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Thread: probability

  1. #1
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    probability

    According to a survey, of those employees living more than 2 miles from work , 90% travel to work by car . Of the remaining employees, only 50% travel to work by car .

    It's known that 75% of employees live more than 2 miles from work .

    Find the probability of that an employee who travels to work by car more than 2 miles from work .


    The ans given is (0.9 x 0.75 ) / (0.75)(0.9) + (0.25)(0.5) = 0.844

    But , I think it should be (0.9 x 0.75 ) = 0.675

    Correct me if i am wrong .
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  2. #2
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    Re: probability

    You are told $P(\text{drives to work} | \text{lives >2 miles from work}) = 0.9$

    $P(A|B) = \dfrac{P(A\cap B)}{P(B)}$

    so

    $P(A\cap B) = P(A|B)P(B)$

    This means that $P(\text{lives >2 miles from work} \cap \text{drives to work}) = 0.9\cdot 0.75$

    But, you are not asked to find the probability that an employee both travels to work by car and lives more than 2 miles from work. You are asked to find the probability that given the employee drives to work, what is the probability that they also live more than two miles from work. You are looking for:

    $P(\text{lives >2 miles from work} | \text{drives to work}) = \dfrac{P(\text{drives to work} \cap \text{lives >2 miles from work})}{P(\text{drives to work})}$

    Now, the numerator we already found. How about the denominator? How do we figure that out?

    Well, we already found $P(\text{drives to work} \cap \text{lives >2 miles from work})$. Next, we need $P(\text{drives to work} \cap \text{lives <2 miles from work}) = P(\text{drives to work}|\text{lives <2 miles from work})P(\text{lives <2 miles from work}) = 0.5\cdot \left(1-P(\text{lives >2 miles from work}) \right) = 0.5(1-0.75) = 0.5\cdot 0.25$.

    Now, $P(\text{drives to work}) = P(\text{drives to work} \cap \text{lives >2 miles from work}) + P(\text{drives to work} \cap \text{lives <2 miles from work}) = 0.9\cdot 0.75 + 0.5\cdot 0.25$.

    So, the probability you are looking for is:

    $P(\text{lives >2 miles from work} | \text{drives to work}) = \dfrac{0.9\cdot 0.75}{0.9\cdot 0.75 + 0.5 \cdot 0.25}$
    Last edited by SlipEternal; Sep 18th 2017 at 06:26 AM.
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  3. #3
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    Re: probability

    Imagine that there 1000 employees. 25% of them, 250, live within 2 miles. Of those 250, 90% of them, 225, travel by car. Of the other 750 employees, 50% of them, 375, travel by car. So there are 225+ 375= 600 employees who travel by car and, of those, 375 travel more than two miles to work. That is \frac{375}{600}= 0.625. So the probability that a person who travels to work by car lives more than 2 miles from work is 0.625.
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    Re: probability

    Quote Originally Posted by HallsofIvy View Post
    Imagine that there 1000 employees. 25% of them, 250, live within 2 miles. Of those 250, 90% of them, 225, travel by car. Of the other 750 employees, 50% of them, 375, travel by car. So there are 225+ 375= 600 employees who travel by car and, of those, 375 travel more than two miles to work. That is \frac{375}{600}= 0.625. So the probability that a person who travels to work by car lives more than 2 miles from work is 0.625.
    You got some of those numbers backwards.
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  5. #5
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    Re: probability

    Quote Originally Posted by HallsofIvy View Post
    Imagine that there 1000 employees. 25% of them, 250, live within 2 miles. Of those 250, 90% of them, 225, travel by car. Of the other 750 employees, 50% of them, 375, travel by car. So there are 225+ 375= 600 employees who travel by car and, of those, 375 travel more than two miles to work. That is \frac{375}{600}= 0.625. So the probability that a person who travels to work by car lives more than 2 miles from work is 0.625.
    Quote Originally Posted by SlipEternal View Post
    You got some of those numbers backwards.
    It is more than and less than that are swapped.
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  6. #6
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    Re: probability

    Quote Originally Posted by SlipEternal View Post
    You are told $P(\text{drives to work} | \text{lives >2 miles from work}) = 0.9$

    $P(A|B) = \dfrac{P(A\cap B)}{P(B)}$

    so

    $P(A\cap B) = P(A|B)P(B)$

    This means that $P(\text{lives >2 miles from work} \cap \text{drives to work}) = 0.9\cdot 0.75$

    But, you are not asked to find the probability that an employee both travels to work by car and lives more than 2 miles from work. You are asked to find the probability that given the employee drives to work, what is the probability that they also live more than two miles from work. You are looking for:

    $P(\text{lives >2 miles from work} | \text{drives to work}) = \dfrac{P(\text{drives to work} \cap \text{lives >2 miles from work})}{P(\text{drives to work})}$

    Now, the numerator we already found. How about the denominator? How do we figure that out?

    Well, we already found $P(\text{drives to work} \cap \text{lives >2 miles from work})$. Next, we need $P(\text{drives to work} \cap \text{lives <2 miles from work}) = P(\text{drives to work}|\text{lives <2 miles from work})P(\text{lives <2 miles from work}) = 0.5\cdot \left(1-P(\text{lives >2 miles from work}) \right) = 0.5(1-0.75) = 0.5\cdot 0.25$.

    Now, $P(\text{drives to work}) = P(\text{drives to work} \cap \text{lives >2 miles from work}) + P(\text{drives to work} \cap \text{lives <2 miles from work}) = 0.9\cdot 0.75 + 0.5\cdot 0.25$.

    So, the probability you are looking for is:

    $P(\text{lives >2 miles from work} | \text{drives to work}) = \dfrac{0.9\cdot 0.75}{0.9\cdot 0.75 + 0.5 \cdot 0.25}$
    Why do you inteperet Find the probability of that an employee who travels to work by car more than 2 miles from work

    as $P(\text{lives >2 miles from work} | \text{drives to work}) $ ???

    Why shouldnt it be (0.9 x 0.75 ) = 0.675 ??
    Last edited by xl5899; Sep 19th 2017 at 01:03 AM.
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  7. #7
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    Re: probability

    Quote Originally Posted by xl5899 View Post
    Why do you inteperet Find the probability of that an employee who travels to work by car more than 2 miles from work

    as $P(\text{lives >2 miles from work} | \text{drives to work}) $ ???

    Why shouldnt it be (0.9 x 0.75 ) = 0.675 ??
    Because an employee "who travels to work by car" has a 100% chance of travelling to work by car, not a 90% chance as you used in your calculation. So, I thought, what is the problem asking? It is asking, among employees who drive to work by car, what is the probability that they live more than 2 miles from work?
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    Re: probability

    Quote Originally Posted by SlipEternal View Post
    Because an employee "who travels to work by car" has a 100% chance of travelling to work by car, not a 90% chance as you used in your calculation. So, I thought, what is the problem asking? It is asking, among employees who drive to work by car, what is the probability that they live more than 2 miles from work?
    By saying that 0.75 * 0.9 , i mean the percentage of the people who drive car and also live more than 2 miles
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  9. #9
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    Re: probability

    Quote Originally Posted by xl5899 View Post
    By saying that 0.75 * 0.9 , i mean the percentage of the people who drive car and also live more than 2 miles
    You are not asked to find the percentage of people who drive to work and also live more than 2 miles from work. The word "and" was not included in the question. You were asked, "Find the probability of that an employee who travels to work by car more than 2 miles from work ."

    It does not say, "Find the probability that an employee both travels to work by car AND lives more than 2 miles from work." If the problem was stated with the word "and", then your answer would make sense. But, it asks about employees "who" travel to work by car, so every employee being considered is already known to travel to work by car. Now, given that they travel to work by car, what is the probability that they also live more than 2 miles from work? That is known as a conditional probability. You can read more about it here:

    https://en.wikipedia.org/wiki/Conditional_probability
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