
Probability and Statistics Basic Probability and Statistics Math Forum 
 LinkBack  Thread Tools  Display Modes 
May 2nd, 2018, 02:59 AM  #1 
Newbie Joined: May 2018 From: Serbia Posts: 2 Thanks: 0  Bayes' Theorem  problem
Two archers ($S_1$ and $S_2$) are shooting target. In period of time $t$ archer $S_1$ shoots $9$ arrows, while archer $S_2$ shoots $10$ arrows. It's known that from $10$ arrows archer $S_1$ hits target with $8$ arrows, and from $10$ arrows archer $S_2$ hits target with $7$ arrows. While shooting target was hit. What is probability that target was hit by archer $S_2$? I tried to solve this problem with Bayes' theorem and I did get right answer, which is $P(H_2A)=0.49$ but in formula $P(H_2A)=\frac{P(H_2)P(AH_2)}{P(A)}$ where $P(A)$ is total probability of even $A$ (target was hit) occurring and $P(A)=1.42>1$ (which is not possible) Other parameters are: $P(H_1) = 0.9, P(H_2) = 1, P(AH_1) = 0.8, P(AH_2)=0.7$ Last edited by andrijaada99; May 2nd, 2018 at 03:02 AM. 
May 2nd, 2018, 10:01 AM  #2 
Senior Member Joined: Sep 2015 From: USA Posts: 2,042 Thanks: 1063 
please define the terms you are using. What do $H_1,~H_2,~A$ stand for? These additional parameters you list at the bottom. You are given these as part of the problem statement? 
May 2nd, 2018, 01:03 PM  #3 
Newbie Joined: May 2018 From: Serbia Posts: 2 Thanks: 0 
$A$  event that archers will hit target Hypothesis: $H_1$  event that archer $S_1$ will shoot target $H_2$  event that archer $S_2$ will shoot target "Other parameters" are not given as part of the problem, they are ones that I calculated from the text of problem. I manange to find error. It was when I defined $P(H_1)$ and $P(H_2)$. Since $H_1 + H_2 = E \implies P(H_1) + P(H_2) = 1$ Now since there is $10$ arrows and $S_1$ shoots $9$ for some time $t$, and there are $2$ archers then, \begin{equation} P(H_1) = \frac{\frac{9}{10}}{2} + x = 0.45 + x \end{equation} and same for other archer: \begin{equation} P(H_2) = \frac{\frac{10}{10}}{2} + x = 0.5 + x \end{equation} Now we have: \begin{equation} 0.45 + x + 0.5 + x = 1 \implies x = 0.025 \end{equation} That gives us $P(H_1) = 0.475, P(H_2) = 0.525$ Since there are $10$ arrows and $S_1$ hit target with $8$ this means that: $P(AH_1) = 0.8$ and same for $S_2$: $P(AH_2)=0.7$ Now from the law of total probability: $P(A) = P(H_1)P(AH_1) + P(H_2)P(AH_2) = 0.7475$ From Bayes' theorem, we have probability that $S_2$ will hit the target: \begin{equation} P(H_2A) = \frac{P(H_2)P(H_1)}{P(A)} = \frac{0.525*0.7}{0.7475} \approx 0.492 \end{equation} Can someone clarify that I'm not making mistake right now? Last edited by skipjack; May 2nd, 2018 at 02:15 PM. Reason: to correct typos 

Tags 
bayes, problem, theorem 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Bayes' Theorem  amphinomos  Probability and Statistics  2  January 20th, 2015 11:43 PM 
Bayes Theorem Equation problem!  PinkMarill  Probability and Statistics  10  July 16th, 2014 03:17 PM 
Bayes' theorem word problem  Damoo  Algebra  1  November 21st, 2012 04:21 PM 
Bayes theorem  FreaKariDunk  Algebra  2  September 21st, 2012 08:23 AM 
bayes theorem  hejmatematik  Algebra  2  December 8th, 2011 11:00 PM 