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I can not remember how to solve this problem. Can someone help me? Tom and Alice work independently in an attempt to solve a certain problem, i.e. whether one of them solves it does not affect the chances that the other will solve it. The probability that Tom solves it is 0.15 and the probability that Alice solves it is 0.5. What is the probability that the problem will be solved? If the problem is solved, what is the probability that Tom solves it?
 
buckweat_86 said:
I can not remember how to solve this problem. Can someone help me? Tom and Alice work independently in an attempt to solve a certain problem, i.e. whether one of them solves it does not affect the chances that the other will solve it. The probability that Tom solves it is 0.15 and the probability that Alice solves it is 0.5. What is the probability that the problem will be solved? If the problem is solved, what is the probability that Tom solves it?
Click to expand...
Hate probabilities; anyway, here goes:
same as:
Tom picks 3 balls from a box containing 3 blue balls and 17 red balls;
if Tom didn't pick at least 1 blue ball, then:
Alice picks 1 ball from a box containing 1 blue ball and 19 red balls.
What is probability that at least 1 blue ball has been picked?

3/20 + 17/20(1/20) = 17/400 ? Probably wrong...
 
What is the probability that the problem will be solved?
P(S)=P(T)+P(A)−P(T)P(A)=(.15)+(.5)−(.15)(.5)=0.575

If the problem is solved, what is the probability that Tom solves it?
P(T|S)=P(T)/P(S)=0.261
 
G'day buckweat_86,

Code: 
Let T be Tom solves the problem; T' be Tom not solve the problem.
Let A be Alice solves the problem; A' be Alice not solve the problem.
   
             0.5    A
                 *  
              * 
           T     0.5
   0.15  *    *
      /          * 
    *              A'    
      *               A
    0.85 *          *
            *    *  0.5
              T'  
                 *
                0.5 *
                      A'
So to find the probability of the problem being solved, multiply down all the branches with a T or A (or both) in it, and add:
0.15 * 0.5 +
0.15 * 0.5 +
0.85 * 0.5
= 0.575
(which agrees with pKa).

If S is the solution being solved:

P(T | S) = P(T and S) / P(S)
but if Tom solves the problem, then the solution must have been solved too so P(T and S) = P(T), as pka has it.
 
Thanks pka and Unco.
Copy of your solutions now in my rather skinny "probability file" :wink:

Plus made note on file flap NEVER to try one late at night :idea:
 
Geeesh...checking further as to why my result was so low, I realised that I
erroneously used .05 instead of .5 for Alice :
you 2 "probability experts" should have noticed that!

My calculations should show:
same as:
Tom picks a ball from a box containing 3 blue balls and 17 red balls;
if Tom didn't pick a blue ball, then:
Alice picks a ball from a box containing 10 blue ball and 10 red balls.
What is probability that at least 1 blue ball has been picked?

3/20 + 17/20(1/2) = 3/20 + 17/40 = 23/40 = .575 :lol:

So I was (by luck cause I wasn't 100% sure) correct !

You guys see anything wrong with my amateur approach?
 
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