## Black or White 3

Posted by Chris on August 29, 2012 – 3:44 am

John provided the following problem on a comment on the “Black or White” page:

There are a number of cards in a hat, 20% of which are white on both sides, 50% of which are black on one side and white on the other and 30% of which are black on both sides.

One card is drawn and only one side is inspected. If that side is black, what is the probability that the other side is white?

August 29th, 2012 at 6:00 am

Sorry it’s been so long since I’ve visited!

Am I looking at this too head on?

If one side is black, that means only 80% of the cards are in play, 5/8ths of which can have a white side.

I’m betting probability of .375!

August 29th, 2012 at 6:08 am

5/8 = 0.625

other than that I’d agree.

I imagine part 2 will ask:

If you don’t look at the card drawn…. what is the probability the reverse side is white?

much more interesting…….

August 29th, 2012 at 6:33 am

DOH … yes … 5/8 = .625

August 29th, 2012 at 7:06 am

agree… 62.5% (since the question was written in percentage)

I’m glad to see a small amount of action on this site. It would be great for some new person to come in and post a lot of problems we’ve never seen before. Or maybe Chris or someone can grab a handfull of problems for the “old site” (whatever that is) and post them here as well. Just to give us something to work on. I suppose I could even try re-posting some older ones even from this site, although that seems a little boring.

August 29th, 2012 at 7:25 am

Hi DP. I’m sure you used to visit the old ToM site. Here’s a link to a problem on it: http://trickofmind.com/2010/04/b-100000.html#comments

Hi curtis. If pick but don’t look, the probability of the reverse side being white is 9/20 = 45%.

That answer is very close to the one required for the posted problem (unless I’ve goofed). i.e. no correct answers so far.

August 29th, 2012 at 2:48 pm

There are 11 sides colored black. Five of them have white on the other side, so the probability is 5/11 = .4545…

August 29th, 2012 at 4:25 pm

Mark has the right answer.

I have seen similar answers (for similar problems) before, and usually don’t like them, but on this occasion, I think Mark has expressed it very well.

August 30th, 2012 at 6:36 am

Wow… me again. I can tell that it’s been a while.

I’ll go with Mark on this one now. 5/11 is the answer. I obviously forgot to account for either side of the black/black cards being used to start with.

August 30th, 2012 at 6:41 am

Wow…”tricked” me again. I can tell that it’s been a while.

I’ll go with Mark on this one now. 5/11 is the answer. I obviously forgot to account for either side of the black/black cards being used to start with.

By the way, Chris, I had actually never seen that site before. I never started visiting until I got a new PC at work with Windows 7. I got the little gadget that goes on my desktop that shows the 10 most recent problems. Sort of a random find whild searing the gadget store (or whatever it’s called).

August 30th, 2012 at 7:46 am

Hi DP. I might be confusing you with SP (!!!).

I was a bit surprised that you got it wrong. You gave a very good explanation for the previous “Black or White” problem.

I’ve got that gadget too. I found it when I was trying to find some info on C#. I stumbled upon Rajesh Lal’s csharptricks website. One thing led to another…

August 30th, 2012 at 1:33 pm

Yeah, maybe so. I knew the solution and how to get there, I just answered faster than I should have.

By the way, thanks for the link to the old site. I found a problem by ZAUX called Pool Que Missing that inspired a new logic based problem I just posted. I sort of made it up as I went, and found 2 soultions to the problem (both leading to the correct answer). Have fun!

http://trickofmind.com/?p=1669

January 13th, 2013 at 11:34 pm

Could somebody explain how they came to the conclusion of 5/11. A little confused. Thanks much.

January 16th, 2013 at 7:35 am

Hi FJ. There are a lot of ways to solve this problem. Mark (post 6) gave a quite nice answer, although his logic required him to say there were 10 cards in the hat.

What’s being asked for is the probability of picking the BW card, given that you pick a B face.

It is equally likely that you’ll pick any of the cards, and it is equally likely that the card will be either way up, so:

The prob of picking a WW card and seeing a B face is (20/100)(0/2) = 0.

The prob of picking a BW card and seeing a B face is (50/100)(1/2) = 25/100.

The prob of picking a BB card and seeing a B face is (30/100)(2/2) = 30/100.

So the probability of picking a B face is (0 + 25 + 30)/100 = 55/100 and so the prob of picking the BW card and seeing a B face given that we picked a B face is (25/100)/(55/100) = 25/55 = 5/11.

Just for clarity (I hope), if we pick a card and see a W face, we have not complied with the condition that we saw a B face, so we completley ignore that trial.

In fact, we could remove the WW cards from the hat – if you pick a WW card, you cannot see a B face, so the WW cards are irrelevant. Assume we’ve removed the WW cards, then 5/8 of the cards are BW, and 3/8 are BB. Then the prob of picking the BW card and seeing a B face is (5/8)(1/2) = 5/16 and the prob of picking a BB card and seeing a B face is (3/8)(2/2) = 6/16. So the prob of picking a B face is 5/16 + 6/16 = 11/16 and so the prob of picking a BW card and seeing a B face given that we saw a B face is (5/16)/(11/16) = 5/11.