question:
$a)\dfrac{1}{14}$
$b)\dfrac{13}{14}$
$c)\dfrac{1}{5}$
$d)\dfrac{2}{7}$
my attempt:
for probability of result to be correct both A and B have to solve wrong and make mistake or both have to solve right without making mistake
from question one can infer probability of making no mistake is $\dfrac{1000}{1001}$ and making mistake is $\dfrac{1}{1001}$
after that i don't know how to proceed further should i use bye's theoram here
or use probability tree method. i'm confused please help and if possible please solve by tree method . thanks in advance .
Let $A$ be the event that student $A$ answers the question correctly; let $B$ be the event that student $B$ answers the question correctly; and let $C$ be the event that $A$ and $B$ coincidentally obtain the same wrong answer. We are given \begin{align*} \Pr(A) & = \frac{1}{8}\\ \Pr(B) & = \frac{1}{12} \end{align*} from which we can determine that \begin{align*} \Pr(A^C) & = 1 - \frac{1}{8} = \frac{7}{8}\\ \Pr(B^C) & = 1 - \frac{1}{12} = \frac{11}{12} \end{align*} We are also given that the odds against $A$ and $B$ obtaining the same wrong answer are $1000:1$, which means that $$\Pr(C) = \frac{1}{1000 + 1} = \frac{1}{1001}$$ The probability that $A$ and $B$ obtain the correct result given that they obtain the same answer is $$\frac{\Pr(A \cap B)}{\Pr(A \cap B) + \Pr(A^C \cap B^C)\Pr(C)}$$ Assuming the probability that student $A$ solves the problem correctly is independent of the probability that student $B$ solves the problem correctly, we obtain \begin{align*} \frac{\Pr(A \cap B)}{\Pr(A \cap B) + \Pr(A^C \cap B^C)\Pr(C)} & = \frac{\Pr(A)\Pr(B)}{\Pr(A)\Pr(B) + \Pr(A^C)\Pr(B^C)\Pr(C)}\\ & = \frac{\frac{1}{8} \cdot \frac{1}{12}}{\frac{1}{8} \cdot \frac{1}{12} + \frac{7}{8} \cdot \frac{11}{12} \cdot \frac{1}{1001}}\\ & = \frac{\frac{1}{96}}{\frac{1}{96} + \frac{77}{96096}}\\ & = \frac{1}{1 + \frac{77}{1001}}\\ & = \frac{1}{\frac{1078}{1001}}\\ & = \frac{1001}{1078}\\ & = \frac{13}{14} \end{align*}