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by tmontgomery » Fri Sep 04, 2015 1:40 pm
A box contains 100 balls, numbered 1 to 100. If three balls are selected at random and with replacement from the box, what is the probability that the sum of three numbers on the balls selected from the box will be odd?
A. 1/4
B. 3/8
C. 1/2
D. 5/8
E. 3/4
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by Matt@VeritasPrep » Fri Sep 04, 2015 4:47 pm
A few ways of thinking about this:

1:: Since each ball is equally likely to be odd or even -- you have the same number of odds as you do of evens -- each sum is equally likely to be odd or even. So there's a 1/2 chance the sum is odd and a 1/2 chance it's even.

2:: Case by case, the three numbers we pick could be: Odd/Odd/Odd, Odd/Odd/Even, Odd/Even/Even, Even/Even/Even. The first and third of these (OOO and OEE) will have odd sums, the second and fourth will have even sums. Since the probability of three Odds = the probability of three Evens, and the probability of two Odds = the probability of two Evens, we again have equally likely outcomes.
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by tmontgomery » Fri Sep 04, 2015 11:17 pm
That helps a lot, thank you!
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by [email protected] » Sat Sep 05, 2015 9:23 am
Hi tmontgomery,

These types of questions can be solved in a couple of different ways, depending on how you like to organize your information and how you "see" the question.

Since half the numbered balls are 'odd' and half are 'even', and we're replacing each ball after selecting it, on any given selection we have a 50/50 chance of getting an odd or even. This means that there are....

(2)(2)(2) = 8 possible arrangements when selecting 3 balls from the box:

EEE
EEO
EOE
OEE

OOO
OOE
OEO
EOO

Since we have the same number of Odds and Evens, each of these options has the same 1/8 probability of happening. Now you just have to find the ones that give us the ODD SUM that we're asked for. They are:

EEO
EOE
OEE
OOO

Four of the eight options. 4/8 = 1/2

Final Answer: C

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by Max@Math Revolution » Sat Sep 05, 2015 9:43 pm
Forget conventional ways of solving math questions. In PS, IVY approach is the easiest and quickest way to find the answer.


A box contains 100 balls, numbered 1 to 100. If three balls are selected at random and with replacement from the box, what is the probability that the sum of three numbers on the balls selected from the box will be odd?
A. 1/4
B. 3/8
C. 1/2
D. 5/8
E. 3/4

==> Probability = specific cases of event/total number of cases
total number of cases:100*100*100
specific cases of event : e+e+o, e+o+e, o+e+e, or o+o+o are all the cases we have, and are 50.
Thus we have 4 cases of 50*50*50. The numerator is 50*50*50*4.
The probability that the addition of three numbers is odd is =50*50*50*4/100*100*100=1/2. Thus the answer is C.


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