Finding probability with Law of Total Probability and Bayes Theorem
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1
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The problem is as such below
I will try solve (i) with total probability:
$A -$ car fail test
$B_1-$ emit excessive pollutants that fail
$B_2-$ 17% not emit excessive pollutants that fail
Then $P(A)=P(B_1)P(A|B_1)+P(B_2)P(A|B_2)=(.25)(.99)+(.25)(.17)=0.29$
and (ii): $P(B_1|A)=frac{P(B_1)P(A|B_1)}{P(A)}=frac{(.25)(.99)}{0.29}=0.85$(2 decimal places)
My first answer of 0.29 seems about right as if 99% of some population makes up close to 25% then the remaining 17% of those that fail but shouldn't fail would make the probability go a bit higher. And if a car were to fail the test than it is likely that it is from $B_i$ so a probability of 0.85 makes sense. Did I make any mistakes?
probability proof-verification
asked 8 hours ago
glockm15
32518
add a comment |
up vote
1
down vote
favorite
The problem is as such below
I will try solve (i) with total probability:
$A -$ car fail test
$B_1-$ emit excessive pollutants that fail
$B_2-$ 17% not emit excessive pollutants that fail
Then $P(A)=P(B_1)P(A|B_1)+P(B_2)P(A|B_2)=(.25)(.99)+(.25)(.17)=0.29$
and (ii): $P(B_1|A)=frac{P(B_1)P(A|B_1)}{P(A)}=frac{(.25)(.99)}{0.29}=0.85$(2 decimal places)
My first answer of 0.29 seems about right as if 99% of some population makes up close to 25% then the remaining 17% of those that fail but shouldn't fail would make the probability go a bit higher. And if a car were to fail the test than it is likely that it is from $B_i$ so a probability of 0.85 makes sense. Did I make any mistakes?
probability proof-verification
asked 8 hours ago
glockm15
32518
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
The problem is as such below
I will try solve (i) with total probability:
$A -$ car fail test
$B_1-$ emit excessive pollutants that fail
$B_2-$ 17% not emit excessive pollutants that fail
Then $P(A)=P(B_1)P(A|B_1)+P(B_2)P(A|B_2)=(.25)(.99)+(.25)(.17)=0.29$
and (ii): $P(B_1|A)=frac{P(B_1)P(A|B_1)}{P(A)}=frac{(.25)(.99)}{0.29}=0.85$(2 decimal places)
My first answer of 0.29 seems about right as if 99% of some population makes up close to 25% then the remaining 17% of those that fail but shouldn't fail would make the probability go a bit higher. And if a car were to fail the test than it is likely that it is from $B_i$ so a probability of 0.85 makes sense. Did I make any mistakes?
probability proof-verification
asked 8 hours ago
glockm15
32518
The problem is as such below
I will try solve (i) with total probability:
$A -$ car fail test
$B_1-$ emit excessive pollutants that fail
$B_2-$ 17% not emit excessive pollutants that fail
Then $P(A)=P(B_1)P(A|B_1)+P(B_2)P(A|B_2)=(.25)(.99)+(.25)(.17)=0.29$
and (ii): $P(B_1|A)=frac{P(B_1)P(A|B_1)}{P(A)}=frac{(.25)(.99)}{0.29}=0.85$(2 decimal places)
My first answer of 0.29 seems about right as if 99% of some population makes up close to 25% then the remaining 17% of those that fail but shouldn't fail would make the probability go a bit higher. And if a car were to fail the test than it is likely that it is from $B_i$ so a probability of 0.85 makes sense. Did I make any mistakes?
probability proof-verification
probability proof-verification
asked 8 hours ago
glockm15
32518
asked 8 hours ago
glockm15
32518
asked 8 hours ago
glockm15
32518
asked 8 hours ago
glockm15
32518
asked 8 hours ago
glockm15
32518
32518
add a comment |
add a comment |
1 Answer
1
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oldest
votes
up vote
0
down vote
accepted
In (i) it should have been $mathbb{P}(A) = dots = 0.25 times 0.99 + 0.75 times 0.17 = 0.375$. This results a minor mistake in (ii):
$$
mathbb{P}(B_1 ~|~ A) = dots = frac{0.25 times 0.99}{0.375} = 0.66.
$$
answered 8 hours ago
pointguard0
1,277821
I see, but how did you get 0.75?
– glockm15
8 hours ago
the portion of cars that do not emit excessive amounts
– pointguard0
8 hours ago
Oh I see now thanks!
– glockm15
8 hours ago
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
In (i) it should have been $mathbb{P}(A) = dots = 0.25 times 0.99 + 0.75 times 0.17 = 0.375$. This results a minor mistake in (ii):
$$
mathbb{P}(B_1 ~|~ A) = dots = frac{0.25 times 0.99}{0.375} = 0.66.
$$
answered 8 hours ago
pointguard0
1,277821
I see, but how did you get 0.75?
– glockm15
8 hours ago
the portion of cars that do not emit excessive amounts
– pointguard0
8 hours ago
Oh I see now thanks!
– glockm15
8 hours ago
add a comment |
up vote
0
down vote
accepted
In (i) it should have been $mathbb{P}(A) = dots = 0.25 times 0.99 + 0.75 times 0.17 = 0.375$. This results a minor mistake in (ii):
$$
mathbb{P}(B_1 ~|~ A) = dots = frac{0.25 times 0.99}{0.375} = 0.66.
$$
answered 8 hours ago
pointguard0
1,277821
I see, but how did you get 0.75?
– glockm15
8 hours ago
the portion of cars that do not emit excessive amounts
– pointguard0
8 hours ago
Oh I see now thanks!
– glockm15
8 hours ago
add a comment |
up vote
0
down vote
accepted
up vote
0
down vote
accepted
In (i) it should have been $mathbb{P}(A) = dots = 0.25 times 0.99 + 0.75 times 0.17 = 0.375$. This results a minor mistake in (ii):
$$
mathbb{P}(B_1 ~|~ A) = dots = frac{0.25 times 0.99}{0.375} = 0.66.
$$
answered 8 hours ago
pointguard0
1,277821
In (i) it should have been $mathbb{P}(A) = dots = 0.25 times 0.99 + 0.75 times 0.17 = 0.375$. This results a minor mistake in (ii):
$$
mathbb{P}(B_1 ~|~ A) = dots = frac{0.25 times 0.99}{0.375} = 0.66.
$$
answered 8 hours ago
pointguard0
1,277821
answered 8 hours ago
pointguard0
1,277821
answered 8 hours ago
pointguard0
1,277821
answered 8 hours ago
pointguard0
1,277821
1,277821
I see, but how did you get 0.75?
– glockm15
8 hours ago
the portion of cars that do not emit excessive amounts
– pointguard0
8 hours ago
Oh I see now thanks!
– glockm15
8 hours ago
add a comment |
I see, but how did you get 0.75?
– glockm15
8 hours ago
the portion of cars that do not emit excessive amounts
– pointguard0
8 hours ago
Oh I see now thanks!
– glockm15
8 hours ago
I see, but how did you get 0.75?
– glockm15
8 hours ago
I see, but how did you get 0.75?
– glockm15
8 hours ago
the portion of cars that do not emit excessive amounts
– pointguard0
8 hours ago
the portion of cars that do not emit excessive amounts
– pointguard0
8 hours ago
Oh I see now thanks!
– glockm15
8 hours ago
Oh I see now thanks!
– glockm15
8 hours ago
add a comment |
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