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Probability question.


Guest SNJERIN

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Guest SNJERIN

Can someone help me understanding this question. I just don't seen to understand what they mean. They gave me the probability that she flies or drives when visiting her relatives, so what is point of having this 13/18 ?? 

post-131733-0-84687300-1454763264_thumb.

Edited by Haitham Wahid
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I think (not sure) that the 13/18 is the probability of her driving to visit relatives in general, and it seems the question says that she has them in both Nashville and St Louis. I might be wrong, though, this question is not overly clear to me. 

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Can someone help me understanding this question. I just don't seen to understand what they mean. They gave me the probability that she flies or drives when visiting her relatives, so what is point of having this 13/18 ?? 

 

I'm no expert at probability. In fact, it's the topic I hate most because it's always very confusing. But here's my attempt at the question.

 

Let Prob, D, F, N, S stand for, 'probability', 'driving', 'flying', 'Nashville', & 'St Louis', respectively. Now if I understand everything correctly, the question states that:

  • Prob of her driving anywhere is gif.latex? P(D)=\frac{13}{18}. So prob of her flying anywhere is gif.latex? P(F)=\frac{5}{18}
  • Prob of driving given that she's going to Nashville is gif.latex? P(D|N)=\frac{4}{5}. So prob of her flying given that she's going to Nashville is gif.latex? P(F|N)=\frac{1}{5}
  • Prob of flying given that she's going to St Louis is gif.latex? P(F|S)=\frac{1}{3}. So prob of driving given that she's going to St Louis is gif.latex? P(D|S)=\frac{2}{3}

For (a), total probability of driving is equal to probability of driving to Nashville plus probability of driving to St Louis. In other words:

gif.latex? P(D)=P(N)P(D|N) + P(S)P(D|S)

gif.latex?\Leftrightarrow P(N) = \frac{5

 

For (b), total probability of going to Nashville is equal to probability of going to Nashville by driving plus probability of going to Nashville by flying, which is exactly equal to the answer in part (a). In other words:

gif.latex? P(N)=\frac{5}{12}= P(D)P(N|D)

gif.latex?=P(D)(1-P(N|F))+P(F)P(N|F)

gif.latex?\Leftrightarrow P(N|F) =\frac{

 

EDIT: My reasoning for part (b) is completely wrong. See my next post in this thread for explanation.

Edited by Vioh
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Guest SNJERIN

 

Can someone help me understanding this question. I just don't seen to understand what they mean. They gave me the probability that she flies or drives when visiting her relatives, so what is point of having this 13/18 ?? 

 

I'm no expert at probability. In fact, it's the topic I hate most because it's always very confusing. But here's my attempt at the question.

 

Let Prob, D, F, N, S stand for, 'probability', 'driving', 'flying', 'Nashville', & 'St Louis', respectively. Now if I understand everything correctly, the question states that:

  • Prob of her driving anywhere is gif.latex? P(D)=\frac{13}{18}. So prob of her flying anywhere is gif.latex? P(F)=\frac{5}{18}
  • Prob of driving given that she's going to Nashville is gif.latex? P(D|N)=\frac{4}{5}
  • Prob of flying given that she's going to St Louis is gif.latex? P(F|S)=\frac{1}{3}. So prob of driving given that she's going to St Louis is gif.latex? P(D|S)=\frac{2}{3}

For (a), total probability of driving is equal to probability of driving to Nashville plus probability of driving to St Louis. In other words:

gif.latex? P(D)=P(N)P(D|N) + P(S)P(D|S)

gif.latex?\Leftrightarrow P(N) = \frac{5

 

For (b), total probability of going to Nashville is equal to probability of going to Nashville by driving plus probability of going to Nashville by flying, which is exactly equal to the answer in part (a). In other words:

gif.latex? P(N)=\frac{5}{12}= P(D)P(N|D)

gif.latex?=P(D)(1-P(N|F))+P(F)P(N|F)

gif.latex?\Leftrightarrow P(N|F) =\frac{

 

I'm not sure that these are correct. Do you have the answers from the book?

 

 

I was able to follow your working here, and I would agree with you that it is a really confusing question. Unfortunately I don't have the answer to this problem but it is from May 15 paper 2 TZ2, however I don't have the mark scheme.  

Edited by Haitham Wahid
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I was able to follow your working here, and I would agree with you that it is a really confusing question. Unfortunately I don't have the answer to this problem but it is from May 15 paper 2 TZ2, however I don't have the mark scheme.  

 

the answer to (a) is 5/12 and (b) is 3/10

 

Thanks a lot for giving the answers, Jai Chandak. I now understand the mistake I made in the previous post. What I did in part (a) was correct, but my reasoning in part (b) was wrong because:

gif.latex?P(N|D) \neq 1 - P(N|F)

That seems so obvious, but unfortunately I missed that.

 

Now I have thought of a much easier way to solve this problem in part (b). All we need to do is figure out what the probability of her flying to Nashville is. Another way of asking is what is the probability of her flying AND going to Nashville. And we have 2 equivalent ways to express that probability, thus I will set up the equation as follow:

gif.latex?P(F)P(N|F) = P(N)P(F|N)

 

Now let's substitute gif.latex?P(F) = \frac{5}{18}, and gif.latex?P(N) = \frac{5}{12}, and gif.latex?P(F|N)= 1 - P(D|N) =\frac{1}{5

 

So: gif.latex?P(N|F) = \frac{P(N)P(F|N)}{P(F

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Vioh provided a valid, logical explanation. But I would like to recommunicate his solution so it's more practical to solve on a test/paper.

Here's a diagram. Each branch represent a "choice" and the probability of choosing that choice. This is a effective and common way to visualize conditional probability. To find the probability of any event, we multiply all the probabilities leading to that branch.

 

post-151681-0-89654200-1455344221.jpg

 

It should be fairly to see the condition probabilities P(drive | Nashville), P(fly | Nashville), P(drive | St. Louis), P(fly | St. Louis).

From the main branch (Nashville vs St. Louis), we are told that x(4/5) + (1-x)(2/3) = 13/18

Solve for x, we get 5/12 = p(Nashville), p(St. Louis) = 7/12. So far this is EXACTLY vioh's logic.

b) There are two event leading to flying. Nashville --> Fly is (5/12)*(1/5) = (1/12).
St. Louis --> Fly is (7/12)*(1/3) = 7/36

Total probability is 1/12 + 7/36 = 10/36.
The probability of flying to Nashville is thus (1/12) / (10/36) = 3/10

I have shown the same method, but the diagram is what I am trying to prove is very helpful.

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