Question

A box contains 4 red and 5 green marbles. If 5 marbles are drawn without replacement, what is the expected number of red marbles?

Answer 1

multiply

Answer 2

@satellite73 Multiply what?

Answer 3

total number of marbles = 9
4 are red
so first draw = 4/5
2nd draw 3/8 ( we already picked one red marble that gives us 4-1 =3 subtract one from total number of marbles 9-1= 8)
3rd draw = 2/7 ^
4th = 1/6
5th = 0/5
now multiply all fraction
i hope that's correct :D

Answer 4

oh wait no :D

Answer 5

on the first draw how is it 4/5?

Answer 6

typo 4/9**

Answer 7

hmm that looks incorrect hmm

Answer 8

@Nnesha so will it be right after the 4/9 correction?

Answer 9

hmm i'm not sure about 0/5
that will give us 0 as a final answer hmm

Answer 10

i got 24/3024 as the answer but that doesn't seem right because we aren't asking for probability

Answer 11

ok i lied, this is more complicated that in looks

Answer 12

the probability you get 0 red is 0 (one has to be red)

Answer 13

the probabiliy you get exactly one red is \[\frac{\binom{4}{1}\binom{5}{4}}{\binom{9}{5}}\]

Answer 14

ok

Answer 15

then you do that for 2 and 3 and 4

Answer 16

that is not really so hard to compute as it looks \[\frac{4\times 5}{126}\]

Answer 17

yeah then repeat for 2, 3, 4
then multiply, then add

Answer 18

1 times the probability you get 1, plus 2 times the probability you get 2 etc

Answer 19

there may be a simpler way to do it but i don't know it
i would cheat

Answer 20

@satellite73 for the denominator would it be going down by one every time because it says without replacement?

Answer 21

no

Answer 22

that is taken in to consideration when you use the formula

Answer 23

oh okay

Answer 24

why am i multiplying 2 times the probability of getting 2?

Answer 25

the pobabilty you get for example 3 red and 2 white is \[\frac{\binom{4}{3}\binom{5}{2}}{\binom{9}{5}}\]

Answer 26

because that is how you get an expected value
the number you get times the probability you get it

Answer 27

oh

Answer 28

\[P(x=1)+2 P(x=2)+3P(x=3)+4P(x=4)\]

Answer 29

in this case

Answer 30

more if there are more possibilities

Answer 31

I get a huge number when I do that

Answer 32

i got this, but i wouldn't bet more that $7 on it
not at all
http://www.wolframalpha.com/input/?i=20%2F126%2B2%284+choose+2%29%285+choose+3%29%2F126%2B3%284+choose+3%29%285+choose+2%29%2F126%2B+4*5%2F126

Answer 33

it is not huge

Answer 34

we can do them one at a time if you like

Answer 35

that was right :D

Answer 36

imagine!