1. Tardigrade
  2. Question
  3. Mathematics
  4. Two universities A and B write questions and their corresponding solutions for a high school mathematics tournament. University A writes 10 questions every hour but makes a mistake in their solutions 10 % of the time. The university B writes 20 questions every hour and makes a mistake 20 % of the time. Each university works for 10 hours and then sends all problems to a Miss ' C ' for checking. However only 75 % of the problems which she thinks are wrong are actually incorrect. Further she thinks that 20 % of the questions from the university A have incorrect solutions, and that 10 % of the questions from the university B have incorrect solutions. If the probability that a problem definitely written and solved correctly, randomly chosen by her, was thought of as having incorrectly solved, is ( p / q ) where p and q coprimes, then find the value of (p+q).

Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

Q. Two universities A and B write questions and their corresponding solutions for a high school mathematics tournament. University A writes 10 questions every hour but makes a mistake in their solutions $10 \%$ of the time. The university B writes 20 questions every hour and makes a mistake $20 \%$ of the time. Each university works for 10 hours and then sends all problems to a Miss ' $C$ ' for checking. However only $75 \%$ of the problems which she thinks are wrong are actually incorrect. Further she thinks that $20 \%$ of the questions from the university A have incorrect solutions, and that $10 \%$ of the questions from the university $B$ have incorrect solutions.
If the probability that a problem definitely written and solved correctly, randomly chosen by her, was thought of as having incorrectly solved, is $\frac{ p }{ q }$ where $p$ and $q$ coprimes, then find the value of $(p+q)$.

Probability - Part 2

Solution:

Let $ C$ : problem is solved correctly $P ( C )=\frac{9 \cdot 10+10 \cdot 16}{300}=\frac{5}{6}$ ....(i)
W: problem is solved wrongly $P ( W )=\frac{1}{6}$
$C _1$ : She thinks that the problem is correct
$W _1$ : She thinks that the problems is wrong
A: Problems from the universityA; $P(A)=\frac{1}{3}$
B : Problems from the university B; $P(B)=\frac{2}{3}$
$P \left( W / W _1\right)=\frac{3}{4} ; P \left( W _1 / A \right)=\frac{1}{5} ; P \left( C _1 W _1\right)=\frac{1}{4} ; P \left( W _1 / B \right)=\frac{1}{10}$
we have to find $P \left( W _1 / C \right)=$ ?
$\text { now } P \left( W _1 / C \right) =\frac{ P \left( W _1 \cap C \right)}{ P ( C )}=\frac{ P \left( W _1\right) \cdot P \left( C / W _1\right)}{ P ( C )} \ldots(2) $
$\text { now } P \left( W _1\right) = P \left( A \cap W _1\right)+ P \left( B \cap W _1\right) $
image
$ = P ( A ) \cdot P \left( W _1 / A \right)+ P ( B ) \cdot P \left( W _1 / B \right)$
$=\frac{1}{3} \cdot \frac{1}{5}+\frac{2}{3} \cdot \frac{1}{10}=\frac{2}{15} $
Hence from (1) and (2)
$P \left( W _1 / C \right)=\frac{(2 / 15) \cdot(1 / 4)}{(5 / 6)}=\frac{1}{30} \cdot \frac{6}{5}=\frac{1}{25} \Rightarrow p + q =1+25=26 $