Probability question

[jacksitt]

The probability of a certain runner finishing in the top 5 of a race is 20%. The probability of him finishing in the top 10 is 45%.

What is the probability of him finishing anywhere from 6-9th. Is it: 45%-20% = 25%?
If there is not enough information to answer this question then what other info would we need?

 

[Deleted member 4993]

The probability of a certain runner finishing in the top 5 of a race is 20%. The probability of him finishing in the top 10 is 45%.

What is the probability of him finishing anywhere from 6-9th. Is it: 45%-20% = 25%?
If there is not enough information to answer this question then what other info would we need?

Please show us what you have tried and exactly where you are stuck.

Please follow the rules of posting in this forum, as enunciated at:

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Please share your work/thoughts about this problem.

 

[Dr.Peterson]

The probability of a certain runner finishing in the top 5 of a race is 20%. The probability of him finishing in the top 10 is 45%.

What is the probability of him finishing anywhere from 6-9th. Is it: 45%-20% = 25%?
If there is not enough information to answer this question then what other info would we need?

Which places count as the top 5? Which are in the top 10? Now compare this with what was said.

 

[jacksitt]

Which places count as the top 5? Which are in the top 10? Now compare this with what was said.

Tried to edit my question but don't know how to:

Top 5 includes: 1,2,3,4,5
Top 10 includes the top 5 as well as 6,7,8,9,10

Trying to find the probability that finishes either 6,7,8,9, or 10

 

[Dr.Peterson]

Tried to edit my question but don't know how to:

Top 5 includes: 1,2,3,4,5
Top 10 includes the top 5 as well as 6,7,8,9,10

Trying to find the probability that finishes either 6,7,8,9, or 10

Well, no, not as the problem was stated:

The probability of a certain runner finishing in the top 5 of a race is 20%. The probability of him finishing in the top 10 is 45%.

What is the probability of him finishing anywhere from 6-9th. Is it: 45%-20% = 25%?
If there is not enough information to answer this question then what other info would we need?

Are you saying the bold part was wrong? (You can't edit what you've written after the first 30 minutes, so you have to post an explicit correction.)

What are your answers to the two questions asked there? (And are they part of the question as presented to you, or your own questions about it?)

 

[Steven G]

If A and B are disjoint (can't happen at the same time), then P( A U B) = P(A) + P(B).
This is all you need to verify if your answer is correct.

In your problem what is AUB, A and B? Are A and B disjoint?

 

[jacksitt]

Well, no, not as the problem was stated:


Are you saying the bold part was wrong? (You can't edit what you've written after the first 30 minutes, so you have to post an explicit correction.)

What are your answers to the two questions asked there? (And are they part of the question as presented to you, or your own questions about it?)

Yes the bold part was wrong. It should state 6-10th instead of 6-9th.

My question is what is the probability of finishing anywhere from 6-10th. My solution was 45%-20% = 25% using the conditional probability law of (P A and not B) = P(A) - P(A and B). I wanted to see what others thought of this and if it was correct.

 

[jacksitt]

If A and B are disjoint (can't happen at the same time), then P( A U B) = P(A) + P(B).
This is all you need to verify if your answer is correct.

In your problem what is AUB, A and B? Are A and B disjoint?

P(A) is finishing in the top 5. P(B) is finishing in the top 10.
A and B can happen at the same time in my case because if a racer were to finish in the top 5 then it would also qualify as a top 10 finish.

 

[Dr.Peterson]

Yes the bold part was wrong. It should state 6-10th instead of 6-9th.

My question is what is the probability of finishing anywhere from 6-10th. My solution was 45%-20% = 25% using the conditional probability law of (P A and not B) = P(A) - P(A and B). I wanted to see what others thought of this and if it was correct.

Yes, your work is correct. I wouldn't call what you state here a "conditional probability law", as it is not about conditional probability; rather, it is just an application of the rule for disjoint union.

If (naming things differently than you did) A = "finishing in place 1, 2, 3, 4, or 5" and B = "finishing in place 6, 7, 8, 9, or 10", which are disjoint, then you are told that P(A) = P(top 5 places) = 20% and P(A U B) = P(top 10 places) = 45%. Since P(A U B) = P(A) + P(B), P(B) = P(A U B) - P(A) = 45% - 20% = 25%.

 

[Steven G]

P(A) is finishing in the top 5. P(B) is finishing in the top 10.
A and B can happen at the same time in my case because if a racer were to finish in the top 5 then it would also qualify as a top 10 finish.

You should have taken my hint and made B = finishing 6-10. Then A and B are disjoint. Hence P(AUB) = P(A) + P(B). You are given two out of the three terms so you can solve for the third just as Dr Peterson did.