Question

A carton contains 12 light bulbs of which 3 are defective. if a random sample of 4 light bulbs is chosen without replacement, what is the probability distribution of defective light bulbs?

Answer 1

Pr(defective bulbs)=3/12=1/4
Pr(light bulbs)=9/12=3/4

Answer 2

can someone please help me find the answer?

Answer 3

@satellite73

Answer 4

probability distribution means you want the probability that
0, 1, 2, 3, or 4, bulbs are defective right?

Answer 5

probability that none are defective is
\[\frac{9}{12}\times \frac{8}{11}\times \frac{7}{10}\times \frac{6}{9}\] is one way of computing it

Answer 6

i didnt get it. what do they mean by the sentence "chosen without replacement"?

Answer 7

it means pick one
put it aside
then pick another one

Answer 8

ok

Answer 9

here is another way to do it

Answer 10

could u please teach me both ways

Answer 11

probability distribution means you want the probability that
0, 1, 2, 3, or 4, bulbs are defective right?

Answer 12

why 4?

Answer 13

the number of ways you can pick 4 out of 12 is called "12 choose 4" written either as
\[\binom{12}{4}\] or as \[_{12}C_4\] and computes as
\[\binom{12}{4}=\frac{12\times 11\times 10\times 9}{4\times 3\times 2}\]
\[=11\times 5\times 9=495\]

Answer 14

ok

Answer 15

oh duh, yeah, you are right it can only be 0, 1, 2, or 3, i was being stupid

Answer 16

ok so \(495\) will be the denominator for all your answers

Answer 17

now lets compute the probability you pick no defective ones
that is 3 out of the 9 not defective, so \(\binom{9}{3}\) is your numerator

Answer 18

therefore
\[P(x=0)=\frac{\binom{9}{3}}{\binom{12}{4}}\]

Answer 19

could you pless tell me by what they mean by "probability distribution of defective light bulbs"?

Answer 20

please*

Answer 21

yes

Answer 22

you pick 4 light bulbs total out of the 12 right?

Answer 23

right

Answer 24

out of those 4, how many can be defective?
you answered that already, it can be 0, 1, 2, or 3

Answer 25

yes

Answer 26

if you put say \(x\) as the number of defective light bulbs chosen then there are 4 possibilities for \(x\) it could be
\[x=0,x=1,x=2,x=3\]

Answer 27

the "probability distribution of defective light bulbs" are the four numbers
\[P(x=0), P(x=1), P(x=2), P(x=3)\]

Answer 28

im writing these down as notes :)

Answer 29

in other words you have to compute four probabilities
the probability none are defective
the probability one is defective
the probability two are defective
the probability three are defective

Answer 30

man your the best :)

Answer 31

lol thanks
now we still have to compute the 4 numbers, right?

Answer 32

ill do it myself and you check my answer if thats alright?

Answer 33

yeah that if fine but there is a quick way to do it, more or less

Answer 34

so if you are knocking yourself out, you are doing too much work
let me know what you get

Answer 35

great. ill finish writing this down and ill get back to you my friend

Answer 36

ok

Answer 37

can we use the general formula to get the answer? through k, n, p?

Answer 38

no

Answer 39

you could if the bulbs were being replaced, because then the probabilities would always be the same
but they are NOT being replaced (remember is says "without replacement") so the probabilities change

Answer 40

by which i mean, if you pick a defective one first, the probability the next one is defective is no longer \(\frac{3}{12}\) it is now \(\frac{2}{11}\)

Answer 41

i'll be happy to show you the more or less snappy way to do this if you like

Answer 42

please im finding it a bit confusing

Answer 43

ok lets pick a middle one, say
\[P(x=2)\] i.e. the probability that exactly 2 are defective
we keep our denominator as \(_{12}C_4=495\)

Answer 44

now we are picking 4 so 2 defective and 2 not defective
the number of ways to pick the 2 out of 3 defective ones is \(_3C_2=3\)
the number of ways to pick the 2 out of 9 not defective ones is \(_9C_2=36\)

Answer 45

answer is therefore
\[P(x=2)=\frac{3\times 36}{495}\]

Answer 46

we can write the answer immediately although we still have to compute the numbers
2 out of 4 defective is
\[P(x=2)=\frac{\binom{3}{2}\binom{9}{2}}{\binom{12}{4}}\]

Answer 47

clear i hope
numerator says number of ways to choose 2 out of 3 times number of way to choose 2 out of 9, denominator is the number of ways to choose 4 out of 12

Answer 48

ill have a look at it once ive finished writing everything down

Answer 49

ok
armed with this, the rest should be easy because they work the same way

Answer 50

thank you soooooooooooooooooooooooo much

Answer 51

will you be around?

Answer 52

here is another quick example
i have a cooler with 6 cans of bud, 4 of miller and 5 of heineken
i pick 3 cans at random
probability i get 2 buds and a heiny is
\[\frac{\binom{6}{2}\binom{5}{1}}{\binom{15}{3}}\]that snappy

Answer 53

for a while at least

Answer 54

where did you get 2 out of 9 not defective?

Answer 55

you are picking 4 total right?

Answer 56

yes

Answer 57

if two are defective, how many are not defective?

Answer 58

the remaining 9

Answer 59

can you please draw me an image :)

Answer 60

it would really help

Answer 61

hold on
you are picking 4, and we want the probability that you get 2 defective
if you have 2 out of the 4 chosen defective, how many out of the 4 chosen are not defective?

Answer 62

2 not defective

Answer 63

right

Answer 64

and the number of ways to pick 2 defective and 2 not defective in this example is
\[\binom{3}{2}\times \binom{9}{2}\]

Answer 65

ohhhhhhhhhhhh now i get it

Answer 66

the one on the left is the number of ways to pick two out of three, the one on the right is the number of ways to pick 2 out of 9

Answer 67

eureka

Answer 68

ok good

Answer 69

so suppose x=1

Answer 70

so without any trouble you should be able in a second to write down how to find
\[P(x=1)\]

Answer 71

1 defective, 3 not defective

Answer 72

right
how many ways?

Answer 73

(3 1) (9 3) ?

Answer 74

yes

Answer 75

if it was 3 defective, 1 not defective

Answer 76

so
\[P(x=1)=\frac{\binom{3}{1}\binom{9}{3}}{\binom{12}{4}}\]

Answer 77

(3 3) (9 1)

Answer 78

yes

Answer 79

yes including the denominator

Answer 80

now why the denominator?

Answer 81

which is trivial to compute since \(\binom{3}{3}=1\) and \(\binom{9}{1}=9\)

Answer 82

why the denominator?

Answer 83

yes

Answer 84

how many light bulbs total?

Answer 85

12

Answer 86

how many are you choosing out of that 12?

Answer 87

4

Answer 88

so how many ways can you choose 4 out of 12?

Answer 89

alright so to find the denominator we use binomial coefficient

Answer 90

???

Answer 91

yes

Answer 92

it is therefore \(\binom{12}{4}\) for all of them

Answer 93

that is why i computed \(\binom{12}{4}\) first

Answer 94

man this thing was giving me a nightmare until u came along :)

Answer 95

i have another nightmare coming soon :)

Answer 96

but now it is more or less easy right?

Answer 97

you still have to compute the numbers, but i would wolfram for that

Answer 98

yeah bit after bit im getting to understand the material thanks to you