ind21 Posted December 9, 2003 Share Posted December 9, 2003 Hi !! I got these question from one of the groups . I don't have the answers. Could anyone please solve them ASAP.Thanks in advance. 1.P can hit a target 4 times in 5 shots, Q, 3 times in 4 shots and R, twice in 3 shots,. They fire simultaneously. What is the probability that at least 2 shots hit? 2.A purse contains 2 silver coins and 4 copper coins and another contains 4 silver coins and 3 copper coins. If a coin is selected at random from one of the two purses what the probability that it is a silver coin? 3.The probability that A can solve a problem is 4/5, that B can solve it is 2/3 and that C can solve it is 3/7. If all of them try independently, find the probability that the problem will be solved. 4. A committee of 4 boys and 3 girls is to be formed by lots from 8 boys and 5 girls. One of the boys is brother of one of the girls. Find the probability that both are included in the committee. 5. An urn contains four white and five black balls, a second urn contains five white and four black balls. One ball is transferred from the first to the second urn, then a ball is drawn from the second urn. What the probability that it is white? 6.In a class of 100 students there are 60 boys and 40 girls, 20 boys and 10 girls failed in Mathematics. A roll number selected at random is found to be that of a student who has failed in Mathematics. What is the probability that it is of a girl? Regards, Indhu. Quote Link to comment Share on other sites More sharing options...
rumen Posted December 9, 2003 Share Posted December 9, 2003 Q1:Prob P =4/5, probQ=3/4, prob R=2/3 prob all of them to hit the target is 4/5x3/4x2/3=2/5 prob that two of them hit the target is P(P& Q)=4/5x3/4x1/3=1/5( 1/3 means that R does not hit, ),prob (P&R)=4/5x2/3x1/4=2/15 Prob (Q&R)=3/4x2/3x1/5=1/10 or probability of at least 2 is 2/5+1/5+2/15+1/10=5/6..it is complicated...the rest later:) Ok! Q2...Prob silver from first purse is 2/6 , from the other is 4/7 or Pa+Pb= 38/42:) Quote Link to comment Share on other sites More sharing options...
vknittala Posted December 9, 2003 Share Posted December 9, 2003 Isolved the Q1 this way: I subtracted the prob of hitting 1 shot and 0 shots from 1. Probabilit of hitting 1 shot is: [3c1(4/5)(1/4)(1/3) + 3c1(1/5)(3/4)(1/3) + 3c1(1/5)(1/4)(2/3)] = (27/60) Probabilit of hitting no shots is: 3c0(1/5)(1/4)(1/3) = 1/60 Now tht probability of hitting atleast two shots is: 1 - [(27/60) + (1/60)] = 32/60 = 8/15 May be wrong:p, if so please let know. 2Q) I agree with Rumen!!!:)2c1/6c1 + 4c1/7c1 = 19/21 3Q) P(A) * P(B) * P© = (4/5) * (2/3) * (3/7) 4Q) (11C5/13C7) = 7/26 (Plz check the calculation:)) 5Q) (6c1/10c1) + (4c1/10c1) = 1 (Plz check ) 6Q) P(G) = 40/100 = 2/5 P(B) = 60/100 = 3/5 P(G'sfailed) = 10/30 P(B'sFailed) = 20/30 (G&Failed in math) = (1/3)*(2/5) = 2/15 I am waiting for others reply to taly my answers Thanks:) Quote Link to comment Share on other sites More sharing options...
lenz Posted December 9, 2003 Share Posted December 9, 2003 hey rumen..for Q2 what about selecting one of the purses...shouldnt the ans be multilied by 1/2 that is probability of selecting one of the two purse :shy:..thus ans being 19/42 ?? Quote Link to comment Share on other sites More sharing options...
manish2003 Posted December 9, 2003 Share Posted December 9, 2003 Originally posted by ind21 Hi !! 1.P can hit a target 4 times in 5 shots, Q, 3 times in 4 shots and R, twice in 3 shots,. They fire simultaneously. What is the probability that at least 2 shots hit? P (atleast 2 targets are hit) = (P & Q & not R) + (P & R & notQ) + (R & Q & notP) + (P & Q & R) = (4/5 x 3/4 x 1/3) + (4/5 x 2/3 x 1/4) + (3/4 x 2/3 x 1/5) + (4/5 x 3/4 x 2/3) = 5/6 - Manish Quote Link to comment Share on other sites More sharing options...
manish2003 Posted December 9, 2003 Share Posted December 9, 2003 Originally posted by ind21 2.A purse contains 2 silver coins and 4 copper coins and another contains 4 silver coins and 3 copper coins. If a coin is selected at random from one of the two purses what the probability that it is a silver coin? P = P (selecting A and getting a silver coin) + (selecting B and getting a silver coin) = (1/2)x(2/4) + (1/2)x(4/7) = 19/42 - Manish Quote Link to comment Share on other sites More sharing options...
manish2003 Posted December 9, 2003 Share Posted December 9, 2003 Originally posted by ind21 3.The probability that A can solve a problem is 4/5, that B can solve it is 2/3 and that C can solve it is 3/7. If all of them try independently, find the probability that the problem will be solved. P (problem will be solved) = 1 - p (problem is not solved) = 1 - (1/5 x 1/3 x 4/7) = 1 - 4/105 = 101/105 - Manish Quote Link to comment Share on other sites More sharing options...
manish2003 Posted December 10, 2003 Share Posted December 10, 2003 Originally posted by ind21 Hi !! 4. A committee of 4 boys and 3 girls is to be formed by lots from 8 boys and 5 girls. One of the boys is brother of one of the girls. Find the probability that both are included in the committee. I am not very sure about this but this is my take on it: P = (7C3 x 4C2)/ (8C4 x 5C3) = 3/10 - Manish Quote Link to comment Share on other sites More sharing options...
lenz Posted December 10, 2003 Share Posted December 10, 2003 exactly manish, even I'm getting the asn same as urs..but for the last one I got 3/16 (10/40)/(40/130)..3/16 but not sure.... hey ind21...please post the ans Quote Link to comment Share on other sites More sharing options...
manish2003 Posted December 10, 2003 Share Posted December 10, 2003 Originally posted by ind21 5. An urn contains four white and five black balls, a second urn contains five white and four black balls. One ball is transferred from the first to the second urn, then a ball is drawn from the second urn. What the probability that it is white? P = p( white ball taken from first Urn)x P(selecting a white ball from the second Urn) + P (black ball is chosen from first urn)x P(selecting a while ball from second Urn) = (4/9 x 6/10) + (5/9 x 5/10) = 49/90 - Manish Quote Link to comment Share on other sites More sharing options...
manish2003 Posted December 10, 2003 Share Posted December 10, 2003 Originally posted by ind21 6.In a class of 100 students there are 60 boys and 40 girls, 20 boys and 10 girls failed in Mathematics. A roll number selected at random is found to be that of a student who has failed in Mathematics. What is the probability that it is of a girl? I am not sure but this looks pretty straight forward: P = P(failed girl)/P(all failed students) = 10/30 = 1/3 - Manish Quote Link to comment Share on other sites More sharing options...
ind21 Posted December 10, 2003 Author Share Posted December 10, 2003 Hi , Lenz , As I have mentioned previously , I have the questions alone . I don't have the answere for the questions. I got it from a group. Indhu. Originally posted by lenz hey rumen..for Q2 what about selecting one of the purses...shouldnt the ans be multilied by 1/2 that is probability of selecting one of the two purse :shy:..thus ans being 19/42 ?? Quote Link to comment Share on other sites More sharing options...
rumen Posted December 10, 2003 Share Posted December 10, 2003 question 2 , my fault , you are right , it is 19/42 Quote Link to comment Share on other sites More sharing options...
vknittala Posted December 10, 2003 Share Posted December 10, 2003 Originally posted by manish2003 Originally posted by ind21 Hi !! 1.P can hit a target 4 times in 5 shots, Q, 3 times in 4 shots and R, twice in 3 shots,. They fire simultaneously. What is the probability that at least 2 shots hit? P (atleast 2 targets are hit) = (P & Q & not R) + (P & R & notQ) + (R & Q & notP) + (P & Q & R) = (4/5 x 3/4 x 1/3) + (4/5 x 2/3 x 1/4) + (3/4 x 2/3 x 1/5) + (4/5 x 3/4 x 2/3) = 5/6 - Manish Are we not supposed to count the position of the P, Q and R? I mean first P, second Q and third R hits and they change their positions as Q, R, and P and then R, P and Q and so on....and in the first three P hitting and secoand in the second three positions Q hitting, and in the third set R hitting etc. Could you please clarify my dobut Thanks Quote Link to comment Share on other sites More sharing options...
vknittala Posted December 10, 2003 Share Posted December 10, 2003 Originally posted by manish2003 Originally posted by ind21 3.The probability that A can solve a problem is 4/5, that B can solve it is 2/3 and that C can solve it is 3/7. If all of them try independently, find the probability that the problem will be solved. P (problem will be solved) = 1 - p (problem is not solved) = 1 - (1/5 x 1/3 x 4/7) = 1 - 4/105 = 101/105 - Manish Please explain why P(A)*P(B)*P© is not correct in solving the problem? After looking at yours I felt your approach is correct buy why is this wrong?:):) Quote Link to comment Share on other sites More sharing options...
vknittala Posted December 10, 2003 Share Posted December 10, 2003 Originally posted by manish2003 Originally posted by ind21 5. An urn contains four white and five black balls, a second urn contains five white and four black balls. One ball is transferred from the first to the second urn, then a ball is drawn from the second urn. What the probability that it is white? P = p( white ball taken from first Urn)x P(selecting a white ball from the second Urn) + P (black ball is chosen from first urn)x P(selecting a while ball from second Urn) = (4/9 x 6/10) + (5/9 x 5/10) = 49/90 - Manish I agree with yours and !!:) Quote Link to comment Share on other sites More sharing options...
vknittala Posted December 10, 2003 Share Posted December 10, 2003 Originally posted by manish2003 Originally posted by ind21 6.In a class of 100 students there are 60 boys and 40 girls, 20 boys and 10 girls failed in Mathematics. A roll number selected at random is found to be that of a student who has failed in Mathematics. What is the probability that it is of a girl? I am not sure but this looks pretty straight forward: P = P(failed girl)/P(all failed students) = 10/30 = 1/3 - Manish Here dont you think P(Failed girl) = (number of failed girls()/total failed students) P(all failed students) = (number of failed students)/(Total number of students) now P (FG & FS) = (1/3)*(3/10) = 1/10 Plz tell me where am I lost?:) Quote Link to comment Share on other sites More sharing options...
manish2003 Posted December 10, 2003 Share Posted December 10, 2003 Originally posted by vknittala Originally posted by manish2003 Originally posted by ind21 Hi !! 1.P can hit a target 4 times in 5 shots, Q, 3 times in 4 shots and R, twice in 3 shots,. They fire simultaneously. What is the probability that at least 2 shots hit? P (atleast 2 targets are hit) = (P & Q & not R) + (P & R & notQ) + (R & Q & notP) + (P & Q & R) = (4/5 x 3/4 x 1/3) + (4/5 x 2/3 x 1/4) + (3/4 x 2/3 x 1/5) + (4/5 x 3/4 x 2/3) = 5/6 - Manish Are we not supposed to count the position of the P, Q and R? I mean first P, second Q and third R hits and they change their positions as Q, R, and P and then R, P and Q and so on....and in the first three P hitting and secoand in the second three positions Q hitting, and in the third set R hitting etc. Could you please clarify my dobut Thanks Problem says "simultaneously". Order does not matter here. - Manish Quote Link to comment Share on other sites More sharing options...
manish2003 Posted December 10, 2003 Share Posted December 10, 2003 Originally posted by vknittala Originally posted by manish2003 Originally posted by ind21 3.The probability that A can solve a problem is 4/5, that B can solve it is 2/3 and that C can solve it is 3/7. If all of them try independently, find the probability that the problem will be solved. P (problem will be solved) = 1 - p (problem is not solved) = 1 - (1/5 x 1/3 x 4/7) = 1 - 4/105 = 101/105 - Manish Please explain why P(A)*P(B)*P© is not correct in solving the problem? After looking at yours I felt your approach is correct buy why is this wrong?:):) P(A)*P(B)*P© means problem is solved by all of them, but problem asks for probability of problem getting solved. If any one of them can solve the problem, problem will be solved. - Manish Quote Link to comment Share on other sites More sharing options...
manish2003 Posted December 10, 2003 Share Posted December 10, 2003 Originally posted by vknittala Originally posted by manish2003 Originally posted by ind21 6.In a class of 100 students there are 60 boys and 40 girls, 20 boys and 10 girls failed in Mathematics. A roll number selected at random is found to be that of a student who has failed in Mathematics. What is the probability that it is of a girl? I am not sure but this looks pretty straight forward: P = P(failed girl)/P(all failed students) = 10/30 = 1/3 - Manish Here dont you think P(Failed girl) = (number of failed girls()/total failed students) P(all failed students) = (number of failed students)/(Total number of students) now P (FG & FS) = (1/3)*(3/10) = 1/10 Plz tell me where am I lost?:) Note that problem clearly says that the number was found to be of a failed student. That means we dont have to take in to account the probability of a failed student. - Manish Quote Link to comment Share on other sites More sharing options...
vknittala Posted December 10, 2003 Share Posted December 10, 2003 Originally posted by manish2003 Originally posted by vknittala Originally posted by manish2003 Originally posted by ind21 Hi !! 1.P can hit a target 4 times in 5 shots, Q, 3 times in 4 shots and R, twice in 3 shots,. They fire simultaneously. What is the probability that at least 2 shots hit? P (atleast 2 targets are hit) = (P & Q & not R) + (P & R & notQ) + (R & Q & notP) + (P & Q & R) = (4/5 x 3/4 x 1/3) + (4/5 x 2/3 x 1/4) + (3/4 x 2/3 x 1/5) + (4/5 x 3/4 x 2/3) = 5/6 - Manish Are we not supposed to count the position of the P, Q and R? I mean first P, second Q and third R hits and they change their positions as Q, R, and P and then R, P and Q and so on....and in the first three P hitting and secoand in the second three positions Q hitting, and in the third set R hitting etc. Could you please clarify my dobut Thanks Problem says "simultaneously". Order does not matter here. - Manish So if the question does not mention the word simultaneously then is my approach correct? Thanks Quote Link to comment Share on other sites More sharing options...
manish2003 Posted December 11, 2003 Share Posted December 11, 2003 Originally posted by vknittala So if the question does not mention the word simultaneously then is my approach correct? Thanks I am not really sure. This does not look like a question where order matters. - Manish Quote Link to comment Share on other sites More sharing options...
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