Question

There are three cabinets, A, B, and C,each of which has 2 drawers. Each drawer has 1 coin: A has 2 gold coins, B has 2 silver coins and C has 1 silver and one gold coin. A cabinet is chosen at random, one drawer is opened and a silver coin is found. What is the probability that the other drawer in that cabinet contains a silver coin? The back of the book says "2/3" I cannot reason through to that answer! Ideas?

Answer 1

only the drawer B and C has silver coins . so the opened drawer maybe any one of B or C. Now total no. of unopened drawer is 3, and remained silver coin is 2 as there were 4 drawers and 3 silver coins and u opened one. so the probability of getting silver coin is 2/3

Answer 2

@ john farmer got it??

Answer 3

if u get it then let me know.

Answer 4

Isn't 2/3 the probability of getting silver if i open a random drawer from B or C? i think that the probability of getting silver by opening the other drawer in the same cabinst is .5!

Answer 5

probability 5! ???

Answer 6

point 5

Answer 7

see the problem says u have found a silver coin, that means u have opened either B or C . do u agree??

Answer 8

Yes

Answer 9

i think u might wanna use bayes thm

Answer 10

ok then in B and C there are total 4 drawers and in these Drawers there are total 3 silver coins. do u agree??

Answer 11

Yes

Answer 12

and u have opened one of them so remaining unopened drawer is 3 and the no. of remaining silver coin is 2 . do u agree??

Answer 13

it asks for the prob that the OTHER drawer in the same cabinet has silver

Answer 14

But the questions is not the prob of getting a silver coin by opening a random drawer You have to open the same cabinet

Answer 15

yes so what??

Answer 16

i think uve got to use bayes thm to show how much prob of the silver coin coming from B

Answer 17

yes use it i think it should give 2/3

Answer 18

@ him use bayes theorem...

Answer 19

yes i got it

Answer 20

[0.33 x 1]/[0.33 x 1 + 0.33x0.5]

Answer 21

its 2/3 man

Answer 22

this comes out to 2/3

Answer 23

use fractions and ull see

Answer 24

yes i know it...

Answer 25

cancel 0.33 and u get 1/1.5 = 2/3

Answer 26

Why are you multiplying by 1?

Answer 27

bcoz in B the prob of finding a silver coin is 1

Answer 28

\[P(A|B) = \frac{P(B|A) P(A)}{P(B|A) P(A) + P(B|A^{^c}) P(A^{^c})} \]he used this

Answer 29

wah yaar kya mehnat ki hai

Answer 30

did u get it john?

Answer 31

Still puzzled.

Answer 32

do u know bayes theorem?

Answer 33

Yes, though I can't do it without making a tree diagram

Answer 34

make a tree diagram.///thats the best way to do it

Answer 35

Can you tell me where each value came from in your eq? I am not seeing the tree!

Answer 36

lets take event A is for choosing box B and event B is finding 1st silver coin. so P(A|B) u want to find. got it??

Answer 37

respond me...otherwise i could not understand that whether u got it or not.

Answer 38

Sorry - had 2 go.
Restating the question " What is prob of cabinet B give silver?" totally helped. Thanks.
That gives: (Ways to get silver from B)/(Total ways to get silver)
\[((1/3)(1/2) +(1/3)(1/2))/((1/3)(1/2)+(1/3)(1/2) +(1/3)(1/2))=2/3\]

I still think that the way the question is phrased does not match the question " What is prob of Cabinet B given silver?" even though the author intended it to be solved using that concept. I think it is saying"What is the prob that <your next move> which has to be opening the other drawer in the same cabinet will get you silver? That probability is 50-50.

Thanks again!