Question

HELP PLEASE :( i've been working on this for almost the whole night and i'm still stumped :(

An urn contains five red and three blue chips. Suppose that four of these chips are selected at
random and transferred to a second urn, which was originally empty. If a random chip from
this second urn is blue, what is the probability that two red and two blue chips were
transferred from the first urn to the second urn?

Answer 1

@.Sam. @ash2326 @ParthKohli @TuringTest @apoorvk @Carl_Pham @Coolsector @estudier @hartnn @helder_edwin @Hero @matricked @satellite73 @sauravshakya @Zekarias
help please :(

Answer 2

i'm reviewing for my upcoming midterms and i got this from a past exam... it's scaring me :( i'm familiar with conditional probability and i could answer majority of the questions in my book just fine, but i find this too complicated. :(

Answer 3

We have to use Bayes' theorem here. Have you studied that?

Answer 4

oh! really? we weren't taught that. maybe there's a way of solving it without the need of that? like a tree diagram? or is it impossible without that theorem?

Answer 5

Umm, I don't know any other way
It's conditional probability, refer this first. Then we'll work on the problem
http://en.wikipedia.org/wiki/Bayes'_theorem

Answer 6

oh! i'm aware of the simple form of the bayes' theorem! i wasn't aware it was called that. i actually tried to use it in this, but i had a hard time identifying the components:(

Answer 7

Ok, let's work :)

Answer 8

i tried to solve for P(2 red and 2 blue chips were transferred from the first urn to the 2nd urn | random chip from second urn is blue) but i just got so lost :( i really appreciate your patience in helping me through this @ash2326 !!!! thank you already!!

Answer 9

We need to find probability of 2 R and 2 Blue being transferred from urn A to B, provided random chip is blue.

Possible combinations when 4 chips are transferred from A to B. Let the combinations be
2R and 2B= X
1R and 3B= Y
3R and 1B=Z
4R=U
Do you get this?

Answer 10

I'm here to help, you're welcome.

Answer 11

I've just given aliases to the combinations

Answer 12

@dydlf are you here?

Answer 13

yes i am! yes i get you so far.

Answer 14

Ok, let's continue. Feel free to interrupt where you have doubt

Answer 15

sure!

Answer 16

So we need to find
\[P(X/B)=\text{ probability of 2 R and 2B being transferred when a random chip is blue}\]
\[\small {P(X/B)=\frac{P(B/X)\times P(X)}{P(B/X)\times P(X)+P(B/Y)\times P(Y)+P(B/Z)\times P(Z)+P(B/U)\times P(U)}}\]
Do you get this? @dydlf

Answer 17

hmmm let me refer to the wiki page you sent. wait

Answer 18

sure

Answer 19

sorry, lost connection. but okay! i see what you mean now!

Answer 20

Now let's find the individual probabilities
\[P(B/X)\times P(X)\]
Can you tell me the value of this?

Answer 21

or How do we find it?

Answer 22

hmm doesn't
P(B|X)×P(X) mean the intersection of B and X? when i tried solving this, that's where i got stuck since i didn't know how to get this.
hmmm P(X) is 5C2×3C2/10C4 ?
P(B|X) is like P(B intersection X)/P(X) right? hmmm 2C1/(5C2×3C2/10C4) ?

Answer 23

Let's work step by step
probability of getting blue when 2 blue and 2 red are transferred =P(B/X)
It's just the probability of getting one blue, when there are 2 red and 2 blue chips
\[P(B/X)=\frac {2}{4}\]

Answer 24

gosh i meant 5C2×3C2/8C4 not 5C2×3C2/10C4

Answer 25

Now we'll find P(X), that's choosing 2 red and 2 blue from 5 red and 3 blue
\[P(X)=\frac{^5C_2 \times ^3C_2}{^8C_4}\]
Yeah you got P(X) right

Answer 26

@dydlf Do you get how I found P(B/X) ?

Answer 27

Take your time, I'm here :)

Answer 28

hmm yup! but i'm just wondering, what's the diff of let's say P(B|X) and P(B intersection X)? like in words?

Answer 29

P(B intersection X) is probability of occurrence of B and X event simultaneously
P(B/X) is the conditional probability of occurrence of B when X has already occurred

Answer 30

Infact P(B/X)=P( B inter X)/P(X)

Answer 31

yes i understand that, but i don't really understand the concept like for solving P(B|X) earlier, if you noticed, i used that equation you just mentioned up there, but i had a hard time looking for P(B inter X) since i think i don't really understand the concept

Answer 32

i know it has to be in both B and X but... gosh i just got confused there.

Answer 33

We don't need to find P( B inter X) here
Finding P(B/X ) is straightforward . See how I found it above

Answer 34

Here is one way to do it without explicitly invoking Bayes Theorem

Answer 35

oh, i was just wondering. because i usually mix the 2 up....

Answer 36

Do you get till here?

Answer 37

Given that my solution does not match Bayes Thm approach, I conclude there is a flaw in my argument, and you really do have to grind through the problem.

Answer 38

Is it 6/13 ??? The answer... I'm getting that only... @dydlf

Answer 39

No, the answer, if you finish up Ash's approach, is 4/7

the numerator is (2/4)* (30/70)
the denominator is (2/4)*(30/70) + (3/4)*(5/70) + (1/4)*(30/70) + 0*(5/70)

Answer 40

I believe it comes down to how to interpret
"If a random chip from this second urn is blue..."

If this means "A chip selected from the 2nd urn is blue..."
then we get 4/7

If we interpret it as "At least one chip in the 2nd urn is blue..."
we get 6/13

Answer 41

I don't think... my approach makes it 6/13 only... maybe some flaw is there in his or mine...
I think he doesn't need to consider the case when 4 red balls are chosen... because it is mentioned in the question that...
*a random chip from this second urn is blue*
which makes me think there's at least one blue ball in the second urn...

Answer 42

*chip*

Answer 43

Yes, I agree it is a subtle question. This is a question on an old exam, and I would like to see its solution write-up.

Answer 44

Even if I consider your 1st approach... I get 3/7 as against yours 4/7...

Answer 45

How ?

Answer 46

2R, 2B ===> 5C2* 3C2 = 30
3R, 1B ===> 5C3*3C1 = 30
1R, 3B ===> 5C1*3C3 = 5
4 Red ===> 5C4 = 5
P(2R, 2B) = 30/ (30+30+5+5) = 3/7
Please check if there's some flaw...