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 is 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 $\dfrac{1}{{10}}$
A) $\dfrac{7}{{26}}$
B) $\dfrac{8}{{26}}$
C) $\dfrac{9}{{26}}$
D) $\dfrac{{10}}{{26}}$

Answer 1

Hint: According to the question we have to find the probability that people were actually born within 25 miles of Manchester. So, first of all we have to the let that the probability of a person who was born 25 miles of Manchester is an event ${E_1}$ and same as the probability of a person who was not born 25 miles of Manchester is an event ${E_2}$.
Now, we have to find the probability that they (people) support Manchester or not after that to find the required probability we have to apply the formula as mentioned below:

Formula used: $ P\left( {\dfrac{E}{{{E_1}}}} \right) = \dfrac{{P({E_1})P\left( {\dfrac{A}{{{E_1}}}} \right)}}{{2\left[ {P({E_1})P\left( {\dfrac{A}{{{E_1}}}} \right) + P({E_2})P\left( {\dfrac{A}{{{E_2}}}} \right)} \right]}}.........(1)$
Where, A is actually born with-in 25 miles, and $P({E_1})$ is the probability that the randomly selected born with-in 25 miles person in a typical local bar environment is born with-in 25 miles of Manchester and \[P({E_2})\] is the probability that a person not born with-in 25 miles of Manchester supports United.

Complete step by step solution:
Step 1: First of all we have to let that the probability of a person who was born 25 miles from Manchester is an event ${E_1}$ and the probability of a person who was not born 25 miles of Manchester is an event ${E_2}$.
Step 2: Now, we have to find the probability of an event that they (people) support Manchester or not.
Hence,
Probability of an event that they (people) support merchant or not $ = \dfrac{1}{2}$
Step 2: Now, we have to find that the chance that a person not born with-in 25 miles of Manchester actually supports United with the help of the chance that a person born with-in 25 miles of Manchester actually supports United is $\dfrac{7}{{10}}$by subtracting with 1. Hence,
$
   = 1 - \dfrac{1}{{10}} \\
   = \dfrac{{10 - 1}}{{10}} \\
   = \dfrac{9}{{10}}
 $
Step 3: Now, with the help of the formula (1) as mentioned in the solution hint we can obtain the value of required probability. Hence, On substituting all the values into formula (1)
$ \Rightarrow P\left( {\dfrac{{{E_1}}}{{{E_2}}}} \right) = \dfrac{{\dfrac{1}{{20}} \times \dfrac{7}{{10}}}}{{2\left[ {\dfrac{1}{{20}} \times \dfrac{7}{{10}} + \dfrac{1}{{10}} \times \dfrac{3}{{10}}} \right]}}$
On solving the expression obtained just above,
$ \Rightarrow P\left( {\dfrac{{{E_1}}}{{{E_2}}}} \right) = \dfrac{{\dfrac{7}{{200}}}}{{2\left[ {\dfrac{7}{{200}} + \dfrac{3}{{100}}} \right]}}$
To solve the terms obtained in denominator we have to solve the L.C.M,
$ \Rightarrow P\left( {\dfrac{{{E_1}}}{{{E_2}}}} \right) = \dfrac{{\dfrac{7}{{200}}}}{{2\left[ {\dfrac{{7 + 6}}{{200}}} \right]}}$
Step 4: Now, to solve the expression as obtained just above we have to multiply terms obtained.
$
   \Rightarrow P\left( {\dfrac{{{E_1}}}{{{E_2}}}} \right) = \dfrac{{\dfrac{7}{{200}}}}{{2 \times \dfrac{{13}}{{200}}}} \\
   \Rightarrow P\left( {\dfrac{{{E_1}}}{{{E_2}}}} \right) = \dfrac{{7 \times 200}}{{200 \times 26}} \\
 $
On eliminating 200 from the expression obtained above, we can obtain the required probability which is the probability that they were actually born within 25 miles of Manchester. Hence,
$ \Rightarrow P\left( {\dfrac{{{E_1}}}{{{E_2}}}} \right) = \dfrac{7}{{26}}$
Final solution: Hence, with the help of the formula (1) as mentioned in the solution hint we have obtained the probability that they were actually born within 25 miles of Manchester is $ = \dfrac{7}{{26}}$

Hence, the given option (A) is correct.

Note: If the probability of an event to be occur is $P(E)$then the probability of an event not to be occur $P(\overline E )$ can be calculated by subtracting the probability of an event to be occur is P(E) by 1 hence, $P(\overline E ) = 1 - P(E)$
To find the probability of an event we have to divide the required event to occur by the total number of outcomes.