logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
Home  >>  CBSE XII  >>  Math  >>  Probability
0 votes

Suppose that 5% of men and 0.25% of women have grey hair. A grey haired person is selected at random. What is the probability of this person being male? Assume that there are equal number of males and females.

Can you answer this question?
 
 

1 Answer

0 votes
Toolbox:
  • $P \large(\frac{B}{A}) =$$ \large (\frac{P(A\;\cap\; B)}{P(A)})$
  • According to Bayes Theorem, if $E_1,E_2,E_3.....E_n$ are a set of mutually exclusive and exhaustive events, then $P\left(\large \frac{E_i}{E}\right ) = \Large \frac{P\left(\frac{E}{E_i}\right ). P(E_i)} {\sum_{i=1}^{n} (P\left(\frac{E}{E_i}\right ).P(E_i))}$
Let $E_1$ be the event that the selected person is male and $E_2$ be the event that its a female.
$ P (E_1) = \large\frac{1}{2} = $$P (E_2)$
$E_1$ and $E_2$ are mutually exclusive and exhaustive, so we can apply Bayes theorem.
Let E: be the event that the person has grey hair. 5% of men picked at random have grey hair $\rightarrow P \large (\frac{E}{E_1}) = \frac{5}{100} = \frac{1}{20}$
We can calcuate the probability: $P \large \left(\frac{E_1}{E}\right) = \Large \frac{P\left(\frac{E}{E_1}\right ). P(E_1)} { (P\left(\frac{E}{E_1}\right ).P(E_1)) + (P\left(\frac{E}{E_2}\right ).P(E_2))}$
$P \large \left(\frac{E_1}{E}\right) = \Large \frac{\frac{1}{20} \frac{1}{2}} { \frac{1}{20}\frac{1}{2} + \frac{1}{400}\frac{1}{2}} = \frac{20}{21}$
answered Jan 30, 2014 by balaji
 

Related questions

Ask Question
student study plans
x
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App
...