# Probability

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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?

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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#2

(Original post by

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?

**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?

p(bicycle | on time)

Edited to change "late" to "on time"

Last edited by mqb2766; 10 months ago

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#4

(Original post by

thanks, do you not need p(late|no bike) when calculating it

**Sheldor29**)thanks, do you not need p(late|no bike) when calculating it

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(Original post by

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.

**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.

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#6

(Original post by

Have you got an answer?

**Sheldor29**)Have you got an answer?

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(Original post by

What have you tried? Happy to help you get to an answer, as is the forum guidelines.

**mqb2766**)What have you tried? Happy to help you get to an answer, as is the forum guidelines.

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#8

(Original post by

I've got 0.214 as my answer. I did this using Bayes theorem

**Sheldor29**)I've got 0.214 as my answer. I did this using Bayes theorem

If you want to upload your working, happy to give it the once over.

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(Original post by

Bayes and conditional probability go hand in hand.

If you want to upload your working, happy to give it the once over.

**mqb2766**)Bayes and conditional probability go hand in hand.

If you want to upload your working, happy to give it the once over.

W = walking

C = car

B = bike

L' = not late

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#10

(Original post by

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

**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

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(Original post by

Obviously the right formula, so assuming you've put the numbers in correctly, it should be fine.

**mqb2766**)Obviously the right formula, so assuming you've put the numbers in correctly, it should be fine.

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?

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#12

(Original post by

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?

**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?

How can you express

p(A|AuB)

as the joint (bayes) and ...

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(Original post by

Independent means p(AnB) = p(A)p(B) = p^2

How can you express

p(A|AuB)

as the joint (bayes) and ...

**mqb2766**)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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#14

p(An(AUB)) = p(...)

p(AuB) = p(...) + p(...) - p(...)

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(Original post by

Yes, think of (sketch) a venn diagram for both these expressions

p(An(AUB)) = p(...)

p(AuB) = p(...) + p(...) - p(...)

**mqb2766**)Yes, think of (sketch) a venn diagram for both these expressions

p(An(AUB)) = p(...)

p(AuB) = p(...) + p(...) - p(...)

This gets an answer of p as the 2p-p^2 cancels out right?

Last edited by Sheldor29; 10 months ago

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#16

(Original post by

P(A|AUB) = p X (2p - p^2) divided by (2p-p^2)?

**Sheldor29**)P(A|AUB) = p X (2p - p^2) divided by (2p-p^2)?

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(Original post by

Agree with the denominator, but how did you get the numerator? Upload a venn diagram sketch?

**mqb2766**)Agree with the denominator, but how did you get the numerator? Upload a venn diagram sketch?

Is this not right

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#18

The intersection between A and AuB

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#19

**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?

0

reply

oh ok, so if p(A) = p

and the intersection of P(AUB) and P(A) is 2p then P(An(AUB)) = 3p?

and the intersection of P(AUB) and P(A) is 2p then P(An(AUB)) = 3p?

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