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What is the probability that the second number that is divisible by 5 is drawn before the third number that is not divisible by 5?
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An urn contains 25 tickets numbered from 1 to 25.Tickets are drawn
from the urn in succession, with replacement.
What is the probability that the second number that is divisible by 5
is drawn before the third number that is not divisible by 5?
I understand that I should use the negative binomial however I don’t know how and what event should I assess as success and what event should I assess as failure and when I apply the formula I always get different answer which, the correct one, must be $0.181$. can you please help me?
Thanks you for all your detailed and correct answers? Can I find the same answer using the negative binomial?
probability
asked Jul 14 at 17:56
Rayri
485
 |Â
show 3 more comments
up vote
0
down vote
favorite
An urn contains 25 tickets numbered from 1 to 25.Tickets are drawn
from the urn in succession, with replacement.
What is the probability that the second number that is divisible by 5
is drawn before the third number that is not divisible by 5?
I understand that I should use the negative binomial however I don’t know how and what event should I assess as success and what event should I assess as failure and when I apply the formula I always get different answer which, the correct one, must be $0.181$. can you please help me?
Thanks you for all your detailed and correct answers? Can I find the same answer using the negative binomial?
probability
asked Jul 14 at 17:56
Rayri
485
1
Note that since the drawing is with replacement, the tickets numbered 5, 10, 15, 20, 25 are indistinguishable from one another, and the other twenty tickets are indistinguishable from one another. So it suffices to pretend that there are only the tickets 1, 2, 3, 4, 5 in the urn. (Or, that there are only two tickets, "divisible-by-5" and "not-divisible-by-5", drawn with probabilities $frac15$ and $frac45$, respectively.)
– Greg Martin
Jul 14 at 17:59
2
This is perhaps best done with a tree, since there will be very few branches to evaluate.
– joriki
Jul 14 at 18:02
How many times do you draw out the tickets?
– Mostafa Ayaz
Jul 14 at 18:07
1
Think about the maximum number of tickets that can drawn before one of the events occurs. This is a small problem.
– saulspatz
Jul 14 at 18:07
can you please elaborate more on that?
– Rayri
Jul 14 at 18:08
 |Â
show 3 more comments
up vote
0
down vote
favorite
up vote
0
down vote
favorite
An urn contains 25 tickets numbered from 1 to 25.Tickets are drawn
from the urn in succession, with replacement.
What is the probability that the second number that is divisible by 5
is drawn before the third number that is not divisible by 5?
I understand that I should use the negative binomial however I don’t know how and what event should I assess as success and what event should I assess as failure and when I apply the formula I always get different answer which, the correct one, must be $0.181$. can you please help me?
Thanks you for all your detailed and correct answers? Can I find the same answer using the negative binomial?
probability
asked Jul 14 at 17:56
Rayri
485
An urn contains 25 tickets numbered from 1 to 25.Tickets are drawn
from the urn in succession, with replacement.
What is the probability that the second number that is divisible by 5
is drawn before the third number that is not divisible by 5?
I understand that I should use the negative binomial however I don’t know how and what event should I assess as success and what event should I assess as failure and when I apply the formula I always get different answer which, the correct one, must be $0.181$. can you please help me?
Thanks you for all your detailed and correct answers? Can I find the same answer using the negative binomial?
probability
asked Jul 14 at 17:56
Rayri
485
edited Jul 14 at 19:19
asked Jul 14 at 17:56
Rayri
485
asked Jul 14 at 17:56
Rayri
485
asked Jul 14 at 17:56
Rayri
485
485
1
Note that since the drawing is with replacement, the tickets numbered 5, 10, 15, 20, 25 are indistinguishable from one another, and the other twenty tickets are indistinguishable from one another. So it suffices to pretend that there are only the tickets 1, 2, 3, 4, 5 in the urn. (Or, that there are only two tickets, "divisible-by-5" and "not-divisible-by-5", drawn with probabilities $frac15$ and $frac45$, respectively.)
– Greg Martin
Jul 14 at 17:59
2
This is perhaps best done with a tree, since there will be very few branches to evaluate.
– joriki
Jul 14 at 18:02
How many times do you draw out the tickets?
– Mostafa Ayaz
Jul 14 at 18:07
1
Think about the maximum number of tickets that can drawn before one of the events occurs. This is a small problem.
– saulspatz
Jul 14 at 18:07
can you please elaborate more on that?
– Rayri
Jul 14 at 18:08
 |Â
show 3 more comments
1
Note that since the drawing is with replacement, the tickets numbered 5, 10, 15, 20, 25 are indistinguishable from one another, and the other twenty tickets are indistinguishable from one another. So it suffices to pretend that there are only the tickets 1, 2, 3, 4, 5 in the urn. (Or, that there are only two tickets, "divisible-by-5" and "not-divisible-by-5", drawn with probabilities $frac15$ and $frac45$, respectively.)
– Greg Martin
Jul 14 at 17:59
2
This is perhaps best done with a tree, since there will be very few branches to evaluate.
– joriki
Jul 14 at 18:02
How many times do you draw out the tickets?
– Mostafa Ayaz
Jul 14 at 18:07
1
Think about the maximum number of tickets that can drawn before one of the events occurs. This is a small problem.
– saulspatz
Jul 14 at 18:07
can you please elaborate more on that?
– Rayri
Jul 14 at 18:08
1
1
Note that since the drawing is with replacement, the tickets numbered 5, 10, 15, 20, 25 are indistinguishable from one another, and the other twenty tickets are indistinguishable from one another. So it suffices to pretend that there are only the tickets 1, 2, 3, 4, 5 in the urn. (Or, that there are only two tickets, "divisible-by-5" and "not-divisible-by-5", drawn with probabilities $frac15$ and $frac45$, respectively.)
– Greg Martin
Jul 14 at 17:59
Note that since the drawing is with replacement, the tickets numbered 5, 10, 15, 20, 25 are indistinguishable from one another, and the other twenty tickets are indistinguishable from one another. So it suffices to pretend that there are only the tickets 1, 2, 3, 4, 5 in the urn. (Or, that there are only two tickets, "divisible-by-5" and "not-divisible-by-5", drawn with probabilities $frac15$ and $frac45$, respectively.)
– Greg Martin
Jul 14 at 17:59
2
2
This is perhaps best done with a tree, since there will be very few branches to evaluate.
– joriki
Jul 14 at 18:02
This is perhaps best done with a tree, since there will be very few branches to evaluate.
– joriki
Jul 14 at 18:02
How many times do you draw out the tickets?
– Mostafa Ayaz
Jul 14 at 18:07
How many times do you draw out the tickets?
– Mostafa Ayaz
Jul 14 at 18:07
1
1
Think about the maximum number of tickets that can drawn before one of the events occurs. This is a small problem.
– saulspatz
Jul 14 at 18:07
Think about the maximum number of tickets that can drawn before one of the events occurs. This is a small problem.
– saulspatz
Jul 14 at 18:07
can you please elaborate more on that?
– Rayri
Jul 14 at 18:08
can you please elaborate more on that?
– Rayri
Jul 14 at 18:08
 |Â
show 3 more comments
2 Answers
2
active
oldest
votes
up vote
1
down vote
accepted
It may be easier to think about the probability that three numbers not divisible by $5$ come first. Call it a success if the number isn't divisible by $5$ and failure if it is. The probability of success is $.8$ on any draw. If we have three success before two failures, then we got at most one failure so the game lasts $4$ turns at most.
If we got no failures, then the first three draws were successes; the probability of this is $.8^3.$ If we had one failure, it must have happened on one of the first $3$ draws; the probability of this is $3cdot.2cdot.8^3.$ The probability that we had three successes before two failures is the sum of these, but the problem asks for the complement, so the answer is $$1-.8^3-3cdot.2cdot.8^3=.1808$$
answered Jul 14 at 18:34
saulspatz
10.7k21323
So as I can infer, the probability of having 3 successes before 2 failures is the complement of that having two failures before 3 successes?
– Rayri
Jul 14 at 18:50
@Rayen Yes, that's what I mean.
– saulspatz
Jul 14 at 18:53
And what if we had 4 successes or more before we get the first failure
– Rayri
Jul 14 at 18:54
The game is over as soon as we get $3$ successes.
– saulspatz
Jul 14 at 18:55
Sorry if I am asking you a lot of questions, but I found the notion of complement really intriguing . how in general do I determine the complement of an event is there like a rule or something because sometimes I can find it obvious and sometimes it’s intriguing. I know it’s a bit off-topic but I would really appreciate your answer
– Rayri
Jul 14 at 19:15
 |Â
show 1 more comment
up vote
0
down vote
Hint:
If $E$ denotes the event that the second number that is divisible by $5$ is drawn before the third number that is not divisible by $5$ then:$$P(E)=P(DD)+P(DND)+P(DNND)+P(NDD)+P(NDND)+P(NNDD)$$where e.g. $DND$ denotes the event that first a number is drawn divisible by $5$, then a number not divisible by $5$ and then a number divisible by $5$.
answered Jul 14 at 18:38
drhab
86.7k541118
How can I get the same answer using the negative binomial?
– Rayri
Jul 14 at 20:38
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
It may be easier to think about the probability that three numbers not divisible by $5$ come first. Call it a success if the number isn't divisible by $5$ and failure if it is. The probability of success is $.8$ on any draw. If we have three success before two failures, then we got at most one failure so the game lasts $4$ turns at most.
If we got no failures, then the first three draws were successes; the probability of this is $.8^3.$ If we had one failure, it must have happened on one of the first $3$ draws; the probability of this is $3cdot.2cdot.8^3.$ The probability that we had three successes before two failures is the sum of these, but the problem asks for the complement, so the answer is $$1-.8^3-3cdot.2cdot.8^3=.1808$$
answered Jul 14 at 18:34
saulspatz
10.7k21323
So as I can infer, the probability of having 3 successes before 2 failures is the complement of that having two failures before 3 successes?
– Rayri
Jul 14 at 18:50
@Rayen Yes, that's what I mean.
– saulspatz
Jul 14 at 18:53
And what if we had 4 successes or more before we get the first failure
– Rayri
Jul 14 at 18:54
The game is over as soon as we get $3$ successes.
– saulspatz
Jul 14 at 18:55
Sorry if I am asking you a lot of questions, but I found the notion of complement really intriguing . how in general do I determine the complement of an event is there like a rule or something because sometimes I can find it obvious and sometimes it’s intriguing. I know it’s a bit off-topic but I would really appreciate your answer
– Rayri
Jul 14 at 19:15
 |Â
show 1 more comment
up vote
1
down vote
accepted
It may be easier to think about the probability that three numbers not divisible by $5$ come first. Call it a success if the number isn't divisible by $5$ and failure if it is. The probability of success is $.8$ on any draw. If we have three success before two failures, then we got at most one failure so the game lasts $4$ turns at most.
If we got no failures, then the first three draws were successes; the probability of this is $.8^3.$ If we had one failure, it must have happened on one of the first $3$ draws; the probability of this is $3cdot.2cdot.8^3.$ The probability that we had three successes before two failures is the sum of these, but the problem asks for the complement, so the answer is $$1-.8^3-3cdot.2cdot.8^3=.1808$$
answered Jul 14 at 18:34
saulspatz
10.7k21323
So as I can infer, the probability of having 3 successes before 2 failures is the complement of that having two failures before 3 successes?
– Rayri
Jul 14 at 18:50
@Rayen Yes, that's what I mean.
– saulspatz
Jul 14 at 18:53
And what if we had 4 successes or more before we get the first failure
– Rayri
Jul 14 at 18:54
The game is over as soon as we get $3$ successes.
– saulspatz
Jul 14 at 18:55
Sorry if I am asking you a lot of questions, but I found the notion of complement really intriguing . how in general do I determine the complement of an event is there like a rule or something because sometimes I can find it obvious and sometimes it’s intriguing. I know it’s a bit off-topic but I would really appreciate your answer
– Rayri
Jul 14 at 19:15
 |Â
show 1 more comment
up vote
1
down vote
accepted
up vote
1
down vote
accepted
It may be easier to think about the probability that three numbers not divisible by $5$ come first. Call it a success if the number isn't divisible by $5$ and failure if it is. The probability of success is $.8$ on any draw. If we have three success before two failures, then we got at most one failure so the game lasts $4$ turns at most.
If we got no failures, then the first three draws were successes; the probability of this is $.8^3.$ If we had one failure, it must have happened on one of the first $3$ draws; the probability of this is $3cdot.2cdot.8^3.$ The probability that we had three successes before two failures is the sum of these, but the problem asks for the complement, so the answer is $$1-.8^3-3cdot.2cdot.8^3=.1808$$
answered Jul 14 at 18:34
saulspatz
10.7k21323
It may be easier to think about the probability that three numbers not divisible by $5$ come first. Call it a success if the number isn't divisible by $5$ and failure if it is. The probability of success is $.8$ on any draw. If we have three success before two failures, then we got at most one failure so the game lasts $4$ turns at most.
If we got no failures, then the first three draws were successes; the probability of this is $.8^3.$ If we had one failure, it must have happened on one of the first $3$ draws; the probability of this is $3cdot.2cdot.8^3.$ The probability that we had three successes before two failures is the sum of these, but the problem asks for the complement, so the answer is $$1-.8^3-3cdot.2cdot.8^3=.1808$$
answered Jul 14 at 18:34
saulspatz
10.7k21323
answered Jul 14 at 18:34
saulspatz
10.7k21323
answered Jul 14 at 18:34
saulspatz
10.7k21323
answered Jul 14 at 18:34
saulspatz
10.7k21323
10.7k21323
So as I can infer, the probability of having 3 successes before 2 failures is the complement of that having two failures before 3 successes?
– Rayri
Jul 14 at 18:50
@Rayen Yes, that's what I mean.
– saulspatz
Jul 14 at 18:53
And what if we had 4 successes or more before we get the first failure
– Rayri
Jul 14 at 18:54
The game is over as soon as we get $3$ successes.
– saulspatz
Jul 14 at 18:55
Sorry if I am asking you a lot of questions, but I found the notion of complement really intriguing . how in general do I determine the complement of an event is there like a rule or something because sometimes I can find it obvious and sometimes it’s intriguing. I know it’s a bit off-topic but I would really appreciate your answer
– Rayri
Jul 14 at 19:15
 |Â
show 1 more comment
So as I can infer, the probability of having 3 successes before 2 failures is the complement of that having two failures before 3 successes?
– Rayri
Jul 14 at 18:50
@Rayen Yes, that's what I mean.
– saulspatz
Jul 14 at 18:53
And what if we had 4 successes or more before we get the first failure
– Rayri
Jul 14 at 18:54
The game is over as soon as we get $3$ successes.
– saulspatz
Jul 14 at 18:55
Sorry if I am asking you a lot of questions, but I found the notion of complement really intriguing . how in general do I determine the complement of an event is there like a rule or something because sometimes I can find it obvious and sometimes it’s intriguing. I know it’s a bit off-topic but I would really appreciate your answer
– Rayri
Jul 14 at 19:15
So as I can infer, the probability of having 3 successes before 2 failures is the complement of that having two failures before 3 successes?
– Rayri
Jul 14 at 18:50
So as I can infer, the probability of having 3 successes before 2 failures is the complement of that having two failures before 3 successes?
– Rayri
Jul 14 at 18:50
@Rayen Yes, that's what I mean.
– saulspatz
Jul 14 at 18:53
@Rayen Yes, that's what I mean.
– saulspatz
Jul 14 at 18:53
And what if we had 4 successes or more before we get the first failure
– Rayri
Jul 14 at 18:54
And what if we had 4 successes or more before we get the first failure
– Rayri
Jul 14 at 18:54
The game is over as soon as we get $3$ successes.
– saulspatz
Jul 14 at 18:55
The game is over as soon as we get $3$ successes.
– saulspatz
Jul 14 at 18:55
Sorry if I am asking you a lot of questions, but I found the notion of complement really intriguing . how in general do I determine the complement of an event is there like a rule or something because sometimes I can find it obvious and sometimes it’s intriguing. I know it’s a bit off-topic but I would really appreciate your answer
– Rayri
Jul 14 at 19:15
Sorry if I am asking you a lot of questions, but I found the notion of complement really intriguing . how in general do I determine the complement of an event is there like a rule or something because sometimes I can find it obvious and sometimes it’s intriguing. I know it’s a bit off-topic but I would really appreciate your answer
– Rayri
Jul 14 at 19:15
 |Â
show 1 more comment
up vote
0
down vote
Hint:
If $E$ denotes the event that the second number that is divisible by $5$ is drawn before the third number that is not divisible by $5$ then:$$P(E)=P(DD)+P(DND)+P(DNND)+P(NDD)+P(NDND)+P(NNDD)$$where e.g. $DND$ denotes the event that first a number is drawn divisible by $5$, then a number not divisible by $5$ and then a number divisible by $5$.
answered Jul 14 at 18:38
drhab
86.7k541118
How can I get the same answer using the negative binomial?
– Rayri
Jul 14 at 20:38
add a comment |Â
up vote
0
down vote
Hint:
If $E$ denotes the event that the second number that is divisible by $5$ is drawn before the third number that is not divisible by $5$ then:$$P(E)=P(DD)+P(DND)+P(DNND)+P(NDD)+P(NDND)+P(NNDD)$$where e.g. $DND$ denotes the event that first a number is drawn divisible by $5$, then a number not divisible by $5$ and then a number divisible by $5$.
answered Jul 14 at 18:38
drhab
86.7k541118
How can I get the same answer using the negative binomial?
– Rayri
Jul 14 at 20:38
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Hint:
If $E$ denotes the event that the second number that is divisible by $5$ is drawn before the third number that is not divisible by $5$ then:$$P(E)=P(DD)+P(DND)+P(DNND)+P(NDD)+P(NDND)+P(NNDD)$$where e.g. $DND$ denotes the event that first a number is drawn divisible by $5$, then a number not divisible by $5$ and then a number divisible by $5$.
answered Jul 14 at 18:38
drhab
86.7k541118
Hint:
If $E$ denotes the event that the second number that is divisible by $5$ is drawn before the third number that is not divisible by $5$ then:$$P(E)=P(DD)+P(DND)+P(DNND)+P(NDD)+P(NDND)+P(NNDD)$$where e.g. $DND$ denotes the event that first a number is drawn divisible by $5$, then a number not divisible by $5$ and then a number divisible by $5$.
answered Jul 14 at 18:38
drhab
86.7k541118
answered Jul 14 at 18:38
drhab
86.7k541118
answered Jul 14 at 18:38
drhab
86.7k541118
answered Jul 14 at 18:38
drhab
86.7k541118
86.7k541118
How can I get the same answer using the negative binomial?
– Rayri
Jul 14 at 20:38
add a comment |Â
How can I get the same answer using the negative binomial?
– Rayri
Jul 14 at 20:38
How can I get the same answer using the negative binomial?
– Rayri
Jul 14 at 20:38
How can I get the same answer using the negative binomial?
– Rayri
Jul 14 at 20:38
add a comment |Â
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1
Note that since the drawing is with replacement, the tickets numbered 5, 10, 15, 20, 25 are indistinguishable from one another, and the other twenty tickets are indistinguishable from one another. So it suffices to pretend that there are only the tickets 1, 2, 3, 4, 5 in the urn. (Or, that there are only two tickets, "divisible-by-5" and "not-divisible-by-5", drawn with probabilities $frac15$ and $frac45$, respectively.)
– Greg Martin
Jul 14 at 17:59
2
This is perhaps best done with a tree, since there will be very few branches to evaluate.
– joriki
Jul 14 at 18:02
How many times do you draw out the tickets?
– Mostafa Ayaz
Jul 14 at 18:07
1
Think about the maximum number of tickets that can drawn before one of the events occurs. This is a small problem.
– saulspatz
Jul 14 at 18:07
can you please elaborate more on that?
– Rayri
Jul 14 at 18:08