whitelight Posted June 3, 2007 Share Posted June 3, 2007 1. The probabilities of A, B, C solving a problem are 1/3, 2/7, and 3/8 respectively. If all the three try to solve the problem simultaneously, fine the probability that a) exactly one of them solves the problem ans: 25/26 2. The odds against a husband who is 45 years old, living till he is 70 are 7:5 and the odds against his wife who is now 36, living till she is 61 are 5:3 . Find the probability that a) the couple will alive 25 years hence b) exactly one of them will be alive 25 years hence c) none of them will be alive 25 years hence d) at least one of them will be alive 25 years hence Ans: a)5/32 b)23/48 c)35/96 d)61/96 3. Two persons A and B throw a die alternately till one of them gets a three and wins the game. Find their respective probability of winning, if A begins. Ans: 6/11&5/11 4. Two cards are drawn from a well shuffled pack of 52 cards without replacement. Find the probability that neither a jack nor a card of spade is drawn. Ans: 105/221 5. A and B toss a coin alternately till one of them tosses a head and wins the game. If A starts the game , find their respective probability of winning. Ans: 2/3&1/3 6. A and B appear for an interview for two posts. The probability of A’s selection is 1/3 and that of B ‘s selection is 2/5 . Find the probability that only one of them will be selected. Ans: 7/5 Please try to explain the solution as well... Thanx a tonn... Quote Link to comment Share on other sites More sharing options...
thankont Posted June 3, 2007 Share Posted June 3, 2007 Q2) don't agree with ans. husband p living is P(H) = 7/12, and for wife P(W) = 5/8 a) p=7/12*5/8 = 35/96 b) p= 7/12*3/8+5/12*5/8 = 23/48 (agree with this one) c) p=5/12*3/8 = 5/32 d) p=1-© Q3) P(A) = 1/6 + 5/6*5/6*1/6 + 5/6*5/6*5/6*5/6*1/6+... this is an infinite GP with a=1/6 and r=25/36 so S=a/(1-r) and P(A) = 6/11 and P(B) = 1-P(A) = 5/11 Quote Link to comment Share on other sites More sharing options...
whitelight Posted June 3, 2007 Author Share Posted June 3, 2007 Thankont, for question # 2 i was also skeptical of the answers, however, if take the probabilities of husb and wife to 5/12 and 3/8, respectively, the answers will match. Plz check and let everyone know what does it mean that - odds against ... Quote Link to comment Share on other sites More sharing options...
mr_sriramesh Posted June 3, 2007 Share Posted June 3, 2007 I think Qn1 is 25/56 Quote Link to comment Share on other sites More sharing options...
whitelight Posted June 3, 2007 Author Share Posted June 3, 2007 I think Qn1 is 25/56 Yeah, i also got 25/56 for ques 1., just wanted to confirm it with someone else - it might be a print mistake in the sol section... try cracking the questions 4 and 6 - quest 5 is same as 3. Quote Link to comment Share on other sites More sharing options...
whitelight Posted June 3, 2007 Author Share Posted June 3, 2007 Hi to everybody... If anyone wants to post some tough questions on probability (only), please continue the serial number of questions starting from 7; and the other person should post accordingly. This will create lesser confusion in this thread... Thanx. Quote Link to comment Share on other sites More sharing options...
ravindra_iit Posted June 4, 2007 Share Posted June 4, 2007 What is the source? Get the correct answers!! Q6 ans is given as 7/5!!! I don't think any GMAT prep book will give a number > 1 as answer for probability!! Quote Link to comment Share on other sites More sharing options...
ravindra_iit Posted June 4, 2007 Share Posted June 4, 2007 13 spades and 4 jacks one is common so total 16 cards from which we can not draw! the probability that neither a jack nor a card of spade is drawn = 36C2/52C2 = 36*35/52*51 = 3 * 35 / 13 * 17 = 105/221 1. 4. Two cards are drawn from a well shuffled pack of 52 cards without replacement. Find the probability that neither a jack nor a card of spade is drawn. Ans: 105/221 Quote Link to comment Share on other sites More sharing options...
thankont Posted June 4, 2007 Share Posted June 4, 2007 Q2) ans. are correct (odds against as you correctly mention whitelight) (I did it the other way around...thanx) Quote Link to comment Share on other sites More sharing options...
whitelight Posted June 4, 2007 Author Share Posted June 4, 2007 ravindra_iit, u banged the question 4 - this shows we really need to stick to basics for getting the solutions... abt the source - one of my friends e-mailed me his problems; i posted those which i could not solve... you are right that prob cannot be greater than 1, in GMAT... Quote Link to comment Share on other sites More sharing options...
hillery779 Posted July 8, 2007 Share Posted July 8, 2007 1. If all the three try to solve the problem simultaneously, fine the probability that a) exactly one of them solves the problem: so the answer is: The probabilities of A, B, C solving a problem are 1/3, 2/7, and 3/8 respectively. for only one of them solving the problem only one answers correctly so: (1/3)*(1-2/7 [probability for wrong answer])*(1-3/8)+(1-1/3)*(2/7)*(1-3/8)+(1-1/3)*(1-2/7)*(3/8)=25/56 2. The odds against a husband who is 45 years old, living till he is 70 are 7:5 and the odds against his wife who is now 36, living till she is 61 are 5:3 . Find the probability that a) the couple will alive 25 years hence housband: 7:5 [(7/12) against (5/12) infavor) wife : 5:3 [(5/8) against (3/8) infavor) answer for one: (3/8)*(5/12)=5/32 exactly one of them will be alive 25 years hence : (5/12)*(5/8)+(3/8)*(7/12)=23/48 none of them will be alive 25 years hence: (7/12)*(5/8)=35/96 at least one of them will be alive 25 years hence: it is (only one will live)+(both of them will live)=23/48+5/32=61/96 3. Two persons A and B throw a die alternately till one of them gets a three and wins the game. Find their respective probability of winning, if A begins. answer: A has 5 options to win (b draws all other numbers except 3) ; b has 5 win as well : and there an option to win (get 3) on the first time when A gets number 3 first draw. hence: A has probability of winning 6/11 B 5/11 5.A and B toss a coin alternately till one of them tosses a head and wins the game. If A starts the game , find their respective probability of winning. answer: A has 1 option to win after B toses tail B has 1 option to win after A toses tail and then there is the first toss which can grant A a win. hence: there are 3 options of winning A: 2/3 B:1/3 A and B appear for an interview for two posts. The probability of A’s selection is 1/3 and that of B ‘s selection is 2/5 . Find the probability that only one of them will be selected. answer: only one wins hence one wins other losses (probability to loss = (1- probability to win) than: (1/3)*(1-2/5)+(1-1/3)*(2/5)=(1/3)*(3/5)+(2/3)*(2/5)=7/15 Quote Link to comment Share on other sites More sharing options...
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