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Probability

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?

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Reply 1
Avatar for mqb2766
mqb2766
21
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"
(edited 4 years ago)
Reply 2
Avatar for Sheldor29
Sheldor29
OP
8
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
Reply 3
Avatar for mqb2766
mqb2766
21
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.
Reply 4
Avatar for Sheldor29
Sheldor29
OP
8
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?
Reply 5
Avatar for mqb2766
mqb2766
21
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.
Reply 6
Avatar for Sheldor29
Sheldor29
OP
8
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
Reply 7
Avatar for mqb2766
mqb2766
21
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.
Reply 8
Avatar for Sheldor29
Sheldor29
OP
8
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
Reply 9
Avatar for mqb2766
mqb2766
21
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.
Reply 10
Avatar for Sheldor29
Sheldor29
OP
8
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?
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 ...
Reply 12
Avatar for Sheldor29
Sheldor29
OP
8
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)
?
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(...)
Reply 14
Avatar for Sheldor29
Sheldor29
OP
8
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?
(edited 4 years ago)
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?
Reply 16
Avatar for Sheldor29
Sheldor29
OP
8
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
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
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?
Reply 19
Avatar for Sheldor29
Sheldor29
OP
8
oh ok, so if p(A) = p
and the intersection of P(AUB) and P(A) is 2p then P(An(AUB)) = 3p?

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