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two boxes contain 3 balls 1,2,3. IF ONE BALL IS TO BE [#permalink]
01 Sep 2008, 03:16

two boxes contain 3 balls 1,2,3. IF ONE BALL IS TO BE SELECTED FROM EACH BOX, what is the prob that the sum of the numbers on the two balls will be even?

1/9 2/9 4/9 5/9 3/3

I HAVE A DOUBT HERE. WHILE CALCULATING POSS CASES, IS THE CASE ( 1,4) DIFFERENT FROM (4,1)?? i always have a confusion regarding this... i read that if the events are occuring simultaneously ,then order does not matter , eg, if we are throwing 2 dice together. but if we throw one after another it does matter.

since in the above case , we are lifting both together, should we take (a,b) = (b,a)?

Just think logically out here. You don't need to worry which way to have like (1,4) or (4,1).

For this problem. we Have A which has 1,2,3 nos. we have B which has 1,2,3 nos.

No. of ways a no. can be selected from A = 3. No. of ways a no. can be selected from B = 3.

Total No. of ways = 3*3 = 9.

Now what are the possibilities : Case 1 : all combination of odd nos will add to give u an even no. So (1,1,), (1,3), (3,1), (3,3). Here it means (no. picked from A, no. picked from B). So you see why we have (1,3) and (3,1) here.

Case 2 : All even nos will add to give you an even no. = (2,2)

Therefore how many possible combinations we can have = 4 + 1 = 5

Therefore Probability = 5/9.

Just a note, The process above looks tedious because of the long explanation. But if you think the steps, it will be faster. For this problem you can have this approach and complete the problem in < 1min. However I would also like to add that if the nos are big, this approach is going to take more than 3 mins. For eg: if you had 3 bags and each had 10 balls numbered 1 to 10, and you are asked to find the same result. In cases like these you should know the basics of permutation and combination. That can help you reduce time drastically and still have the answer in less than 1 min.

two boxes contain 3 balls 1,2,3. IF ONE BALL IS TO BE SELECTED FROM EACH BOX, what is the prob that the sum of the numbers on the two balls will be even?

1/9 2/9 4/9 5/9 3/3

I HAVE A DOUBT HERE. WHILE CALCULATING POSS CASES, IS THE CASE ( 1,4) DIFFERENT FROM (4,1)?? i always have a confusion regarding this... i read that if the events are occuring simultaneously ,then order does not matter , eg, if we are throwing 2 dice together. but if we throw one after another it does matter.

since in the above case , we are lifting both together, should we take (a,b) = (b,a)?

Originally posted on MIT Sloan School of Management : We are busy putting the final touches on our application. We plan to have it go live by July 15...