# Probability

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Thread starter 10 months ago
#1
Josie has three ways of getting to school. 30% of the time she travels by car, 20% of the
time she rides her bicycle and 50% of the time she walks.
When travelling by car, Josie is late 5% of the time. When riding her bicycle she is
late 10%of the time. When walking she is late 25% of the time. Given that she was on
time, find the probability that she rides her bicycle.

Any ideas?
0
10 months ago
#2
(Original post by Sheldor29)
Josie has three ways of getting to school. 30% of the time she travels by car, 20% of the
time she rides her bicycle and 50% of the time she walks.
When travelling by car, Josie is late 5% of the time. When riding her bicycle she is
late 10%of the time. When walking she is late 25% of the time. Given that she was on
time, find the probability that she rides her bicycle.

Any ideas?
Conditional probability, you want to calculate
p(bicycle | on time)

Edited to change "late" to "on time"
Last edited by mqb2766; 10 months ago
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Thread starter 10 months ago
#3
(Original post by mqb2766)
Conditional probability, you want to calculate
p(bicycle | late)
thanks, do you not need p(late|no bike) when calculating it
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10 months ago
#4
(Original post by Sheldor29)
thanks, do you not need p(late|no bike) when calculating it
THere are a few equivalent ways of calcuating the conditional probability. List what you're given in the question (marginal and conditional) and try writing down the expression.
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Thread starter 10 months ago
#5
(Original post by mqb2766)
THere are a few equivalent ways of calcuating the conditional probability. List what you're given in the question (marginal and conditional) and try writing down the expression.
Have you got an answer?
0
10 months ago
#6
(Original post by Sheldor29)
Have you got an answer?
What have you tried? Happy to help you get to an answer, as is the forum guidelines.
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Thread starter 10 months ago
#7
(Original post by mqb2766)
What have you tried? Happy to help you get to an answer, as is the forum guidelines.
I've got 0.214 as my answer. I did this using Bayes theorem
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10 months ago
#8
(Original post by Sheldor29)
I've got 0.214 as my answer. I did this using Bayes theorem
Bayes and conditional probability go hand in hand.
If you want to upload your working, happy to give it the once over.
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Thread starter 10 months ago
#9
(Original post by mqb2766)
Bayes and conditional probability go hand in hand.
If you want to upload your working, happy to give it the once over.
p(B|L') = p(L'|B) X P(B) divided by (P(L'|B) X P(B)) + (P(L'|C) X P(C)) + (P(L'|W) X p(W))
W = walking
C = car
B = bike
L' = not late
0
10 months ago
#10
(Original post by Sheldor29)
p(B|L') = p(L'|B) X P(B) divided by (P(L'|B) X P(B)) + (P(L'|C) X P(C)) + (P(L'|W) X p(W))
W = walking
C = car
B = bike
L' = not late
Obviously the right formula, so assuming you've put the numbers in correctly, it should be fine.
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Thread starter 10 months ago
#11
(Original post by mqb2766)
Obviously the right formula, so assuming you've put the numbers in correctly, it should be fine.
OK, thank you very much.
Sorry to take up your time but i found another one that I am struggling with.

A and B are independent events such that p(A) = p(B) = p Find p(A|AUB) in simplest form.
Any ideas?
0
10 months ago
#12
(Original post by Sheldor29)
OK, thank you very much.
Sorry to take up your time but i found another one that I am struggling with.

A and B are independent events such that p(A) = p(B) = p Find p(A|AUB) in simplest form.
Any ideas?
Independent means p(AnB) = p(A)p(B) = p^2
How can you express
p(A|AuB)
as the joint (bayes) and ...
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Thread starter 10 months ago
#13
(Original post by mqb2766)
Independent means p(AnB) = p(A)p(B) = p^2
How can you express
p(A|AuB)
as the joint (bayes) and ...
p(A|AUB) = p(An(AUB)) / p(AUB)
?
0
10 months ago
#14
(Original post by Sheldor29)
p(A|AUB) = p(An(AUB)) / p(AUB)
?
Yes, think of (sketch) a venn diagram for both these expressions
p(An(AUB)) = p(...)
p(AuB) = p(...) + p(...) - p(...)
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Thread starter 10 months ago
#15
(Original post by mqb2766)
Yes, think of (sketch) a venn diagram for both these expressions
p(An(AUB)) = p(...)
p(AuB) = p(...) + p(...) - p(...)
P(A|AUB) = p X (2p - p^2) divided by (2p-p^2)?
This gets an answer of p as the 2p-p^2 cancels out right?
Last edited by Sheldor29; 10 months ago
0
10 months ago
#16
(Original post by Sheldor29)
P(A|AUB) = p X (2p - p^2) divided by (2p-p^2)?
Agree with the denominator, but how did you get the numerator? Upload a venn diagram sketch?
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Thread starter 10 months ago
#17
(Original post by mqb2766)
Agree with the denominator, but how did you get the numerator? Upload a venn diagram sketch?
I just assumed that p(An(AUB)) = P(A) X P(AUB)
Is this not right
0
10 months ago
#18
(Original post by Sheldor29)
I just assumed that p(An(AUB)) = P(A) X P(AUB)
Is this not right
No. A and B are independent, not A and AuB. It is simpler than you're thinking if you sketch it.
The intersection between A and AuB
0
10 months ago
#19
(Original post by Sheldor29)
Josie has three ways of getting to school. 30% of the time she travels by car, 20% of the
time she rides her bicycle and 50% of the time she walks.
When travelling by car, Josie is late 5% of the time. When riding her bicycle she is
late 10%of the time. When walking she is late 25% of the time. Given that she was on
time, find the probability that she rides her bicycle.

Any ideas?
are you doing probability yet?
0
Thread starter 10 months ago
#20
oh ok, so if p(A) = p
and the intersection of P(AUB) and P(A) is 2p then P(An(AUB)) = 3p?
0
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