Question

PLease help me :(
A shipment of 25 televisions sets contains 3 defective units. in how many ways can a vending company purchase four of these units and receive a. all good units b. 2 good units c. at least 2 good units??????

Answer 1

this is a combination problem

Answer 2

yeah i got that

Answer 3

for a i got 7,315 is that correct?

Answer 4

Can you show your work?

Answer 5

ok.
22 C 4
22!/4!(22-4)!
22!/4!(18)!
=7, 315( i got this by reducing)

Answer 6

I agree with your answer for A.

Answer 7

P (x =4 good) = ( 22C4 * 3C0 ) / 25 C 4

Answer 8

that helps thanks

Answer 9

so is my answer correct?

Answer 10

Yes. They only asked the number of ways. Which is just 22c4, like you did. Chlorophyll showed you how to calculate the probability of choosing 4 good ones.

Answer 11

ok then could u possibly guide me through b?

Answer 12

or anyone?

Answer 13

please?

Answer 14

Sure.
You figured out that the number of ways to choose 4 good units was 22choose4.
How many ways could they choose 2 good units? And how many ways could they choose 2 bad units?

Answer 15

thats what i dont understand quite clearly

Answer 16

You're choosing 4 tvs. 2good and 2 bad. There are 22 total good tvs, so the number of ways I can choose 2 of them is just 22choose2

Answer 17

How many ways can I choose 2 bad?

Answer 18

22choose 2?

Answer 19

There aren't 22 bad tvs to choose from.

Answer 20

u c? i dont really like get this second part 2 the question...im a bit slow

Answer 21

How many bad tvs are there to choose from?

Answer 22

there are 2?

Answer 23

no 3

Answer 24

Right. They tell you that there are 3 bad tvs. And if there are 2 good units in your 4, then that means you've chosen 2 bad tvs from the 3 possible bad tvs.

Answer 25

3choose2 is the number of ways to have 2 bad tvs.

Answer 26

ok i get it so i would solve then 3choose2 3 being total of bad tvs and 2 being the good ones right?

Answer 27

Right.

Answer 28

Wait wait. You might be thinking of it wrong.

Answer 29

y? or how?

Answer 30

can u plz explain?

Answer 31

Sorry, OpenStudy was freezin' on me.

Answer 32

its fine thanks :) so then was i thinking of it wrong or right?

Answer 33

Haha it's hard to say. I'm not sure how you're thinking of it. Tell me, how would you try to calculate part b, number of ways two buy exactly 2 good units out of 4.

Answer 34

ummmm maybe 4choose 2?

Answer 35

is that right or wrong?

Answer 36

come on make math smooth lol

Answer 37

Wrong.
You're buying 4 tvs total. 2 of them are good. 2 of them are bad.
Number of ways to buy 2 good tvs? There are 22 total, you're choosing 2.
Number of ways to buy 2 bad tvs? There are 3 total, you're choosing 2.

Multiply these two results.

Answer 38

ooooooooooooooooooo that makes sense got it

Answer 39

so it would be 693 then? like for the answer?

Answer 40

Right.

Answer 41

thank u! :)

Answer 42

Now, for the last part, it asks number of ways to get at least 2 good units. At least 2 means 2, 3, or 4. You just found number of ways to get exactly 2. You also already calculated number of ways to get 4.
Just do number of ways to get exactly 3 good tvs, and add up the results.

Answer 43

My pleasure =D

Answer 44

so when i would do that i mean 2 get 3 good tvs i would then go just 22choose3? cuz if i try 2 multiply it by 3!/3! it would just b 1?

Answer 45

so am i right or wrong?

Answer 46

(22 choose 3) * (3 choose 1)

Answer 47

how is the second part (3choose1) possible or y?

Answer 48

If I'm choosing 4 tvs and 3 are good, then 1 is bad.
How many ways can I get 1 bad tv? Well there are 3 total, I'm choosing 1.
3 choose 1

\[\frac{3!}{2!*1!}\]

Answer 49

okay but wouldnt there be 4 in total?

Answer 50

never mind my question i got it :)

Answer 51

so then i would get 4620 for my answer

Answer 52

right?