2
   

Maths (probability ) pls help

 
 
Reply Thu 7 Jun, 2012 02:23 am
A problem was given to "A" whose probability to solve it is 2/3. If "A' couldnt solve it, the same is given to "B" whose probability to solve it when "A' could not solve is 5/7. If "B" couldnt solve it then, the same is given to "C" whose probability to solve it then is 1/10. what is the probability that problem gets solved.
  • Topic Stats
  • Top Replies
  • Link to this Topic
Type: Discussion • Score: 2 • Views: 1,851 • Replies: 6
No top replies

 
engineer
 
  1  
Reply Thu 7 Jun, 2012 05:11 am
@sandy111,
The probability of it not being solved is 1/3 x 2/7 x 9/10 = 6/70 so the probability of solving it is 64/70.
sandy111
 
  1  
Reply Thu 7 Jun, 2012 06:47 am
@engineer,
hello....
This is not correct. Your answer is correct if the situation is like a problem is given to 3 persons simultaneously and all are working it together. but here condition is different. If first student is unable to solve, it is given to second (first not involved) and if second is unable to solve, it is given to third (second not involved).....Plss...solve it..
maxdancona
 
  1  
Reply Thu 7 Jun, 2012 07:06 am
@sandy111,
Look at engineer's solution again. It is correct (hint, he multiplied the numbers subracted from 1).
sandy111
 
  1  
Reply Thu 7 Jun, 2012 07:26 am
@maxdancona,
Hii.
i understood how he solved but he understood question differently... Actual problem i already stated......Let me explain...In my problem, B solved only when A is unable to and that time A's role is over.. Same C solved wen B unable to solve. that time B role is over.... But engineer solution presumes that when A solves , B and C also try to solve but unable to solve. When B solves, A and C also try to solve, but unable to solve...............
0 Replies
 
engineer
 
  1  
Reply Thu 7 Jun, 2012 07:35 am
@sandy111,
If you are trying to compute the probability of not solving the problem, there is no difference between them working sequentially or in parallel - they all must still fail to solve the problem. That's why you approach problems like this from the fail perspective instead of the pass perspective. If you want to do it the much harder way, it looks like this:

P = A + A'B + A'B'C

P = 2/3 + (1/3)(5/7) + (1/3)(2/7)(1/10) = 64/70
sandy111
 
  1  
Reply Thu 7 Jun, 2012 07:39 am
@engineer,
superb.........great...........
0 Replies
 
 

Related Topics

Amount of Time - Question by Randy Dandy
logical number sequence riddle - Question by feather
Calc help needed - Question by mjborowsky
HELP! The Product and Quotient Rules - Question by charsha
STRAIGHT LINES - Question by iqrasarguru
Possible Proof of the ABC Conjecture - Discussion by oralloy
Help with a simple math problem? - Question by Anonymous1234567890
How do I do this on a ti 84 calculator? - Question by Anonymous1234567890
 
  1. Forums
  2. » Maths (probability ) pls help
Copyright © 2024 MadLab, LLC :: Terms of Service :: Privacy Policy :: Page generated in 0.03 seconds on 04/25/2024 at 11:45:15