wirelessnz Posted January 27, 2003 Share Posted January 27, 2003 Could some one explain how to do the following Prob questions. Thanks. 1. From a box containing 10 cards numbered 1, 2, 3, ...., 10, four cards are drawn. find the probability that their sum will be even (a) id the cards are drawn together, (b) if each card drawn is replaced before the next is drawn. 2. A and B having equal skill, are playing a game for three points. After A has won 2 points and B has won 1 point, what is the probability that A will win the game? 3. One bag contains 3 white and 2 black balls, and another contains 2 white and 3 black balls. A ball is drawn from the second bag and placed in the first, then a ball is drawn from the first ball and placed in the second. when the pair of operation is repeated, what is the probability that the first bag will contain 5 white balls? 4. Three bags contain respectively 2 white and 3 black ball, 3 white and 3 black balls, 6 white and 2 black balls. Two bags are selected and a ball is drawn from each. Find the probability (a) that both balls are white, (b) that both balls are of the same color. 5. If four trials are made in problem 4, find the probability (a) that the first two will result in pairs of white balls and the other two in pairs of black balls, (b) that a pair of black balls will be obtained at least three times. Quote Link to comment Share on other sites More sharing options...
Lasha Posted January 28, 2003 Share Posted January 28, 2003 My asnwers for question 1 are 13/25 and19/37 and for question 2 3/4. If my answers are correct let me know to provide explanation for all of the questions. P.S. Where did you get these questions? No 3,4,5 are not likely to be from GMAT because solving them requires definitely more than 1,5 minutes. Quote Link to comment Share on other sites More sharing options...
wirelessnz Posted January 28, 2003 Author Share Posted January 28, 2003 Hello Lasha, The ans are: 1. (a)11/21, (b)1/2 2. 3/4 3. 1/225 4. (a) 29/72, (b)19/36 5. (a) 841/331776, (b)125/36864 I got them from another GMAT forum. Only ans were provided with no explanation. So I would be grateful for your explanation as you seem to understand these kind of questions very well. Thank you. Originally posted by Lasha My asnwers for question 1 are 13/25 and19/37 and for question 2 3/4. If my answers are correct let me know to provide explanation for all of the questions. P.S. Where did you get these questions? No 3,4,5 are not likely to be from GMAT because solving them requires definitely more than 1,5 minutes. Quote Link to comment Share on other sites More sharing options...
Erin Posted January 28, 2003 Share Posted January 28, 2003 Originally posted by Lasha P.S. Where did you get these questions? No 3,4,5 are not likely to be from GMAT because solving them requires definitely more than 1,5 minutes. If you're getting a high score on the Quant, you will run into some questions that take five or so minutes to complete. Quote Link to comment Share on other sites More sharing options...
pankajagg Posted January 28, 2003 Share Posted January 28, 2003 3. In first trial a white ball is transfered into a first bag therefore p(2-1) = 2/5 now the number of balls in each bag has changed, Therefore probability of transfering a black ball from bag 1 to bag 2 is 2/6 similarly, when we again transfer a white ball from bag 2 p`(2-1) = 1/5 {we have 1white and 4 black balls} p`(1-2) = 1/6 {black goes into second box} therefore total probability is 2/5 * 2/6 *1/5 *1/6 = 1/225 ----- 5. what is meant by 4 trails ??? how can we have 4 trails in this case ? Quote Link to comment Share on other sites More sharing options...
Lasha Posted January 29, 2003 Share Posted January 29, 2003 1) A)There are three possible ways to have the sum of four integers even. ev+ev+ev+ev, odd+odd+odd+odd, and 2odd+2ev From 1,2,3....10 we have 5even and 5odd numbers. We can have C(5)4=5 combinations of even integers. The same is true for odd integers. As to 2odd and 2even C(5)2 x C(5)4=100. So if we sum up 5+5+100=110. There are C(10)4=210 total 4-integer combinations, so the probability that their sum will be even is 110/210=11/21 B) We have the same combinations here ev+ev+ev+ev, odd+odd+odd+odd, and 2odd+2ev. For only-even combination the probability is 1/2x1/2x1/2x1/2=1/16. The same goes for only-odd combination. For 2-even-2-odd combination teh thing is more complicated. when picking first two cards we have ev-ve, odd-odd,odd-ev or ev-odd possible outcomes. Now we must treat them seperately. The probability up to this point(for the first two draws) is 1/4 for each. For ev-ev(odd-odd) we need odd(ev) as a third draw and again odd(ev) for the fourth draw so the probability becomes 1/4x1/2x1/2=1/16. In case of odd-ev(ev-odd) third outcome does not matter so we have 1/4x1x1/2=1/8So if we sum up 1/16+1/16+1/8+1/8=3/8 now we must sum up all-even all-odd and 2-odd-2-even probabilities 1/16+1/16+3/8=1/2 2) The probabilty that B wins is 1/2x1/2=1/4 1-1/4=3/4 the probability that A wins. 3) Thanks to pankajagg. 4) A) 2w 3b, 3w 3b, 6w 2b we have 3 ways to choose 2 bags out of three 1-2,1-3,2-3. We have to calculate the probabilty of picking white-white combination for each case and sum up. 1-2 bags---2/5x1/2=1/5 1-3 bags---2/5x6/8=3/10 2-3 bags---1/2x6/8=3/8 (1/5+3/10+3/8)x1/3 (I multiplied the sum on 1/3 because there is 1/3 probability of choosing 2bags out of three.) =7/24 My solution may be wrong B) You do the same for black-black combinations and then sum up white-white and black-black probabilities. Quote Link to comment Share on other sites More sharing options...
pankajagg Posted January 29, 2003 Share Posted January 29, 2003 As the matter of fact, i got the same answer for 4th question, 7/24, exactly on the similar lines as by lasha. In the 5th question are we replacing the balls or we kept removing htem in eash trial ? Quote Link to comment Share on other sites More sharing options...
Ulviya Posted January 29, 2003 Share Posted January 29, 2003 Hi Lasha - my question is to you. I am a bit confused. Why in the second test you are finding the probability of B winning first and then subtract that from 1? In what cases that is the way of solving this kind of problems? Could you explain me? Thanks Ulviya Quote Link to comment Share on other sites More sharing options...
Ulviya Posted January 29, 2003 Share Posted January 29, 2003 To Erin Eric, you indicated that the hard questions might take 5 min or so. should we spent that much time on the question or move on? Quote Link to comment Share on other sites More sharing options...
pankajagg Posted January 29, 2003 Share Posted January 29, 2003 Ulviya as for GRE you should not leave any question, you must answer it.Moreover in GRE there is plenty of time on quant section, if you practice well i'm sure you can solve that whole quant section 2 times. :) Quote Link to comment Share on other sites More sharing options...
Lasha Posted January 30, 2003 Share Posted January 30, 2003 Hi Ulviya. There are two ways to solve question 2. 1) The way I did. You find probability that B wins(that means that A looses) and subtruct it from 1(the probability that A wins + the probabilty that B wins). To be honest, however, the question is little bit inacurate since it does not rule out ties. 2) The second way. The probability that A wins the next game is 1/2 but even in case if he does not he still has a chance because the overall score between A and B becomes 2-2. So in order for A to become a winner he should win either the next or after-the-next game. The probability of wining the next game is 1/2 and after-the-next game 1/2x1/2=1/4. If we sum up 1/2+1/4=3/4 note that this way is little longer. Quote Link to comment Share on other sites More sharing options...
Ulviya Posted January 30, 2003 Share Posted January 30, 2003 Pankajagg I am not GRE- I am GMAT. Thanks anyway. Ulviya Quote Link to comment Share on other sites More sharing options...
Ulviya Posted January 30, 2003 Share Posted January 30, 2003 Thanks Lasha The second way is longer but it seems to make more sense to me Ulviya Quote Link to comment Share on other sites More sharing options...
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