Question

While watching a game of championship league football in a café, you observe someone who is clearly supporting Manchester United in the game. What is the probability that they were actually born within $25$miles of Manchester? Assume that: The probability that the randomly selected born with-in $25$miles person in a typical local bar environment in born with-in $25$miles of Manchester is $\dfrac{1}{{20}}$and: The chance that a person born with-in $25$miles of Manchester actually supports united is $\dfrac{7}{{10}}$; The probability that a person not born with-in $25$miles of Manchester supports united with probability $\dfrac{1}{{10}}$.

Answer 1

First we have to define what the terms we need to solve the problem are.
Let us find the probability of the given problem which is people born within twenty-five miles of Manchester. So let us fix some events according to the probability of the person who was born twenty-five miles in Manchester, let it be ${E_1}$(event one). Now the same as above but the person who will not be born in the twenty-five miles of Manchester is ${E_2}$(event two).
Formula used: $P(\dfrac{E}{{{E_1}}}) = \dfrac{{P({E_1})P(\dfrac{A}{{{E_1}}})}}{{2[P({E_1})P(\dfrac{A}{{{E_1}}}) + P({E_2})P(\dfrac{A}{{{E_2}}})]}}$, A is the given actual born twenty-five miles and $P({E_1}),P({E_2})$is the respective probability that we discussed above; which is who born and not born within twenty-five miles of Manchester.

Complete step by step answer:
Since the probability that people support Manchester or not is $\dfrac{1}{2}$(equal supporters and not supporters).
Since ${E_1}$ and ${E_2}$ be the events respectively with supporters or not supporters thus from given the probability of $P({E_1})$ is $\dfrac{1}{{20}}$ and the probability of $P({E_2})$ is $\dfrac{1}{{10}}$ also A is the actual born within twenty-five miles. Thus substitute every known value into the given formula we get $P(\dfrac{E}{{{E_1}}}) = \dfrac{{P({E_1})P(\dfrac{A}{{{E_1}}})}}{{2[P({E_1})P(\dfrac{A}{{{E_1}}}) + P({E_2})P(\dfrac{A}{{{E_2}}})]}} \Rightarrow \dfrac{{\dfrac{1}{{20}} \times \dfrac{7}{{10}}}}{{2[\dfrac{1}{{20}} \times \dfrac{7}{{10}} + \dfrac{1}{{10}} \times \dfrac{3}{{10}}]}}$ (every value of this formula is given above just substitute regarding the formula); thus, further solving we get $P(\dfrac{E}{{{E_1}}}) \Rightarrow \dfrac{{\dfrac{1}{{20}} \times \dfrac{7}{{10}}}}{{2[\dfrac{1}{{20}} \times \dfrac{7}{{10}} + \dfrac{1}{{10}} \times \dfrac{3}{{10}}]}} \Rightarrow \dfrac{7}{{26}}$(by using the operators of multiplication and division we get the simplification). Hence the probability from the given information was born within $25$miles of Manchester is $\dfrac{7}{{26}}$.

Note: Since the probability of the given events to occurs is $P(E)$ then the probability of the non-occurs in the complement of the answer which is $P(\overline E )$ , thus there is a formula that $P(\overline E ) = 1 - P(E)$ is used to find the total outcomes by numbers from the given set of questions.