conditional probability

Adam k

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Hi guys
i have tried my best in this question it is about conditional probability . Q1 Clara has three ways of getting to job office 30% of the she travels by car ,20% she travels by bicycle,and 50% she walks .When travelling by car Clara is late 5% of the time.when riding 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?

 

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First question: what is the probability of Clara being on time when she rides her bike?
 
I must admit I have no idea what you were thinking.

What follows is not the most efficient way to do this, but it may be the clearest.

There are six mutually exclusive but exhaustive possibilities for the conjunction of arrival status and mode of transportation, namely

timely and car
late and car
timely and bike
late and bike
timely and foot
late and foot

What is the probability of each?

What is the sum of those probabilities?

What is the probability of being on time?

What is the answer to your problem?
 
i have tried my best in this question it is about conditional probability . Q1 Clara has three ways of getting to job office 30% of the she travels by car ,20% she travels by bicycle,and 50% she walks .When travelling by car Clara is late 5% of the time.when riding 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?
We want [imath]\bf\mathcal{P}(B|T)[/imath]. Probability she has ridden a bike given that she is on time.
But [imath]\mathcal{P}(B|T)=\dfrac{\mathcal{P}(B\cap T)}{\mathcal{P}(T)}[/imath]. Now [imath]B\text{ bike },C\text{ car },~\&~W\text{ walking,}[/imath] partition the space.
By Bayes' Rule [imath]\mathcal{P}(B|T)=\dfrac{\mathcal{P}(T|B)\mathcal{P}(B)}{\mathcal{P}(T|B)\mathcal{P}(B)+\mathcal{P}(T|C)\mathcal{P}(C)+\mathcal{P}(T|W)\mathcal{P}(W)}[/imath]
Can you fill in the numbers?
 
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