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Seemingly tricky dice question-probability that one event occurs before another event?
Clash Royale CLAN TAG#URR8PPP
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The question is as follows:
You roll two fair dice over and over. Let $A$ be the event you see two even sums. Let $B$ be the event you see a sum of $7$ four times. What is the probability that event $A$ occurs before event $B$?
I know that for mutually exclusive events with independent trials, the probability that event $E$ occurs before event $F$ is
$$fracmathbbP(E)mathbbP(E) + mathbbP(F).$$
I tried using this formula, but I ran into a problem. I calculated
$$mathbbP(A)=frac12cdot frac12=frac14 textand mathbbP(B)=left ( frac16 right )^4=frac11296.$$
However, I then realized that these probabilites are the events that two even sums occur $textitin a row,$ and similarly for my $mathbbP(B).$
Does anyone have any suggestions on how to figure this out, or even if this formula is the one I should be using?
Thanks in advance!
Edit: I know there is a geometric distribution involved.
probability dice
asked 2 hours ago
StatGuy
44719
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up vote
2
down vote
favorite
The question is as follows:
You roll two fair dice over and over. Let $A$ be the event you see two even sums. Let $B$ be the event you see a sum of $7$ four times. What is the probability that event $A$ occurs before event $B$?
I know that for mutually exclusive events with independent trials, the probability that event $E$ occurs before event $F$ is
$$fracmathbbP(E)mathbbP(E) + mathbbP(F).$$
I tried using this formula, but I ran into a problem. I calculated
$$mathbbP(A)=frac12cdot frac12=frac14 textand mathbbP(B)=left ( frac16 right )^4=frac11296.$$
However, I then realized that these probabilites are the events that two even sums occur $textitin a row,$ and similarly for my $mathbbP(B).$
Does anyone have any suggestions on how to figure this out, or even if this formula is the one I should be using?
Thanks in advance!
Edit: I know there is a geometric distribution involved.
probability dice
asked 2 hours ago
StatGuy
44719
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
The question is as follows:
You roll two fair dice over and over. Let $A$ be the event you see two even sums. Let $B$ be the event you see a sum of $7$ four times. What is the probability that event $A$ occurs before event $B$?
I know that for mutually exclusive events with independent trials, the probability that event $E$ occurs before event $F$ is
$$fracmathbbP(E)mathbbP(E) + mathbbP(F).$$
I tried using this formula, but I ran into a problem. I calculated
$$mathbbP(A)=frac12cdot frac12=frac14 textand mathbbP(B)=left ( frac16 right )^4=frac11296.$$
However, I then realized that these probabilites are the events that two even sums occur $textitin a row,$ and similarly for my $mathbbP(B).$
Does anyone have any suggestions on how to figure this out, or even if this formula is the one I should be using?
Thanks in advance!
Edit: I know there is a geometric distribution involved.
probability dice
asked 2 hours ago
StatGuy
44719
The question is as follows:
You roll two fair dice over and over. Let $A$ be the event you see two even sums. Let $B$ be the event you see a sum of $7$ four times. What is the probability that event $A$ occurs before event $B$?
I know that for mutually exclusive events with independent trials, the probability that event $E$ occurs before event $F$ is
$$fracmathbbP(E)mathbbP(E) + mathbbP(F).$$
I tried using this formula, but I ran into a problem. I calculated
$$mathbbP(A)=frac12cdot frac12=frac14 textand mathbbP(B)=left ( frac16 right )^4=frac11296.$$
However, I then realized that these probabilites are the events that two even sums occur $textitin a row,$ and similarly for my $mathbbP(B).$
Does anyone have any suggestions on how to figure this out, or even if this formula is the one I should be using?
Thanks in advance!
Edit: I know there is a geometric distribution involved.
probability dice
probability dice
asked 2 hours ago
StatGuy
44719
asked 2 hours ago
StatGuy
44719
asked 2 hours ago
StatGuy
44719
asked 2 hours ago
StatGuy
44719
asked 2 hours ago
StatGuy
44719
44719
add a comment |Â
add a comment |Â
2 Answers
2
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2
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accepted
You have two Bernoulli processes,$S_A$ and $S_B$, and you are asked about the probability that the 2nd arrival in $S_A$ process occurs before the 4th arrival in $S_B$ process. PMF of the time of $k$-th arrival in a Bernoulli process with probability of success $p$ is
$$p_X_k(t)=binomt-1k-1p^k(1-p)^t-k, t=k,k+1,ldots$$
PS: the events "sum is even" and "sum is 7" are dependent, and it makes the question tricky indeed.
answered 2 hours ago
kludg
914511
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up vote
3
down vote
I would do a Markov chain with the states being the number of even rolls/the number of seven rolls. Work backwards starting from the state that you have seen one even roll and three $7$s. What is the chance that A comes first from there?
answered 2 hours ago
Ross Millikan
284k23193361
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
You have two Bernoulli processes,$S_A$ and $S_B$, and you are asked about the probability that the 2nd arrival in $S_A$ process occurs before the 4th arrival in $S_B$ process. PMF of the time of $k$-th arrival in a Bernoulli process with probability of success $p$ is
$$p_X_k(t)=binomt-1k-1p^k(1-p)^t-k, t=k,k+1,ldots$$
PS: the events "sum is even" and "sum is 7" are dependent, and it makes the question tricky indeed.
answered 2 hours ago
kludg
914511
add a comment |Â
up vote
2
down vote
accepted
You have two Bernoulli processes,$S_A$ and $S_B$, and you are asked about the probability that the 2nd arrival in $S_A$ process occurs before the 4th arrival in $S_B$ process. PMF of the time of $k$-th arrival in a Bernoulli process with probability of success $p$ is
$$p_X_k(t)=binomt-1k-1p^k(1-p)^t-k, t=k,k+1,ldots$$
PS: the events "sum is even" and "sum is 7" are dependent, and it makes the question tricky indeed.
answered 2 hours ago
kludg
914511
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
You have two Bernoulli processes,$S_A$ and $S_B$, and you are asked about the probability that the 2nd arrival in $S_A$ process occurs before the 4th arrival in $S_B$ process. PMF of the time of $k$-th arrival in a Bernoulli process with probability of success $p$ is
$$p_X_k(t)=binomt-1k-1p^k(1-p)^t-k, t=k,k+1,ldots$$
PS: the events "sum is even" and "sum is 7" are dependent, and it makes the question tricky indeed.
answered 2 hours ago
kludg
914511
You have two Bernoulli processes,$S_A$ and $S_B$, and you are asked about the probability that the 2nd arrival in $S_A$ process occurs before the 4th arrival in $S_B$ process. PMF of the time of $k$-th arrival in a Bernoulli process with probability of success $p$ is
$$p_X_k(t)=binomt-1k-1p^k(1-p)^t-k, t=k,k+1,ldots$$
PS: the events "sum is even" and "sum is 7" are dependent, and it makes the question tricky indeed.
answered 2 hours ago
kludg
914511
edited 1 hour ago
answered 2 hours ago
kludg
914511
answered 2 hours ago
kludg
914511
answered 2 hours ago
kludg
914511
914511
add a comment |Â
add a comment |Â
up vote
3
down vote
I would do a Markov chain with the states being the number of even rolls/the number of seven rolls. Work backwards starting from the state that you have seen one even roll and three $7$s. What is the chance that A comes first from there?
answered 2 hours ago
Ross Millikan
284k23193361
add a comment |Â
up vote
3
down vote
I would do a Markov chain with the states being the number of even rolls/the number of seven rolls. Work backwards starting from the state that you have seen one even roll and three $7$s. What is the chance that A comes first from there?
answered 2 hours ago
Ross Millikan
284k23193361
add a comment |Â
up vote
3
down vote
up vote
3
down vote
I would do a Markov chain with the states being the number of even rolls/the number of seven rolls. Work backwards starting from the state that you have seen one even roll and three $7$s. What is the chance that A comes first from there?
answered 2 hours ago
Ross Millikan
284k23193361
I would do a Markov chain with the states being the number of even rolls/the number of seven rolls. Work backwards starting from the state that you have seen one even roll and three $7$s. What is the chance that A comes first from there?
answered 2 hours ago
Ross Millikan
284k23193361
answered 2 hours ago
Ross Millikan
284k23193361
answered 2 hours ago
Ross Millikan
284k23193361
answered 2 hours ago
Ross Millikan
284k23193361
284k23193361
add a comment |Â
add a comment |Â
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