Math Is Fun Forum
  Discussion about math, puzzles, games and fun.   Useful symbols: √ ∞ ≠ ≤ ≥ ≈ ⇒ ∈ Δ θ ∴ ∑ ∫ π -




Not registered yet?

  • Index
  •  » Help Me !
  •  » can i solve conditional probability using n(E)/n(S) approach

Post a reply

Go back

Write your message and submit
:) :| :( :D :o ;) :/ :P :lol: :mad: :rolleyes: :cool: | :dizzy :eek :kiss :roflol :rolleyes :shame :down :up :touched :sleep :wave :swear :tongue :what :faint :dunno

Go back

Topic review (newest first)

bob bundy
2013-01-27 19:31:32

hi ashmath

I'll try.

Firstly, go back to the definition:

So, how to work out the total number of outcomes?

My approach is to imagine the 'experiment' is repeated again and again until every event has had a fair chance to occur.  The key here is what do I mean by a fair chance. 

For bag A there are 5 balls so I can get the probabilities just by saying 'total number of outcomes' = 5.

For bag B in the same way 'total number of outcomes' = 7

So how to fairly decide on the number when either bag may be picked.

If I pick from bag A 35 times (5x7) then P(W) = 14/35 and P(B) = 21/35

and for bag B,  I would get P(W) = 15/35 and P(B) = 20/35

So I can treat A and B fairly by having 35 trials for A and 35 trials for B.

So I'll have 70 trials altogether.

70 splits like this P(A) = 35/70 and P(B) = 35/70

Then use the probabilities above to get P(A & W) = 14/70    P(A & B) = 21/35      P(B & W) = 15/70     P(B & B) = 20/35

Then I can see that 'total number of outcomes' = 70

'Number of ways W happens' = 14 + 15 = 29

So P(W) = 29/70

Hope that helps, smile


2013-01-27 08:04:46

Thanks a lot but can u explain it a bit more

bob bundy
2013-01-25 10:27:03

hi ashmath,

I don't know if my tree diagram will help but let's find out.

There are got fifths and sevenths so I want a number that can be divided into both ... ie 35

And I want equal chances of choosing A or B so I'll double that to 70.

So now imagine I repeat the experiment exactly 70 times and the breaks are exactly according to the 'law of averages' 

The diagram shows the 70 outcomes.  This demonstrates why the formula works and allows you to 'go back to basics'


2013-01-25 08:45:07

Q. there are two bags A and B; bag A contains 2 white and 3 black balls; bag B contains 3 white and 4 black balls. A bag is chosen at random find the probability that the ball drawn is white.

My answer : I am OK with P(W) = P(A,W)+ P(B,W)= 1/2 * 2/5 + 1/2 * 3/7 = 29/70 OK !!

but if I want to solve the same problem using the classical approach,

           n(E)    number of ways of selecting white ball   
P(W) = ---- = ---------------------------------------------- then what ????
           n(s)    total number of ways of drawing a ball

I think! the no. of selecting white balls should be = (number of ways of selecting a bag = 2) * [(either take white ball from A = 2

ways) + (either take white ball from B = 3 ways)]= 10 ways ----------------------- but I know it is wrong.

I am confused and unable about Finding the number of ways of selecting white balls and total number of ways of drawing a balls.


the conditional probability problems can not be solved using n(E)/n(S) approach ??????? ------- but why ??

Please help me soon I am in a great need of it Thanks in advance.

Board footer

Powered by FluxBB