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ind21

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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.

 

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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:)

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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:)

 

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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

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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

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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

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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

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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

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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

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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 ??

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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

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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?:):)

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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 !!:)

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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?:)

 

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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

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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

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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

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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

 

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