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How to calculate the XOR probability
Probabilities of a race outcomeHow do I formalize this probability exercise?Calculating a metric to compare multiple posterior probability distributionsProbability question with bounds?probability over 3 values with dependencyNeed assistance with conditional probability problemConditional probability in a Pakistani cricket tournamentThree shooters shoot a targetIf today is sunny, what is the probability that day after tomorrow will be cloudy?Markov Process and interpretation of the transition matrices
$begingroup$
I have some questions regarding this problem:
Arthur and Dutch are planned together to go out the next Saturday which is predicted to be sunny.
There is a probability of 0.4 that Dutch wears sunglasses the day they are supposed to meet. The probability that Arthur wears sunglasses on the same day is 0.7 if Dutch wears sunglasses and 0.35 if he does not.
The question is:
What is the probability that exactly one of them wear sunglasses on the day they are supposed to meet?
Let A be Arthur wears glasses, and D be Dutch wears glasses.
Is it $P(A oplus D)$? I can think of two formula, but I'm not sure about the difference:
- $P(A cap barD) + P(D cap barA)= P(A cup D)-P(A cap D) = 0.33$
- $P(A cap barD) + P(D cap barA)= P(A | barD)P(barD) + P(D|barA)P(barA) = 0.35*0.6+0.235294...*0.51$
However, they yield different results.
In the second, I used this formula to calculate $P(barA)$
$P(A) = P(A | barD)P(barD) + P(A|D)P(D) = 0.6*0.35+0.4*0.7 = 0.49$
$P(barA) = 0.51$
$P(D|barA) = P(D cap barA)/P(barA) = (P(D)-P(A cap D))/0.51 = (0.4-0.28)/0.51=0.235294...$
Which solution is true, and what is the problem with different results. I doubt the second one is correct because of the result.
probability
asked 2 days ago
AhmadAhmad
23119
$endgroup$
add a comment |
$begingroup$
I have some questions regarding this problem:
Arthur and Dutch are planned together to go out the next Saturday which is predicted to be sunny.
There is a probability of 0.4 that Dutch wears sunglasses the day they are supposed to meet. The probability that Arthur wears sunglasses on the same day is 0.7 if Dutch wears sunglasses and 0.35 if he does not.
The question is:
What is the probability that exactly one of them wear sunglasses on the day they are supposed to meet?
Let A be Arthur wears glasses, and D be Dutch wears glasses.
Is it $P(A oplus D)$? I can think of two formula, but I'm not sure about the difference:
- $P(A cap barD) + P(D cap barA)= P(A cup D)-P(A cap D) = 0.33$
- $P(A cap barD) + P(D cap barA)= P(A | barD)P(barD) + P(D|barA)P(barA) = 0.35*0.6+0.235294...*0.51$
However, they yield different results.
In the second, I used this formula to calculate $P(barA)$
$P(A) = P(A | barD)P(barD) + P(A|D)P(D) = 0.6*0.35+0.4*0.7 = 0.49$
$P(barA) = 0.51$
$P(D|barA) = P(D cap barA)/P(barA) = (P(D)-P(A cap D))/0.51 = (0.4-0.28)/0.51=0.235294...$
Which solution is true, and what is the problem with different results. I doubt the second one is correct because of the result.
probability
asked 2 days ago
AhmadAhmad
23119
$endgroup$
$begingroup$
It seems to me that both results agree. Did you try computing the expression you get from the second approach?
$endgroup$
– Pedro
2 days ago
add a comment |
$begingroup$
I have some questions regarding this problem:
Arthur and Dutch are planned together to go out the next Saturday which is predicted to be sunny.
There is a probability of 0.4 that Dutch wears sunglasses the day they are supposed to meet. The probability that Arthur wears sunglasses on the same day is 0.7 if Dutch wears sunglasses and 0.35 if he does not.
The question is:
What is the probability that exactly one of them wear sunglasses on the day they are supposed to meet?
Let A be Arthur wears glasses, and D be Dutch wears glasses.
Is it $P(A oplus D)$? I can think of two formula, but I'm not sure about the difference:
- $P(A cap barD) + P(D cap barA)= P(A cup D)-P(A cap D) = 0.33$
- $P(A cap barD) + P(D cap barA)= P(A | barD)P(barD) + P(D|barA)P(barA) = 0.35*0.6+0.235294...*0.51$
However, they yield different results.
In the second, I used this formula to calculate $P(barA)$
$P(A) = P(A | barD)P(barD) + P(A|D)P(D) = 0.6*0.35+0.4*0.7 = 0.49$
$P(barA) = 0.51$
$P(D|barA) = P(D cap barA)/P(barA) = (P(D)-P(A cap D))/0.51 = (0.4-0.28)/0.51=0.235294...$
Which solution is true, and what is the problem with different results. I doubt the second one is correct because of the result.
probability
asked 2 days ago
AhmadAhmad
23119
$endgroup$
I have some questions regarding this problem:
Arthur and Dutch are planned together to go out the next Saturday which is predicted to be sunny.
There is a probability of 0.4 that Dutch wears sunglasses the day they are supposed to meet. The probability that Arthur wears sunglasses on the same day is 0.7 if Dutch wears sunglasses and 0.35 if he does not.
The question is:
What is the probability that exactly one of them wear sunglasses on the day they are supposed to meet?
Let A be Arthur wears glasses, and D be Dutch wears glasses.
Is it $P(A oplus D)$? I can think of two formula, but I'm not sure about the difference:
- $P(A cap barD) + P(D cap barA)= P(A cup D)-P(A cap D) = 0.33$
- $P(A cap barD) + P(D cap barA)= P(A | barD)P(barD) + P(D|barA)P(barA) = 0.35*0.6+0.235294...*0.51$
However, they yield different results.
In the second, I used this formula to calculate $P(barA)$
$P(A) = P(A | barD)P(barD) + P(A|D)P(D) = 0.6*0.35+0.4*0.7 = 0.49$
$P(barA) = 0.51$
$P(D|barA) = P(D cap barA)/P(barA) = (P(D)-P(A cap D))/0.51 = (0.4-0.28)/0.51=0.235294...$
Which solution is true, and what is the problem with different results. I doubt the second one is correct because of the result.
probability
probability
asked 2 days ago
AhmadAhmad
23119
asked 2 days ago
AhmadAhmad
23119
asked 2 days ago
AhmadAhmad
23119
asked 2 days ago
AhmadAhmad
23119
asked 2 days ago
AhmadAhmad
23119
23119
$begingroup$
It seems to me that both results agree. Did you try computing the expression you get from the second approach?
$endgroup$
– Pedro
2 days ago
add a comment |
$begingroup$
It seems to me that both results agree. Did you try computing the expression you get from the second approach?
$endgroup$
– Pedro
2 days ago
$begingroup$
It seems to me that both results agree. Did you try computing the expression you get from the second approach?
$endgroup$
– Pedro
2 days ago
$begingroup$
It seems to me that both results agree. Did you try computing the expression you get from the second approach?
$endgroup$
– Pedro
2 days ago
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The results agree. It helps to work in fractions rather than decimals. Then $0.235dots$ becomes $frac1251$, so the second method returns $frac21100+frac1251frac51100=frac33100$.
You might have found $P(DcapoverlineA)$ easier to compute as $P(overlineA|D)P(D)=0.3times 0.4=0.12$.
answered 2 days ago
J.G.J.G.
29.8k22946
$endgroup$
$begingroup$
Thanks, I used the venn diagram for the second but the Bayes rule also works.
$endgroup$
– Ahmad
2 days ago
add a comment |
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1 Answer
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active
oldest
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1 Answer
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oldest
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$begingroup$
The results agree. It helps to work in fractions rather than decimals. Then $0.235dots$ becomes $frac1251$, so the second method returns $frac21100+frac1251frac51100=frac33100$.
You might have found $P(DcapoverlineA)$ easier to compute as $P(overlineA|D)P(D)=0.3times 0.4=0.12$.
answered 2 days ago
J.G.J.G.
29.8k22946
$endgroup$
$begingroup$
Thanks, I used the venn diagram for the second but the Bayes rule also works.
$endgroup$
– Ahmad
2 days ago
add a comment |
$begingroup$
The results agree. It helps to work in fractions rather than decimals. Then $0.235dots$ becomes $frac1251$, so the second method returns $frac21100+frac1251frac51100=frac33100$.
You might have found $P(DcapoverlineA)$ easier to compute as $P(overlineA|D)P(D)=0.3times 0.4=0.12$.
answered 2 days ago
J.G.J.G.
29.8k22946
$endgroup$
$begingroup$
Thanks, I used the venn diagram for the second but the Bayes rule also works.
$endgroup$
– Ahmad
2 days ago
add a comment |
$begingroup$
The results agree. It helps to work in fractions rather than decimals. Then $0.235dots$ becomes $frac1251$, so the second method returns $frac21100+frac1251frac51100=frac33100$.
You might have found $P(DcapoverlineA)$ easier to compute as $P(overlineA|D)P(D)=0.3times 0.4=0.12$.
answered 2 days ago
J.G.J.G.
29.8k22946
$endgroup$
The results agree. It helps to work in fractions rather than decimals. Then $0.235dots$ becomes $frac1251$, so the second method returns $frac21100+frac1251frac51100=frac33100$.
You might have found $P(DcapoverlineA)$ easier to compute as $P(overlineA|D)P(D)=0.3times 0.4=0.12$.
answered 2 days ago
J.G.J.G.
29.8k22946
answered 2 days ago
J.G.J.G.
29.8k22946
answered 2 days ago
J.G.J.G.
29.8k22946
answered 2 days ago
J.G.J.G.
29.8k22946
29.8k22946
$begingroup$
Thanks, I used the venn diagram for the second but the Bayes rule also works.
$endgroup$
– Ahmad
2 days ago
add a comment |
$begingroup$
Thanks, I used the venn diagram for the second but the Bayes rule also works.
$endgroup$
– Ahmad
2 days ago
$begingroup$
Thanks, I used the venn diagram for the second but the Bayes rule also works.
$endgroup$
– Ahmad
2 days ago
$begingroup$
Thanks, I used the venn diagram for the second but the Bayes rule also works.
$endgroup$
– Ahmad
2 days ago
add a comment |
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$begingroup$
It seems to me that both results agree. Did you try computing the expression you get from the second approach?
$endgroup$
– Pedro
2 days ago