Difficult Card Problem - Probablities

[Guest]

1. Tammie has 10 cards numbered 1 through 10. If she deals two to Tarrell without
replacing any of them what is the probability that Tarrell will get both a 2 and a 3?

(A) 1/5
(B) 1/45
(C) 1/50
(D) 1/90
(E) 14/45

The answer is supposed to be 1/45.

Thanks for your answer and explanation!

[a2bhatia]

Tammie has 10 cards numbered 1 through 10. If she deals two to Tarrell without
replacing any of them what is the probability that Tarrell will get both a 2 and a 3?

(A) 1/5
(B) 1/45
(C) 1/50
(D) 1/90
(E) 14/45


You must count how many ways Tarrells cards can be given to him and divide by the number of ways two cards can be handed out.

Since Tammie doesnt replace the cards - the first card can be any of 10 cards, and the second can be any of the remaining 9 cards (the 10th has already been given out). Therefore 10x9 = 90 possible ways to hand out the cards.

Tammie can give Tarrell a 2 first, then a 3. OR a 3 first and then a 2. Two of the aforementioned ways will cound towards this possibility. (you can also use 2! to get this number - 3! for 3 cards etc.).

Therefore 2/90 = 1/45. Answer is B.

[guest]

You can also use conditional probabilities to drive to this answer:

P(Getting either a 2 or 3)*P(Getting the remaining card | Tammy received a 2 or 3 first) =

2/10*1/9 = 2/90 = 1/45

[StaceyKoprince]

Please don't forget to cite the source of the problem - if you do not cite the author, we will have to delete the question. Thanks!