ashmath wrote:

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

| |

and or

take white ball from B = 3 ways)]= 10 ways --------------

--------- but I know it is wrong.

Finding the number of ways of selecting white balls and total number of ways of drawing a balls.

-----------------------------------or-----------------------------------

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

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

great question!

[P(a)=n(a)/total ]

[p(b)=n(b)/total]

.....

[p(k)=n(k)/total] holds true if the following 3 conditions are met:

1.) events a, b, c, .. k are mutually exclusive - i.e. 2 events can not occur simultaneously. (e.g. for a coin, heads and tails can't occur simultaneously)

2.) a, b, c,...k are the only outcomes possible. ie. p(a)+p(b)+...p(k)=1 (for a coin toss, the possibilities are only heads and tails)

3.) Its a fair process. i.e. if a,b,c..k have equal sample sizes - probability that event a occurs = probability that event b occurs = ... probability that k occurs. (like in a fair coin the possibility of landing a head is 0.5 and so is that of landing a tail; unlike a biased coin - where every time a coin is tossed probability of heads is 0.4 or 0.5 or anything but 0.5).

Therefore, because probability of selecting a white ball from the first method is not equal to the probability of selecting a white ball from second method , you cannot calculate probability like the way you did. You need to use conditional probability. (through which you already calculated the answer - probability of selecting bag 1 then white + probability of selecting bag 2 then white)

Because the two whites have different probabilities you cannot just ADD the total cases and you should use conditional.

I hope that clears your doubt. If you are interested in doing more research and waste some more time during office hours, go through conditional probability text on wiki. Obviously, won't help much for GMAT. You won't need to worry about any of these 3 on GMAT questions. In almost all arenas of our lives, for that matter, these 3 hold true.

_________________

My Debrief | MBA Timeline - New! Stay on top of deadlines, receive recommendations for each stage, get reminders.