Question

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 $(p+q)$.

Answer 1

Answer: 26

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 $P\left(W_1/C\right)=$ ?
$\text{ now }P\left(W_1/C\right)=\frac{P\left(W_1\cap C\right)}{P(C)}=\frac{P\left(W_1\right)\cdot P\left(C/W_1\right)}{P(C)}\dots (2)$
$\text{ now }P\left(W_1\right)=P\left(A\cap W_1\right)+P\left(B\cap W_1\right)$

$=P(A)\cdot P\left(W_1/A\right)+P(B)\cdot P\left(W_1/B\right)$
$=\frac{1}{3}\cdot \frac{1}{5}+\frac{2}{3}\cdot \frac{1}{10}=\frac{2}{15}$
Hence from (1) and (2)
$P\left(W_1/C\right)=\frac{(2/15)\cdot (1/4)}{(5/6)}=\frac{1}{30}\cdot \frac{6}{5}=\frac{1}{25}\Rightarrow p+q=1+25=26$