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Probability Question

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by sgraves » Tue Mar 15, 2016 4:28 pm
A box contains 100 balls, numbered from 1 to 100. If three balls are selected at random and with replacement from the box, what is the probability that the sum of the 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

I'm having a really hard time understanding how you come up with 4 possible outcomes of 1/8 which would make the answer 1/2 (4 *1/8). I see how the outcome of one is 1/8 but I only found two possibilities the first time. (E+E+O and O + O + O)
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by MartyMurray » Tue Mar 15, 2016 8:08 pm
sgraves wrote:I'm having a really hard time understanding how you come up with 4 possible outcomes of 1/8 which would make the answer 1/2 (4 *1/8). I see how the outcome of one is 1/8 but I only found two possibilities the first time. (E+E+O and O + O + O)
Since you are replacing the balls, even though there are 100 balls, since half are marked with odd numbers and half are marked with even numbers, this question basically works the same as a coin flip question.

The probability of choosing an odd ball and the probability of choosing an even ball are the same, 1/2.

To get an odd sum, you need either 3 odd balls or two even and one odd.

There is 1 way to choose 3 odd balls. O - O - O. The probability of choosing three odd balls in a row is 1/2 x 1/2 x 1/2 = 1/8

There are three ways to choose one odd ball and two even balls.

E - E - O
E - O - E
O - E - E

The probability of each of those occurring is 1/2 x 1/2 x 1/2 = 1/8

So we have four ways to get an odd sum, each having a probability of 1/8 of occurring.

4 x 1/8 = 1/2

The correct answer is C.
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by Matt@VeritasPrep » Thu Mar 17, 2016 10:01 pm
This is a great question! You'll have eight options since each of our three numbers could be either even or odd. That gives you 2 * 2 * 2 = 8 possible orders of even and odd, as you can see below:

O + O + O = Odd

O + O + E = Even

O + E + E = Odd

E + E + O = Odd

E + O + O = Even

O + E + O = Even

E + O + E = Odd

E + E + E = Even

Each of these is equally likely, and four of the eight give an odd sum.
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