# Probability - Need Help please?

In a game, you role two dice, and then pick a card. IF the card and the dices are the same number, then you win. Aces and kings have been removed

What is the probability that you will role the sum of two and pick a two from the deck?

### 1 Answer

- Anonymous10 years agoFavorite Answer
Hi there...

You could try to picture the probability tree for this problem with your first stage of the tree being the event where a die is rolled (6 branches with the numbers 1,2,3,4,5,6). Second event is the next die being rolled - so of each of your previous branches you have 6 new branches again with the possibilities of 1,2,3,4,5,6. At this point there are 36 branches in the second row. Next event is picking from a pack of cards. Of each of your previous 36 branches there would be 44 new branches with (2,2,2,2,3,3,3,3,4,4,4,4,...,J,J,J,J,Q,Q,Q,Q). There's 44 because 52 minus 4 kings and 4 aces = 44.

So in our final stage of the prob tree there would be a total of 6x6x44 = 1584 branches which means 1584 different combinations of the possible outcomes (eg. "1,1,2" would be where you get ones on the dice and pick a 2 from the cards. "5,3,J" is another possible outcome). Also, 1584 is called the sample space - it's the total number of possible outcomes.

So we need to know the probability of getting a sum of 2 on the dice (which is only possible with two 1's) and a two from the cards.

We need "1,1,2", where the first two are dice and the third is a card.

From probability tree we go along the "1" branch in first event, then along the "1" branch in 2nd event and then we would have four "2" branches to choose from in 3rd event (coz there's four 2's in a pack of cards). This means there are 4 paths that will get us the combination of "1,1,2". So our probability will be 4/1584 = 2/792 = 1/396.

Alternatively, P(A and B) = P(A) x P(B)

So in your case:

P(sum of two and two from deck)

= P(1 and 1 and 2)

= 1/6 x 1/6 x 4/44

= 4/1584

= 1/396

Hope it helps (^_^)