Question

While watching a game of Champions League football in a cafe, 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 a randomly selected person in a typical local bar environment is born within $25$ miles of Manchester is $\frac{1}{20}$, and;

$⟹$the chance that a person born within $25$ miles of Manchester actually supports United is $\frac{7}{10}$;

$⟹$the probability that a person not born within $25$ miles of Manchester supports United with probability $\frac{1}{10}$

• A. $\frac{7}{26}$
• B. $\frac{8}{26}$
• C. $\frac{9}{26}$
• D. $\frac{10}{26}$

Answer 1

The correct option is A $\frac{7}{26}$

Probability that they support Manchest or not $=\frac{1}{2}$

Let $E_1$ and $E_2$ be the events of a person who was born $25$ miles of Manchester and $E_2$ be event of a person not born with in $25$ miles of Manchester

$\therefore P\left(E_1\right)=\frac{1}{20}$ $P\left(E_2\right)=\frac{1}{10}$

$A=$ Actually born with in $25$ miles

$\Rightarrow P\left(\frac{E}{E_1}\right)=\frac{P\left(E_1\right)P\left({\frac{A}{E}}_1\right)}{2[P\left(E_1\right)P\left({\frac{A}{E}}_1\right)+P\left(E_2\right)P\left({\frac{A}{E}}_2\right)]}$

$=\frac{\frac{1}{20}\times \frac{7}{10}}{2[\frac{1}{20}\times \frac{7}{10}+\frac{1}{10}\times \frac{3}{10}]}$

$=\frac{7}{26}.$

Hence, the answer is $\frac{7}{26}.$