Mixing
Probability of rolling first 3 with a fair die before 10th roll and after 4th roll.
$begingroup$
My question
Lets say we have a fair 6-sided die that we roll continuously. We roll this same dice 10 times in a row. What is the probability that we get a 3 after our 4th roll but before our 10th roll?
In other words, given the events
$$R_1,R_2,R_3,R_4,R_5,R_6,R_7,R_8,R_9,R_{10}$$
What is the probability that
$$R_5,R_6,R_7,R_8,R_9$$
We get our first 3?
My attempt
$R_1,R_2,R_3,R_4$ cannot be 3s. This means that they have probabilities $frac{5}{6}$.
The first 3 appearing somewhere here $R_5,R_6,R_7,R_8,R_9$ is probably something like $(5 text{ choose } 1) * 6^4$ (I have no reasoning, just intuition).
I have no idea how to handle the last part of the problem with "before $R_{10}$"...
probability combinatorics
edited Jan 24 at 10:59
N. F. Taussig
44.7k103358
asked Jan 24 at 2:46
AlanSTACKAlanSTACK
1,76421526
$endgroup$
add a comment |
$begingroup$
My question
Lets say we have a fair 6-sided die that we roll continuously. We roll this same dice 10 times in a row. What is the probability that we get a 3 after our 4th roll but before our 10th roll?
In other words, given the events
$$R_1,R_2,R_3,R_4,R_5,R_6,R_7,R_8,R_9,R_{10}$$
What is the probability that
$$R_5,R_6,R_7,R_8,R_9$$
We get our first 3?
My attempt
$R_1,R_2,R_3,R_4$ cannot be 3s. This means that they have probabilities $frac{5}{6}$.
The first 3 appearing somewhere here $R_5,R_6,R_7,R_8,R_9$ is probably something like $(5 text{ choose } 1) * 6^4$ (I have no reasoning, just intuition).
I have no idea how to handle the last part of the problem with "before $R_{10}$"...
probability combinatorics
edited Jan 24 at 10:59
N. F. Taussig
44.7k103358
asked Jan 24 at 2:46
AlanSTACKAlanSTACK
1,76421526
$endgroup$
add a comment |
$begingroup$
My question
Lets say we have a fair 6-sided die that we roll continuously. We roll this same dice 10 times in a row. What is the probability that we get a 3 after our 4th roll but before our 10th roll?
In other words, given the events
$$R_1,R_2,R_3,R_4,R_5,R_6,R_7,R_8,R_9,R_{10}$$
What is the probability that
$$R_5,R_6,R_7,R_8,R_9$$
We get our first 3?
My attempt
$R_1,R_2,R_3,R_4$ cannot be 3s. This means that they have probabilities $frac{5}{6}$.
The first 3 appearing somewhere here $R_5,R_6,R_7,R_8,R_9$ is probably something like $(5 text{ choose } 1) * 6^4$ (I have no reasoning, just intuition).
I have no idea how to handle the last part of the problem with "before $R_{10}$"...
probability combinatorics
edited Jan 24 at 10:59
N. F. Taussig
44.7k103358
asked Jan 24 at 2:46
AlanSTACKAlanSTACK
1,76421526
$endgroup$
My question
Lets say we have a fair 6-sided die that we roll continuously. We roll this same dice 10 times in a row. What is the probability that we get a 3 after our 4th roll but before our 10th roll?
In other words, given the events
$$R_1,R_2,R_3,R_4,R_5,R_6,R_7,R_8,R_9,R_{10}$$
What is the probability that
$$R_5,R_6,R_7,R_8,R_9$$
We get our first 3?
My attempt
$R_1,R_2,R_3,R_4$ cannot be 3s. This means that they have probabilities $frac{5}{6}$.
The first 3 appearing somewhere here $R_5,R_6,R_7,R_8,R_9$ is probably something like $(5 text{ choose } 1) * 6^4$ (I have no reasoning, just intuition).
I have no idea how to handle the last part of the problem with "before $R_{10}$"...
probability combinatorics
probability combinatorics
edited Jan 24 at 10:59
N. F. Taussig
44.7k103358
asked Jan 24 at 2:46
AlanSTACKAlanSTACK
1,76421526
edited Jan 24 at 10:59
N. F. Taussig
44.7k103358
asked Jan 24 at 2:46
AlanSTACKAlanSTACK
1,76421526
edited Jan 24 at 10:59
N. F. Taussig
44.7k103358
edited Jan 24 at 10:59
N. F. Taussig
44.7k103358
edited Jan 24 at 10:59
N. F. Taussig
44.7k103358
44.7k103358
asked Jan 24 at 2:46
AlanSTACKAlanSTACK
1,76421526
asked Jan 24 at 2:46
AlanSTACKAlanSTACK
1,76421526
asked Jan 24 at 2:46
AlanSTACKAlanSTACK
1,76421526
1,76421526
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
I am making the assumption that you're going to roll the dice 10 times, even if you manage to roll a 3 at a desired location (e.g. if the first 3 appears at 7, you continue rolling to the 10th dice.)
To deal with $R_5, R_6, R_7, R_8, R_9$, observe that we can model this with the binomial distribution. Let X denote the number of 3's in the rolls $R_5, R_6, R_7, R_8, R_9$.
$$
X sim BinBig(n=5, p=frac{1}{6}Big)
$$
(Where we have 5 tries and the probability of rolling a $3$ on any of these tries is $frac{1}{6}$)
We want the probability:
$$
Pr(X > 0) = 1- Pr(X=0) = 1 - Big(frac{5}{6}Big)^5
$$
So, multiplying this with the probability that you arrived at for not rolling a 3 in the first 4 rolls, we obtain the final result:
$$
P(text{No 3's in the first four rolls} cap text{at least one 3 in rolls 5 to 9} ) = Big(frac{5}{6}Big)^4 Big( 1 - Big(frac{5}{6}Big)^5 Big)
$$
Which I believe works out to be $approx0.288$
answered Jan 24 at 3:03
Sean LeeSean Lee
533212
$endgroup$
$begingroup$
How does "but before our 10th roll?" factor into your answer?
$endgroup$
– AlanSTACK
Jan 24 at 3:11
1
$begingroup$
I'm assuming that you do not care about the 10th roll - whether or not the 10th roll is a 3 or otherwise, it's irrelevant. Only what happens in the first 9 rolls is relevant.
$endgroup$
– Sean Lee
Jan 24 at 3:13
$begingroup$
Does that part of the question not change the probability?
$endgroup$
– AlanSTACK
Jan 24 at 3:15
2
$begingroup$
To put it another way, success consists in getting a 3 in at least one of the trials $R_5,R_6,R_7,R_8,R_9$ after not getting a 3 in any of the trials $R_1,R_2,R_3,R_4$. So the solution doesn't require information about the outcomes of the tenth or later rolls.
$endgroup$
– hardmath
Jan 24 at 3:19
1
$begingroup$
@hardmath Thank you, that cleared it up for me.
$endgroup$
– AlanSTACK
Jan 24 at 3:25
add a comment |
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1 Answer
1
active
oldest
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1 Answer
1
active
oldest
votes
active
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$begingroup$
I am making the assumption that you're going to roll the dice 10 times, even if you manage to roll a 3 at a desired location (e.g. if the first 3 appears at 7, you continue rolling to the 10th dice.)
To deal with $R_5, R_6, R_7, R_8, R_9$, observe that we can model this with the binomial distribution. Let X denote the number of 3's in the rolls $R_5, R_6, R_7, R_8, R_9$.
$$
X sim BinBig(n=5, p=frac{1}{6}Big)
$$
(Where we have 5 tries and the probability of rolling a $3$ on any of these tries is $frac{1}{6}$)
We want the probability:
$$
Pr(X > 0) = 1- Pr(X=0) = 1 - Big(frac{5}{6}Big)^5
$$
So, multiplying this with the probability that you arrived at for not rolling a 3 in the first 4 rolls, we obtain the final result:
$$
P(text{No 3's in the first four rolls} cap text{at least one 3 in rolls 5 to 9} ) = Big(frac{5}{6}Big)^4 Big( 1 - Big(frac{5}{6}Big)^5 Big)
$$
Which I believe works out to be $approx0.288$
answered Jan 24 at 3:03
Sean LeeSean Lee
533212
$endgroup$
$begingroup$
How does "but before our 10th roll?" factor into your answer?
$endgroup$
– AlanSTACK
Jan 24 at 3:11
1
$begingroup$
I'm assuming that you do not care about the 10th roll - whether or not the 10th roll is a 3 or otherwise, it's irrelevant. Only what happens in the first 9 rolls is relevant.
$endgroup$
– Sean Lee
Jan 24 at 3:13
$begingroup$
Does that part of the question not change the probability?
$endgroup$
– AlanSTACK
Jan 24 at 3:15
2
$begingroup$
To put it another way, success consists in getting a 3 in at least one of the trials $R_5,R_6,R_7,R_8,R_9$ after not getting a 3 in any of the trials $R_1,R_2,R_3,R_4$. So the solution doesn't require information about the outcomes of the tenth or later rolls.
$endgroup$
– hardmath
Jan 24 at 3:19
1
$begingroup$
@hardmath Thank you, that cleared it up for me.
$endgroup$
– AlanSTACK
Jan 24 at 3:25
add a comment |
$begingroup$
I am making the assumption that you're going to roll the dice 10 times, even if you manage to roll a 3 at a desired location (e.g. if the first 3 appears at 7, you continue rolling to the 10th dice.)
To deal with $R_5, R_6, R_7, R_8, R_9$, observe that we can model this with the binomial distribution. Let X denote the number of 3's in the rolls $R_5, R_6, R_7, R_8, R_9$.
$$
X sim BinBig(n=5, p=frac{1}{6}Big)
$$
(Where we have 5 tries and the probability of rolling a $3$ on any of these tries is $frac{1}{6}$)
We want the probability:
$$
Pr(X > 0) = 1- Pr(X=0) = 1 - Big(frac{5}{6}Big)^5
$$
So, multiplying this with the probability that you arrived at for not rolling a 3 in the first 4 rolls, we obtain the final result:
$$
P(text{No 3's in the first four rolls} cap text{at least one 3 in rolls 5 to 9} ) = Big(frac{5}{6}Big)^4 Big( 1 - Big(frac{5}{6}Big)^5 Big)
$$
Which I believe works out to be $approx0.288$
answered Jan 24 at 3:03
Sean LeeSean Lee
533212
$endgroup$
$begingroup$
How does "but before our 10th roll?" factor into your answer?
$endgroup$
– AlanSTACK
Jan 24 at 3:11
1
$begingroup$
I'm assuming that you do not care about the 10th roll - whether or not the 10th roll is a 3 or otherwise, it's irrelevant. Only what happens in the first 9 rolls is relevant.
$endgroup$
– Sean Lee
Jan 24 at 3:13
$begingroup$
Does that part of the question not change the probability?
$endgroup$
– AlanSTACK
Jan 24 at 3:15
2
$begingroup$
To put it another way, success consists in getting a 3 in at least one of the trials $R_5,R_6,R_7,R_8,R_9$ after not getting a 3 in any of the trials $R_1,R_2,R_3,R_4$. So the solution doesn't require information about the outcomes of the tenth or later rolls.
$endgroup$
– hardmath
Jan 24 at 3:19
1
$begingroup$
@hardmath Thank you, that cleared it up for me.
$endgroup$
– AlanSTACK
Jan 24 at 3:25
add a comment |
$begingroup$
I am making the assumption that you're going to roll the dice 10 times, even if you manage to roll a 3 at a desired location (e.g. if the first 3 appears at 7, you continue rolling to the 10th dice.)
To deal with $R_5, R_6, R_7, R_8, R_9$, observe that we can model this with the binomial distribution. Let X denote the number of 3's in the rolls $R_5, R_6, R_7, R_8, R_9$.
$$
X sim BinBig(n=5, p=frac{1}{6}Big)
$$
(Where we have 5 tries and the probability of rolling a $3$ on any of these tries is $frac{1}{6}$)
We want the probability:
$$
Pr(X > 0) = 1- Pr(X=0) = 1 - Big(frac{5}{6}Big)^5
$$
So, multiplying this with the probability that you arrived at for not rolling a 3 in the first 4 rolls, we obtain the final result:
$$
P(text{No 3's in the first four rolls} cap text{at least one 3 in rolls 5 to 9} ) = Big(frac{5}{6}Big)^4 Big( 1 - Big(frac{5}{6}Big)^5 Big)
$$
Which I believe works out to be $approx0.288$
answered Jan 24 at 3:03
Sean LeeSean Lee
533212
$endgroup$
I am making the assumption that you're going to roll the dice 10 times, even if you manage to roll a 3 at a desired location (e.g. if the first 3 appears at 7, you continue rolling to the 10th dice.)
To deal with $R_5, R_6, R_7, R_8, R_9$, observe that we can model this with the binomial distribution. Let X denote the number of 3's in the rolls $R_5, R_6, R_7, R_8, R_9$.
$$
X sim BinBig(n=5, p=frac{1}{6}Big)
$$
(Where we have 5 tries and the probability of rolling a $3$ on any of these tries is $frac{1}{6}$)
We want the probability:
$$
Pr(X > 0) = 1- Pr(X=0) = 1 - Big(frac{5}{6}Big)^5
$$
So, multiplying this with the probability that you arrived at for not rolling a 3 in the first 4 rolls, we obtain the final result:
$$
P(text{No 3's in the first four rolls} cap text{at least one 3 in rolls 5 to 9} ) = Big(frac{5}{6}Big)^4 Big( 1 - Big(frac{5}{6}Big)^5 Big)
$$
Which I believe works out to be $approx0.288$
answered Jan 24 at 3:03
Sean LeeSean Lee
533212
answered Jan 24 at 3:03
Sean LeeSean Lee
533212
answered Jan 24 at 3:03
Sean LeeSean Lee
533212
answered Jan 24 at 3:03
Sean LeeSean Lee
533212
533212
$begingroup$
How does "but before our 10th roll?" factor into your answer?
$endgroup$
– AlanSTACK
Jan 24 at 3:11
1
$begingroup$
I'm assuming that you do not care about the 10th roll - whether or not the 10th roll is a 3 or otherwise, it's irrelevant. Only what happens in the first 9 rolls is relevant.
$endgroup$
– Sean Lee
Jan 24 at 3:13
$begingroup$
Does that part of the question not change the probability?
$endgroup$
– AlanSTACK
Jan 24 at 3:15
2
$begingroup$
To put it another way, success consists in getting a 3 in at least one of the trials $R_5,R_6,R_7,R_8,R_9$ after not getting a 3 in any of the trials $R_1,R_2,R_3,R_4$. So the solution doesn't require information about the outcomes of the tenth or later rolls.
$endgroup$
– hardmath
Jan 24 at 3:19
1
$begingroup$
@hardmath Thank you, that cleared it up for me.
$endgroup$
– AlanSTACK
Jan 24 at 3:25
add a comment |
$begingroup$
How does "but before our 10th roll?" factor into your answer?
$endgroup$
– AlanSTACK
Jan 24 at 3:11
1
$begingroup$
I'm assuming that you do not care about the 10th roll - whether or not the 10th roll is a 3 or otherwise, it's irrelevant. Only what happens in the first 9 rolls is relevant.
$endgroup$
– Sean Lee
Jan 24 at 3:13
$begingroup$
Does that part of the question not change the probability?
$endgroup$
– AlanSTACK
Jan 24 at 3:15
2
$begingroup$
To put it another way, success consists in getting a 3 in at least one of the trials $R_5,R_6,R_7,R_8,R_9$ after not getting a 3 in any of the trials $R_1,R_2,R_3,R_4$. So the solution doesn't require information about the outcomes of the tenth or later rolls.
$endgroup$
– hardmath
Jan 24 at 3:19
1
$begingroup$
@hardmath Thank you, that cleared it up for me.
$endgroup$
– AlanSTACK
Jan 24 at 3:25
$begingroup$
How does "but before our 10th roll?" factor into your answer?
$endgroup$
– AlanSTACK
Jan 24 at 3:11
$begingroup$
How does "but before our 10th roll?" factor into your answer?
$endgroup$
– AlanSTACK
Jan 24 at 3:11
1
1
$begingroup$
I'm assuming that you do not care about the 10th roll - whether or not the 10th roll is a 3 or otherwise, it's irrelevant. Only what happens in the first 9 rolls is relevant.
$endgroup$
– Sean Lee
Jan 24 at 3:13
$begingroup$
I'm assuming that you do not care about the 10th roll - whether or not the 10th roll is a 3 or otherwise, it's irrelevant. Only what happens in the first 9 rolls is relevant.
$endgroup$
– Sean Lee
Jan 24 at 3:13
$begingroup$
Does that part of the question not change the probability?
$endgroup$
– AlanSTACK
Jan 24 at 3:15
$begingroup$
Does that part of the question not change the probability?
$endgroup$
– AlanSTACK
Jan 24 at 3:15
2
2
$begingroup$
To put it another way, success consists in getting a 3 in at least one of the trials $R_5,R_6,R_7,R_8,R_9$ after not getting a 3 in any of the trials $R_1,R_2,R_3,R_4$. So the solution doesn't require information about the outcomes of the tenth or later rolls.
$endgroup$
– hardmath
Jan 24 at 3:19
$begingroup$
To put it another way, success consists in getting a 3 in at least one of the trials $R_5,R_6,R_7,R_8,R_9$ after not getting a 3 in any of the trials $R_1,R_2,R_3,R_4$. So the solution doesn't require information about the outcomes of the tenth or later rolls.
$endgroup$
– hardmath
Jan 24 at 3:19
1
1
$begingroup$
@hardmath Thank you, that cleared it up for me.
$endgroup$
– AlanSTACK
Jan 24 at 3:25
$begingroup$
@hardmath Thank you, that cleared it up for me.
$endgroup$
– AlanSTACK
Jan 24 at 3:25
add a comment |
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