Probability that a 5 occurs first
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Suppose we roll pair of dice until a sum of either 5 or 7 appears.
What is the probability that a sum of 5 occurs first?
Try:
Let $A$ be the event that a sum of $5$ occurs on the ith roll and $B$ that the sum of 7 occurs on the ith roll. We are interested on the event $A | A^c cup B^c $. We have
$$ P(A | A^c cup B^c) = dfrac P(A cap (A^c cup B^c))P(A^c cup B^c) = fracP(A cap B^c)1 - P(A cap B) = fracP(A cap B^c)1-0 = P(A cap B^c)$$
Is this approach correct so far?
probability
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up vote
1
down vote
favorite
Suppose we roll pair of dice until a sum of either 5 or 7 appears.
What is the probability that a sum of 5 occurs first?
Try:
Let $A$ be the event that a sum of $5$ occurs on the ith roll and $B$ that the sum of 7 occurs on the ith roll. We are interested on the event $A | A^c cup B^c $. We have
$$ P(A | A^c cup B^c) = dfrac P(A cap (A^c cup B^c))P(A^c cup B^c) = fracP(A cap B^c)1 - P(A cap B) = fracP(A cap B^c)1-0 = P(A cap B^c)$$
Is this approach correct so far?
probability
Possible duplicate of A pair of unbiased dice are rolled together till a sum of either 5 or 7 is obtained, find the probability that 5 comes before 7
â Rob
47 mins ago
Possible duplicate of Probability sum of 5 before sum of 7
â Key Flex
3 mins ago
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Suppose we roll pair of dice until a sum of either 5 or 7 appears.
What is the probability that a sum of 5 occurs first?
Try:
Let $A$ be the event that a sum of $5$ occurs on the ith roll and $B$ that the sum of 7 occurs on the ith roll. We are interested on the event $A | A^c cup B^c $. We have
$$ P(A | A^c cup B^c) = dfrac P(A cap (A^c cup B^c))P(A^c cup B^c) = fracP(A cap B^c)1 - P(A cap B) = fracP(A cap B^c)1-0 = P(A cap B^c)$$
Is this approach correct so far?
probability
Suppose we roll pair of dice until a sum of either 5 or 7 appears.
What is the probability that a sum of 5 occurs first?
Try:
Let $A$ be the event that a sum of $5$ occurs on the ith roll and $B$ that the sum of 7 occurs on the ith roll. We are interested on the event $A | A^c cup B^c $. We have
$$ P(A | A^c cup B^c) = dfrac P(A cap (A^c cup B^c))P(A^c cup B^c) = fracP(A cap B^c)1 - P(A cap B) = fracP(A cap B^c)1-0 = P(A cap B^c)$$
Is this approach correct so far?
probability
probability
asked 53 mins ago
Jimmy Sabater
1,468115
1,468115
Possible duplicate of A pair of unbiased dice are rolled together till a sum of either 5 or 7 is obtained, find the probability that 5 comes before 7
â Rob
47 mins ago
Possible duplicate of Probability sum of 5 before sum of 7
â Key Flex
3 mins ago
add a comment |Â
Possible duplicate of A pair of unbiased dice are rolled together till a sum of either 5 or 7 is obtained, find the probability that 5 comes before 7
â Rob
47 mins ago
Possible duplicate of Probability sum of 5 before sum of 7
â Key Flex
3 mins ago
Possible duplicate of A pair of unbiased dice are rolled together till a sum of either 5 or 7 is obtained, find the probability that 5 comes before 7
â Rob
47 mins ago
Possible duplicate of A pair of unbiased dice are rolled together till a sum of either 5 or 7 is obtained, find the probability that 5 comes before 7
â Rob
47 mins ago
Possible duplicate of Probability sum of 5 before sum of 7
â Key Flex
3 mins ago
Possible duplicate of Probability sum of 5 before sum of 7
â Key Flex
3 mins ago
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
1
down vote
accepted
Guide:
- Compute the probability of $7$ appears, call it $q$.
- Compute the probability of $5$ appears, call it $p$.
We are interested in
$$sum_i=1^infty (1-p-q)^i-1p$$
Remark about your attempt:
Are you intending to use some indices to denote $i$-th throw?
add a comment |Â
up vote
2
down vote
A cleaner approach would be to consider the experiment until either a sum of $5$ or $7$ is rolled at some step, say $n$. (Note that $n<infty$ almost surely.) Then, the event $A$ is that $5$ was rolled and $B$ is that $7$ was rolled, and you need
$$
mathbbP[A|A cup B]
$$
where $A$ and $B$ are disjoint...
add a comment |Â
up vote
2
down vote
All events are ignored until one in the desired event space are achieved. Thus you can consider this the entire sample space.
$$S=(1,4),(4,1), (2,3), (3,2), (1,6), (6,1), (2,5), (5,2), (3,4), (4,3)$$
Now it's just the law of classical probability.
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Guide:
- Compute the probability of $7$ appears, call it $q$.
- Compute the probability of $5$ appears, call it $p$.
We are interested in
$$sum_i=1^infty (1-p-q)^i-1p$$
Remark about your attempt:
Are you intending to use some indices to denote $i$-th throw?
add a comment |Â
up vote
1
down vote
accepted
Guide:
- Compute the probability of $7$ appears, call it $q$.
- Compute the probability of $5$ appears, call it $p$.
We are interested in
$$sum_i=1^infty (1-p-q)^i-1p$$
Remark about your attempt:
Are you intending to use some indices to denote $i$-th throw?
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Guide:
- Compute the probability of $7$ appears, call it $q$.
- Compute the probability of $5$ appears, call it $p$.
We are interested in
$$sum_i=1^infty (1-p-q)^i-1p$$
Remark about your attempt:
Are you intending to use some indices to denote $i$-th throw?
Guide:
- Compute the probability of $7$ appears, call it $q$.
- Compute the probability of $5$ appears, call it $p$.
We are interested in
$$sum_i=1^infty (1-p-q)^i-1p$$
Remark about your attempt:
Are you intending to use some indices to denote $i$-th throw?
answered 47 mins ago
Siong Thye Goh
85.3k1457107
85.3k1457107
add a comment |Â
add a comment |Â
up vote
2
down vote
A cleaner approach would be to consider the experiment until either a sum of $5$ or $7$ is rolled at some step, say $n$. (Note that $n<infty$ almost surely.) Then, the event $A$ is that $5$ was rolled and $B$ is that $7$ was rolled, and you need
$$
mathbbP[A|A cup B]
$$
where $A$ and $B$ are disjoint...
add a comment |Â
up vote
2
down vote
A cleaner approach would be to consider the experiment until either a sum of $5$ or $7$ is rolled at some step, say $n$. (Note that $n<infty$ almost surely.) Then, the event $A$ is that $5$ was rolled and $B$ is that $7$ was rolled, and you need
$$
mathbbP[A|A cup B]
$$
where $A$ and $B$ are disjoint...
add a comment |Â
up vote
2
down vote
up vote
2
down vote
A cleaner approach would be to consider the experiment until either a sum of $5$ or $7$ is rolled at some step, say $n$. (Note that $n<infty$ almost surely.) Then, the event $A$ is that $5$ was rolled and $B$ is that $7$ was rolled, and you need
$$
mathbbP[A|A cup B]
$$
where $A$ and $B$ are disjoint...
A cleaner approach would be to consider the experiment until either a sum of $5$ or $7$ is rolled at some step, say $n$. (Note that $n<infty$ almost surely.) Then, the event $A$ is that $5$ was rolled and $B$ is that $7$ was rolled, and you need
$$
mathbbP[A|A cup B]
$$
where $A$ and $B$ are disjoint...
answered 49 mins ago
gt6989b
31.2k22349
31.2k22349
add a comment |Â
add a comment |Â
up vote
2
down vote
All events are ignored until one in the desired event space are achieved. Thus you can consider this the entire sample space.
$$S=(1,4),(4,1), (2,3), (3,2), (1,6), (6,1), (2,5), (5,2), (3,4), (4,3)$$
Now it's just the law of classical probability.
add a comment |Â
up vote
2
down vote
All events are ignored until one in the desired event space are achieved. Thus you can consider this the entire sample space.
$$S=(1,4),(4,1), (2,3), (3,2), (1,6), (6,1), (2,5), (5,2), (3,4), (4,3)$$
Now it's just the law of classical probability.
add a comment |Â
up vote
2
down vote
up vote
2
down vote
All events are ignored until one in the desired event space are achieved. Thus you can consider this the entire sample space.
$$S=(1,4),(4,1), (2,3), (3,2), (1,6), (6,1), (2,5), (5,2), (3,4), (4,3)$$
Now it's just the law of classical probability.
All events are ignored until one in the desired event space are achieved. Thus you can consider this the entire sample space.
$$S=(1,4),(4,1), (2,3), (3,2), (1,6), (6,1), (2,5), (5,2), (3,4), (4,3)$$
Now it's just the law of classical probability.
answered 42 mins ago
David Peterson
8,30121834
8,30121834
add a comment |Â
add a comment |Â
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Possible duplicate of A pair of unbiased dice are rolled together till a sum of either 5 or 7 is obtained, find the probability that 5 comes before 7
â Rob
47 mins ago
Possible duplicate of Probability sum of 5 before sum of 7
â Key Flex
3 mins ago