An unbiased coin is tossed. If the result is a head, a pair of unbiased dice is rolled and the sum of the numbers obtained is noted. If the result is a tail, a well shuffled card of eleven cards numbered 2,3,4...12 is picked and the number on the card is noted. The probability that the noted number is either 7 or 8 is...

Question

anonymous · Answer

is the question considering the "noted number" to be either the sum of the dice roll OR the card chosen?

anonymous · Answer

there are 2 cases out of 11 for the card choice, which only occurs 1/2 of the time, i.e. when tails is the coin flip.

But there are a bunch of 7 and 8 options for a 2-dice roll.

Interesting question :)

anonymous · Answer

It can be both, the sum when its head and the card when its tail..

anonymous · Answer

there are 6 ways to roll a 7, and 5 ways to roll an 8.  So out of 36 outcomes, 11 are "winners".

I think you have (1/2)(11/36) for the dice option and (1/2)(2/11) for the cards option.

anonymous · Answer

Yes..that's right for both the parts...Now the final answer? @JakeV8

anonymous · Answer

it's the sum of the two outcomes, since either produces a winner.

P = 11/72 + 2/22
  = 242/1584  +  144/1584   = 386/1584 = 0.2437

anonymous · Answer

Is this your work or just Saturday edition of "fun with probability"?  :)

anonymous · Answer

Won't it be the multiplication of the two probabilities....? :)

anonymous · Answer

I was thinking that through too :)  I might have messed that up.

zarkon · Answer

your answer is correct jake

anonymous · Answer

Yes you are right..thanks..Its not work. I like probability. So, I solve them regularly.

anonymous · Answer

Since either path produces winners with the given probabilities, then they add.

anonymous · Answer

That's my "final answer", Regis.

anonymous · Answer

No its not multiplication...its addition

anonymous · Answer

It was great practice for me :) Thanks for posting it :)

anonymous · Answer

:-)

anonymous · Answer

Ohh thanks .. I too have some probability questions .. Hope to post it soon..

anonymous · Answer

Shoot them sooner..lets try @Miyuru

anonymous · Answer

Ok

anonymous · Answer

I have to drop off in a minute.. will be sorry to miss them :)  Hey @Miyuru, maybe for fun (loosely defined), you can throw an @JakeV8 in the questions... I'd like to read them later.

anonymous · Answer

OK sure @JakeV8 I would mention u on my question hope to see u soon..bye

anonymous · Answer

bye all, thanks again...

anonymous · Answer

U are welcome @JakeV8 
U did a great job for us...