Question

A box contains 20 lightbulbs, of which 5 are defective.
If 4 lightbulbs are picked from the box randomly, the probability that at most 2 of them are defective is ?

Answer 1

@hero

Answer 2

at most two means
0,
1, or
2
you have to compute each of these separately and add them up

Answer 3

the probability that none are defective would be \[\frac{15}{20}\times \frac{14}{19}\times \frac{13}{18}\times \frac{12}{17}\] there may be a more elegant way to do this problem but i can't think of it
or maybe do \[\frac{\binom{15}{4}}{\binom{20}{4}}\]

Answer 4

you seen any of this before?

Answer 5

No, I don't understand math at all. That's why I'm repeating the class.

Answer 6

this question is quite involved, it will take a lot of doing
are you up for it?

Answer 7

Sure. I literally have no choice if i want to be successful, right?

Answer 8

ok so first, have you ever seen this notation before ?

Answer 9

\[\frac{\binom{15}{4}}{\binom{20}{4}}\]

Answer 10

No, never.

Answer 11

how about this \[_{15}C_4\]

Answer 12

it is read "fifteen choose four" are you familiar with that ?

Answer 13

I've seen it before but I've never used it. This is all new to me.

Answer 14

it is the number of ways to choose 4 out of a set of 15
if you have 20 light bulbs, 5 of which are defective, then there are 15 that are not defective
that is clear right?

Answer 15

this is where you say "yes, that is obvious"

Answer 16

yes, sorry. I wasn't at my laptop

Answer 17

ok so the question is, if there are 15 that are not defective, how many ways can you choose 4 of them
the answer is 15 choose 4 computed via \[_{15}C_4=\frac{15\times 14\times 13\times 12}{4\times 3\times 2}\] it is a whole number, so cancel first, multiply last
or else use wolfram

Answer 18

http://www.wolframalpha.com/input/?i=(15+choose+4)

Answer 19

Okay. You've completely lost me. I'm sorry, I've been up for days and working at this for hours. I have to have it done by the 30th this month so late nights, early morning, or just plain no sleep at all. I just never got algebra anyway. Anyway, you've completely lost me there.

Answer 20

then i suggest you give this a rest, because this is only one tiny part of the question, which, as i said, is fairly involved
you have to compute \({15}C_4\) but also \({20}C_4\) and that is just for the probability you get 0 defective ones, so that is only a start

Answer 21

by the way, this is not algebra, it is probability, which involves mostly arithmetic if you understand how to do the problem

Answer 22

I'd love some sleep, but I'm tired of my family griping about my "failing classes and grades". But thank you for helping with what you could. I'll tag you in the morning for some more help if that's okay.

Answer 23

sure
and good luck
just for my own information, how were you going to try to solve this question?

Answer 24

I have no idea, but all the answers that I guessed on were wrong and it doesn't give me any other choices.

the answers they gave me to choose from were:
31/969
70/323
91/323
938/969

Answer 25

then lets just do it, it can't take more than ten minutes

Answer 26

Okay, that works. Thank you

Answer 27

the total number of ways to pick 4 out of 20 is called 20 choose 4
the total number of ways to pick 4 out of 15 is called 15 chose 4
to the probability you get no defective ones is 15 choose 4 over 20 choose 4
forget about how to compute this, use technology
http://www.wolframalpha.com/input/?i=(15+choose+4)%2F(20+choose+4)

Answer 28

copy and paste that, we need that number
now we compute the number of ways to get 1 defective one
the denominator will still be 20 choose 4
the numerator will be 15 choose 3 times 5 chose 1 (3 not defective, one defective)

Answer 29

here is that number
http://www.wolframalpha.com/input/?i=(15+choose+3)*(5+choose+1)%2F(20+choose+4)

Answer 30

now the probabilty you get two defecttive
15 choose 2 times 5 chose 2 over 20 choose 4

Answer 31

here is that number
http://www.wolframalpha.com/input/?i=(15+choose+3)*(5+choose+1)%2F(20+choose+4)

Answer 32

Hey, this is her teacher. Even I don't know what I'm doing, could you please just give us the answer so this can be over with. (Also, the mother.)

Answer 33

now we have to add them up so lets let wolfram do it
http://www.wolframalpha.com/input/?i=((15+choose+4)%2B(15+choose+3)(5+choose+1)+%2B(15+choose+2)(5+choose+2))%2F(20+choose+4)

Answer 34

copy and paste, the answer is there

Answer 35

teacher?

Answer 36

Yes. I'm "BeautifulAngel's" teacher and mother.

Answer 37

What am I copying and pasting?

Answer 38

http://www.wolframalpha.com/input/?i=((15+choose+4)%2B(15+choose+3)(5+choose+1)+%2B(15+choose+2)(5+choose+2))%2F(20+choose+4)
home school?

Answer 39

Note to self, fix the url parser so it does not break with ('s.

Answer 40

i think that can't be helped

Answer 41

by which i mean i think it did the same thing in OS

Answer 42

No, she goes to a public school. But I teach senior honors and cambridge math. She goes to school in Tennessee.

Answer 43

wolfram says \[\frac{938}{969}\] which seems high to me, but whatever

Answer 44

I can fix it though lol I just didn't account for it while writing the function that verifies a url

Answer 45

don't know cambridge math
heard of cambridge though...