Select 2 of 83 good.
Select 1 of 17 bad.
from a selection of 3 from 100.

Answer 6

Sorry :P .

Answer 7

Yeah that's what's confusing me. Why is it 2 from 83? 83 are the bulbs that are NOT broken.

Answer 8

1 bad means 2 good to go with the bad one.

Answer 9

Ahh makes sense!

Answer 10

So really it would be (83 C 2)*(17 C 1)

Answer 11

If i'm correct.

Answer 12

Yep. That is indeed correct. Thanks a lot!

Answer 13

That's actually incorrect :P . . That would be how many samples contain no broken light bulbs.

Answer 14

83*82*81/100*99*98=30627/53900

Answer 15

one is defective so (1-30627/53900)

Answer 16

23273/53900

Answer 17

The binomial approximation that I tried to get you to do up front suggests the correct probability is in the neighborhood of \(3\cdot0.83^{2}\cdot 0.17 = 0.351\)

You should try: \(\dfrac{ _{83} C_{2}*_{17} C_{1} }{_{100} C_{3}}\)

Answer 18

0.4317 is it correct

Answer 19

No. Please be more systematic and make an attempt to explain what it is you are doing.

Answer 20

p(none is defective)=83c3/100c3=30627/53900
p(one is defective)=1-30627/53900

Answer 21

How about p(2 are defective) and p(3 are defective)?

Answer 22

exactly exactly 1 broken light bulb?

Answer 23

No, p(none broken) + p(1 broken) is NOT unity (1).

No, p(none broken) + p(ANY broken) is unity (1). This is what you calculated, ANY, not ONE.

Answer 24

p(0 broken) + p(1 broken) + p(2 broken) + p(3 broken) = p(0 broken) + p(any broken)

Answer 25

Please be more systematic and make an attempt to explain what it is you are doing.

Answer 26

No. That is now farther off than it was before. Look a few posts up. The answer is given correctly as (83C2)(17C1)/(100C3)

Answer 27

p one is defective=17/100*83/99*82/98

Answer 28

Okay, now you're just writing things down. Please stop doing it. Go look at the correct answer and learn from it. Stop trying to guess that it might be something else. In this case, you have entirely forgotten that the three can be drawn in any order.

Answer 29

Really? Then why does it give an incorrect answer?

Take a very hard look at the binomial approximation.

p^3+3qp^2+3pq^2+q^3

This would be exact for WITH replacement. Since there are so few drawn, from such a large population, this is very close and very clear. In particular, why are those middle coefficients "3" instead of "1"?

Answer 30

but u should add the answer i think p(none is defective)=83c3/100c3=30627/53900
+P(one defective)=17/100*83/99*82/98+83/100*82/99*17/98+83/100*17/99*82/98

Answer 31

i.e 0.568+0.378=0.9462

Answer 32

And 1-0.9462 = p(2) + p(3)