Question

Three items are selected at random without replacement from a box containing
ten items, of which four are defective. Calculate the probability distribution for
the number of defectives in the sample. What is the expected number of
defectives in the sample?

Answer 1

X: 1 2 3 4
what is about p(x)???

Answer 2

how many ways can you select 3 items?

Answer 3

ggg

ggb
gbg
bgg

gbb
bgb
bbg

bbb

this looks like all the ways we can pull 3 items, such that they are in some good bad relationship

Answer 4

what is the probability of pulling a good part? the probability of pulling a bad part?

Answer 5

for the ways I will have 2^3=8 as you write.
good=non-defective
bad=defective
so, i think
p(pulling good part) = 7/8
p(pulling bad part)= 7/8

Answer 6

lets see ... without replacement we get denominators of d = 10.9.8

ggg = 6.5.4/d

3 sets of ggb gives us 3 * (6.5)*4/d

3 sets of gbb gives us 3 * (6)*(4.3)/d

bbb = 4.3.2/d

and i see no way to pull 4 bad parts if we are only pulling 3 parts

Answer 7

so:

P(4bad) = 0
P(3bad) = 4.3.2/d
P(2bad) = 3*6.4.3/d
P(1bad) = 3*6.5.4/d
P(0bad) = 6.5.4/d

Answer 8

sorry sir.. still I cannot recognize...

Answer 9

you mean probability distribution will be like this:

X: 0 1 2 3 4
p(x): ? ? ? ? ?

Answer 10

P(4bad) = 0

P(3bad) = 4.3.2/d
(4*3*2)/(10*9*8)=1/6
P(2bad) = 3*6.4.3/d
3(6*4*3)/(10*9*8)
P(1bad) = 3*6.5.4/d
3(6*5*4)/(10*9*8)
P(0bad) = 6.5.4/d
6*5*4/(10*9*8)

Answer 11

yep

Answer 12

the probability of something happening is the sum of the probability that it happens this way, that way, and another way. hence the 1,3,3,1 multipleirs

Answer 13

X: 0 1 2 3 4
p(x): 1/6 1/2 3/10 1/30 0

Answer 14

P(3bad) = 4.3.2/d
1/30

P(2bad) = 3*6.4.3/d
3(6*4*3)/(10*9*8)
3/10

P(1bad) = 3*6.5.4/d
3(6*5*4)/(10*9*8)
1/2

P(0bad) = 6.5.4/d
6*5*4/(10*9*8)
1/6

1/30 + 1/6 + 3/10 + 1/2 = 1 since it contains all the possible ways

Answer 15

so , E(X) = 1.2

Answer 16

lol, id have to look up that to refresh expected value froma weighted averae

Answer 17

E(X) = sum of x*p(x) right?

Answer 18

yes, I did like what ou write

Answer 19

X: 0 1 2 3 4
p(x): 1/6 1/2 3/10 1/30 0
0 .5 .6 .1 0
1.2 seems fair to me

Answer 20

do we want a P(0)?

Answer 21

surly no

Answer 22

sorry

Answer 23

http://www.statlect.com/expval1.htm

this seems like a good review, hence the sum of x p(x) as E(X) for discrete variables.

Answer 24

i mean p(4)=0
we don't need
but p(0)=1/6 ...according to other examples that i solve may be we need

Answer 25

0*1/6 seems redundant :) but thats just me

Answer 26

so on average we expect to get about 1.2 defects in a given 3 part pull

Answer 27

than you sir...
your way in explaining is very excellent

Answer 28

youre welcome :)

Answer 29

this was my teacher solution which I didn't anderstand to study for exam

Answer 30

your way is best and easiest

Answer 31

his way is just notational differences. if you can read the notation, its just shorthand overall

Answer 32

i tend to mess up the notational route so i have to process it the logical route :)

Answer 33

sir 1 question that I recognize now
why we writ 6.4.3 instead of 6.5.4 ???
P(2bad) = 3*6.4.3/d
3(6*4*3)/(10*9*8)
3/10

Answer 34

1 good, 6 out of 10
2 bad, 4 and 3 out of (9 and 8)

Answer 35

we are effectively considering the event:

p(g) * p(b) * p(b) without replacement

6 out of 10, times 4 out of 9, times 3 out of 8

since multiplication is commutative and associative, the 3 cases: gbb, bgb, bbg all have the same probability. so: p(gbb) + p(bgb) + p(bbg) = p(2bad)