Mixing
Bayes' Theorem Question with irregular die.
$begingroup$
I'm having trouble with the following question:
Suppose we have three, six-sided dice. These are the numbers on the sides of the dice:
{1, 2, 3, 4, 5, 6} (regular die)
{2, 2, 3, 4, 5, 5} (irregular die)
{1, 1, 2, 3, 4, 5} (irregular die)
First, we randomly select a die. Then we roll it three times and get 2 on the first roll, 3 on the second roll and 5 on the third roll. Use Bayes’ rule to determine the chance that we were using die 2. Round your answer to four decimal places.
If we call the probability that die 2 is rolled $P(A)$, and the three rolls of {2, 3, 5} $P(B)$, then here's what I have so far:
$$P(A|B) = frac{P(B|A) P(A)}{P(B)} = frac{P(B|A) cdot (1/3)}{(2/6) cdot (1/6) cdot (2/6)} = 1/54$$
And then I'm stuck. Any help?
probability conditional-probability dice bayes-theorem
edited Feb 10 at 11:06
jvdhooft
5,67561641
asked Jan 26 at 4:33
LimaLima
1
$endgroup$
add a comment |
$begingroup$
I'm having trouble with the following question:
Suppose we have three, six-sided dice. These are the numbers on the sides of the dice:
{1, 2, 3, 4, 5, 6} (regular die)
{2, 2, 3, 4, 5, 5} (irregular die)
{1, 1, 2, 3, 4, 5} (irregular die)
First, we randomly select a die. Then we roll it three times and get 2 on the first roll, 3 on the second roll and 5 on the third roll. Use Bayes’ rule to determine the chance that we were using die 2. Round your answer to four decimal places.
If we call the probability that die 2 is rolled $P(A)$, and the three rolls of {2, 3, 5} $P(B)$, then here's what I have so far:
$$P(A|B) = frac{P(B|A) P(A)}{P(B)} = frac{P(B|A) cdot (1/3)}{(2/6) cdot (1/6) cdot (2/6)} = 1/54$$
And then I'm stuck. Any help?
probability conditional-probability dice bayes-theorem
edited Feb 10 at 11:06
jvdhooft
5,67561641
asked Jan 26 at 4:33
LimaLima
1
$endgroup$
add a comment |
$begingroup$
I'm having trouble with the following question:
Suppose we have three, six-sided dice. These are the numbers on the sides of the dice:
{1, 2, 3, 4, 5, 6} (regular die)
{2, 2, 3, 4, 5, 5} (irregular die)
{1, 1, 2, 3, 4, 5} (irregular die)
First, we randomly select a die. Then we roll it three times and get 2 on the first roll, 3 on the second roll and 5 on the third roll. Use Bayes’ rule to determine the chance that we were using die 2. Round your answer to four decimal places.
If we call the probability that die 2 is rolled $P(A)$, and the three rolls of {2, 3, 5} $P(B)$, then here's what I have so far:
$$P(A|B) = frac{P(B|A) P(A)}{P(B)} = frac{P(B|A) cdot (1/3)}{(2/6) cdot (1/6) cdot (2/6)} = 1/54$$
And then I'm stuck. Any help?
probability conditional-probability dice bayes-theorem
edited Feb 10 at 11:06
jvdhooft
5,67561641
asked Jan 26 at 4:33
LimaLima
1
$endgroup$
I'm having trouble with the following question:
Suppose we have three, six-sided dice. These are the numbers on the sides of the dice:
{1, 2, 3, 4, 5, 6} (regular die)
{2, 2, 3, 4, 5, 5} (irregular die)
{1, 1, 2, 3, 4, 5} (irregular die)
First, we randomly select a die. Then we roll it three times and get 2 on the first roll, 3 on the second roll and 5 on the third roll. Use Bayes’ rule to determine the chance that we were using die 2. Round your answer to four decimal places.
If we call the probability that die 2 is rolled $P(A)$, and the three rolls of {2, 3, 5} $P(B)$, then here's what I have so far:
$$P(A|B) = frac{P(B|A) P(A)}{P(B)} = frac{P(B|A) cdot (1/3)}{(2/6) cdot (1/6) cdot (2/6)} = 1/54$$
And then I'm stuck. Any help?
probability conditional-probability dice bayes-theorem
probability conditional-probability dice bayes-theorem
edited Feb 10 at 11:06
jvdhooft
5,67561641
asked Jan 26 at 4:33
LimaLima
1
edited Feb 10 at 11:06
jvdhooft
5,67561641
asked Jan 26 at 4:33
LimaLima
1
edited Feb 10 at 11:06
jvdhooft
5,67561641
edited Feb 10 at 11:06
jvdhooft
5,67561641
edited Feb 10 at 11:06
jvdhooft
5,67561641
5,67561641
asked Jan 26 at 4:33
LimaLima
1
asked Jan 26 at 4:33
LimaLima
1
asked Jan 26 at 4:33
LimaLima
1
1
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Your formula is correct, you just calculated the probabilities of the dice wrong. See the pmf of a multinomial https://en.wikipedia.org/wiki/Multinomial_distribution. So $P(B|A)=frac26*frac16*frac26=frac1{54}$. And by the Law of Total Probability,
$$P(B) = P(dice1)P(B|dice 1)+P(dice2)P(B|dice 2)+P(dice3)P(B|dice 3)$$
$$ = frac13*frac16*frac16*frac16+frac13*frac26*frac16*frac26+frac13*frac16*frac16*frac16 = frac1{748}+frac1{162}+frac1{748} = frac1{108}$$
So you have $P(A|B) = frac{frac1{54}*frac13}{frac1{108}}=frac23approx.6667$.
answered Jan 26 at 4:51
Erik ParkinsonErik Parkinson
1,17519
$endgroup$
$begingroup$
Since the outcomes of the rolls are ordered, I would say $P(Bmid A)=frac{2}{6}cdot frac{1}{6}cdotfrac{2}{6}$, not $P(Bmid A)=6cdot frac{2}{6}cdot frac{1}{6}cdotfrac{2}{6}$. Same for all the other conditional probabilities.
$endgroup$
– kccu
Jan 26 at 5:00
$begingroup$
Oh, I was assuming order didn't matter and we were looking at the $3$ rolls as a whole. In that case you're right; the final answer will be the same as the $6$ factors out of the top and bottom, I'll edit my answer to fix that.
$endgroup$
– Erik Parkinson
Jan 26 at 5:01
add a comment |
$begingroup$
What you wrote for $P(B)$ is actually $P(B|A)$ (or $P(Bmid text{die 2})$ in the notation I use in the rest of this answer). To calculate $P(B)$, you need to decompose it as:
$$P(B)=P(B|text{die 1})P(text{die 1})+P(Bmid text{die 2})P(text{die 2})+P(Bmid text{die 3})P(text{die 3}).$$
answered Jan 26 at 4:38
kccukccu
10.6k11229
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
Your formula is correct, you just calculated the probabilities of the dice wrong. See the pmf of a multinomial https://en.wikipedia.org/wiki/Multinomial_distribution. So $P(B|A)=frac26*frac16*frac26=frac1{54}$. And by the Law of Total Probability,
$$P(B) = P(dice1)P(B|dice 1)+P(dice2)P(B|dice 2)+P(dice3)P(B|dice 3)$$
$$ = frac13*frac16*frac16*frac16+frac13*frac26*frac16*frac26+frac13*frac16*frac16*frac16 = frac1{748}+frac1{162}+frac1{748} = frac1{108}$$
So you have $P(A|B) = frac{frac1{54}*frac13}{frac1{108}}=frac23approx.6667$.
answered Jan 26 at 4:51
Erik ParkinsonErik Parkinson
1,17519
$endgroup$
$begingroup$
Since the outcomes of the rolls are ordered, I would say $P(Bmid A)=frac{2}{6}cdot frac{1}{6}cdotfrac{2}{6}$, not $P(Bmid A)=6cdot frac{2}{6}cdot frac{1}{6}cdotfrac{2}{6}$. Same for all the other conditional probabilities.
$endgroup$
– kccu
Jan 26 at 5:00
$begingroup$
Oh, I was assuming order didn't matter and we were looking at the $3$ rolls as a whole. In that case you're right; the final answer will be the same as the $6$ factors out of the top and bottom, I'll edit my answer to fix that.
$endgroup$
– Erik Parkinson
Jan 26 at 5:01
add a comment |
$begingroup$
Your formula is correct, you just calculated the probabilities of the dice wrong. See the pmf of a multinomial https://en.wikipedia.org/wiki/Multinomial_distribution. So $P(B|A)=frac26*frac16*frac26=frac1{54}$. And by the Law of Total Probability,
$$P(B) = P(dice1)P(B|dice 1)+P(dice2)P(B|dice 2)+P(dice3)P(B|dice 3)$$
$$ = frac13*frac16*frac16*frac16+frac13*frac26*frac16*frac26+frac13*frac16*frac16*frac16 = frac1{748}+frac1{162}+frac1{748} = frac1{108}$$
So you have $P(A|B) = frac{frac1{54}*frac13}{frac1{108}}=frac23approx.6667$.
answered Jan 26 at 4:51
Erik ParkinsonErik Parkinson
1,17519
$endgroup$
$begingroup$
Since the outcomes of the rolls are ordered, I would say $P(Bmid A)=frac{2}{6}cdot frac{1}{6}cdotfrac{2}{6}$, not $P(Bmid A)=6cdot frac{2}{6}cdot frac{1}{6}cdotfrac{2}{6}$. Same for all the other conditional probabilities.
$endgroup$
– kccu
Jan 26 at 5:00
$begingroup$
Oh, I was assuming order didn't matter and we were looking at the $3$ rolls as a whole. In that case you're right; the final answer will be the same as the $6$ factors out of the top and bottom, I'll edit my answer to fix that.
$endgroup$
– Erik Parkinson
Jan 26 at 5:01
add a comment |
$begingroup$
Your formula is correct, you just calculated the probabilities of the dice wrong. See the pmf of a multinomial https://en.wikipedia.org/wiki/Multinomial_distribution. So $P(B|A)=frac26*frac16*frac26=frac1{54}$. And by the Law of Total Probability,
$$P(B) = P(dice1)P(B|dice 1)+P(dice2)P(B|dice 2)+P(dice3)P(B|dice 3)$$
$$ = frac13*frac16*frac16*frac16+frac13*frac26*frac16*frac26+frac13*frac16*frac16*frac16 = frac1{748}+frac1{162}+frac1{748} = frac1{108}$$
So you have $P(A|B) = frac{frac1{54}*frac13}{frac1{108}}=frac23approx.6667$.
answered Jan 26 at 4:51
Erik ParkinsonErik Parkinson
1,17519
$endgroup$
Your formula is correct, you just calculated the probabilities of the dice wrong. See the pmf of a multinomial https://en.wikipedia.org/wiki/Multinomial_distribution. So $P(B|A)=frac26*frac16*frac26=frac1{54}$. And by the Law of Total Probability,
$$P(B) = P(dice1)P(B|dice 1)+P(dice2)P(B|dice 2)+P(dice3)P(B|dice 3)$$
$$ = frac13*frac16*frac16*frac16+frac13*frac26*frac16*frac26+frac13*frac16*frac16*frac16 = frac1{748}+frac1{162}+frac1{748} = frac1{108}$$
So you have $P(A|B) = frac{frac1{54}*frac13}{frac1{108}}=frac23approx.6667$.
answered Jan 26 at 4:51
Erik ParkinsonErik Parkinson
1,17519
edited Jan 26 at 5:03
answered Jan 26 at 4:51
Erik ParkinsonErik Parkinson
1,17519
answered Jan 26 at 4:51
Erik ParkinsonErik Parkinson
1,17519
answered Jan 26 at 4:51
Erik ParkinsonErik Parkinson
1,17519
1,17519
$begingroup$
Since the outcomes of the rolls are ordered, I would say $P(Bmid A)=frac{2}{6}cdot frac{1}{6}cdotfrac{2}{6}$, not $P(Bmid A)=6cdot frac{2}{6}cdot frac{1}{6}cdotfrac{2}{6}$. Same for all the other conditional probabilities.
$endgroup$
– kccu
Jan 26 at 5:00
$begingroup$
Oh, I was assuming order didn't matter and we were looking at the $3$ rolls as a whole. In that case you're right; the final answer will be the same as the $6$ factors out of the top and bottom, I'll edit my answer to fix that.
$endgroup$
– Erik Parkinson
Jan 26 at 5:01
add a comment |
$begingroup$
Since the outcomes of the rolls are ordered, I would say $P(Bmid A)=frac{2}{6}cdot frac{1}{6}cdotfrac{2}{6}$, not $P(Bmid A)=6cdot frac{2}{6}cdot frac{1}{6}cdotfrac{2}{6}$. Same for all the other conditional probabilities.
$endgroup$
– kccu
Jan 26 at 5:00
$begingroup$
Oh, I was assuming order didn't matter and we were looking at the $3$ rolls as a whole. In that case you're right; the final answer will be the same as the $6$ factors out of the top and bottom, I'll edit my answer to fix that.
$endgroup$
– Erik Parkinson
Jan 26 at 5:01
$begingroup$
Since the outcomes of the rolls are ordered, I would say $P(Bmid A)=frac{2}{6}cdot frac{1}{6}cdotfrac{2}{6}$, not $P(Bmid A)=6cdot frac{2}{6}cdot frac{1}{6}cdotfrac{2}{6}$. Same for all the other conditional probabilities.
$endgroup$
– kccu
Jan 26 at 5:00
$begingroup$
Since the outcomes of the rolls are ordered, I would say $P(Bmid A)=frac{2}{6}cdot frac{1}{6}cdotfrac{2}{6}$, not $P(Bmid A)=6cdot frac{2}{6}cdot frac{1}{6}cdotfrac{2}{6}$. Same for all the other conditional probabilities.
$endgroup$
– kccu
Jan 26 at 5:00
$begingroup$
Oh, I was assuming order didn't matter and we were looking at the $3$ rolls as a whole. In that case you're right; the final answer will be the same as the $6$ factors out of the top and bottom, I'll edit my answer to fix that.
$endgroup$
– Erik Parkinson
Jan 26 at 5:01
$begingroup$
Oh, I was assuming order didn't matter and we were looking at the $3$ rolls as a whole. In that case you're right; the final answer will be the same as the $6$ factors out of the top and bottom, I'll edit my answer to fix that.
$endgroup$
– Erik Parkinson
Jan 26 at 5:01
add a comment |
$begingroup$
What you wrote for $P(B)$ is actually $P(B|A)$ (or $P(Bmid text{die 2})$ in the notation I use in the rest of this answer). To calculate $P(B)$, you need to decompose it as:
$$P(B)=P(B|text{die 1})P(text{die 1})+P(Bmid text{die 2})P(text{die 2})+P(Bmid text{die 3})P(text{die 3}).$$
answered Jan 26 at 4:38
kccukccu
10.6k11229
$endgroup$
add a comment |
$begingroup$
What you wrote for $P(B)$ is actually $P(B|A)$ (or $P(Bmid text{die 2})$ in the notation I use in the rest of this answer). To calculate $P(B)$, you need to decompose it as:
$$P(B)=P(B|text{die 1})P(text{die 1})+P(Bmid text{die 2})P(text{die 2})+P(Bmid text{die 3})P(text{die 3}).$$
answered Jan 26 at 4:38
kccukccu
10.6k11229
$endgroup$
add a comment |
$begingroup$
What you wrote for $P(B)$ is actually $P(B|A)$ (or $P(Bmid text{die 2})$ in the notation I use in the rest of this answer). To calculate $P(B)$, you need to decompose it as:
$$P(B)=P(B|text{die 1})P(text{die 1})+P(Bmid text{die 2})P(text{die 2})+P(Bmid text{die 3})P(text{die 3}).$$
answered Jan 26 at 4:38
kccukccu
10.6k11229
$endgroup$
What you wrote for $P(B)$ is actually $P(B|A)$ (or $P(Bmid text{die 2})$ in the notation I use in the rest of this answer). To calculate $P(B)$, you need to decompose it as:
$$P(B)=P(B|text{die 1})P(text{die 1})+P(Bmid text{die 2})P(text{die 2})+P(Bmid text{die 3})P(text{die 3}).$$
answered Jan 26 at 4:38
kccukccu
10.6k11229
answered Jan 26 at 4:38
kccukccu
10.6k11229
answered Jan 26 at 4:38
kccukccu
10.6k11229
answered Jan 26 at 4:38
kccukccu
10.6k11229
10.6k11229
add a comment |
add a comment |
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