Diego Posted May 21, 2004 Share Posted May 21, 2004 Here there are a some tricky probability Qs I - How many different words with up to 4 letters can be made with the letters A and B? a) 4 b) 16 c) 30 d) 31 e) 32 II - What is the probability that you receive a pair when you are dealt two cards from a complete deck? a) 1/17 b) 1/13 c) 1/12 d) 4/17 e) 1/2 Quote Link to comment Share on other sites More sharing options...
bazooka76 Posted May 24, 2004 Share Posted May 24, 2004 For me the first question sounds to be way too general. Given 4 positions and 2 letters A & B. If there is no restriction on the other two letters that form the four letter word, we can say (1)(1)(26)(26) (repition allowed). As I said, this is just one way of looking and interpreting at the question. For the second one. P(getting a pair of cards from a group of 4) = 4/52 * 3/51 P(getting a pair of cards from a pack of cards ie, from 13 sets (1 to A)) = 13*(4/52)*(3/51) = 1/17 (I hope I interpreted the ques correctly though I am not sure) Quote Link to comment Share on other sites More sharing options...
roborob Posted May 24, 2004 Share Posted May 24, 2004 For the first Q, I would solve it like this. This is the super, duper long way, but I would actually sit there and write it all out. AAAA AAAB AABA ABAA BAAA AABB ABAB ABBA BAAB BBAA BABA ABBB BABB BBAB BBBA BBBB Therefore, totalling 16. Now, for the second Q, I would think of it like this... since they ask for a pair dealt from a deck, that would mean it doesn't matter what you got the first card, the second card would just have to match. So, that would mean that the chances your second card matches your first card would be, 3 out of 51. Does that make any sense at all? Hope this helps, Rob Quote Link to comment Share on other sites More sharing options...
Umar Posted May 25, 2004 Share Posted May 25, 2004 Yes it makes sense, moreover, you have also got the answer right(1/17). There must be a fast way to do the first question. Quote Link to comment Share on other sites More sharing options...
Dharmin Posted May 25, 2004 Share Posted May 25, 2004 I - (a,b)^4 + (a,b)^3 + (a,b)^2 + (a,b) = 16+8+4+2 = 30 II - there are 13 * 4 = 52 cards in complete deck: so, required probability, = 13C1 * 4C2 / 52C2 = (13*6) / (26 * 51) = 1/17 Dharmin Quote Link to comment Share on other sites More sharing options...
Diego Posted May 25, 2004 Author Share Posted May 25, 2004 Thank you all for your posts, The correct answers are : I - c) = 30 II - a) = 1/17 bazooka76, please, explain your approach!? I still don't have clear why you stated this: "P(getting a pair of cards from a pack of cards ie, from 13 sets (1 to A)) = 13*(4/52)*(3/51) = 1/17"? Rob, your answer makes a lot of sense, thanks. Dharmin, You got them both right, can you explain me how did you arrive to "4C2 / 52C2" on the second one? and by the way thank you for posting also at the gmatclub. Quote Link to comment Share on other sites More sharing options...
Dharmin Posted May 25, 2004 Share Posted May 25, 2004 Dear Diego, if we need to choose pair of card, we have to select any of the 2 cards from 4 cards. As u know, there are 4 cards of each kind( from A to king) Hope, this will help, Dharmin Quote Link to comment Share on other sites More sharing options...
bazooka76 Posted May 25, 2004 Share Posted May 25, 2004 Diego, For the second question. In a pack of cards there are 13 kinds of 4 card set such as 1 1 1 1 2 2 2 2 . . . K K K K A A A A Consider the P(Getting "2 KING(K)s" from a pack of cards) = 4/52 * 3/51 But a pair can be either 2 kings OR 2 queens OR 2 4's OR 2 5's etc.. ie total 13 kinds) = 13 * 4/52 * 3/51 = 1/17 Hope this helps. Quote Link to comment Share on other sites More sharing options...
Diego Posted May 25, 2004 Author Share Posted May 25, 2004 Now I've got it, thanks a lot Bazooka!!! Quote Link to comment Share on other sites More sharing options...
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