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. $$\frac{1}{4}$$

B. $$\frac{3}{8}$$

C. $$\frac{1}{2}$$

D. $$\frac{5}{8}$$

E. $$\frac{3}{4}$$

## A box contains 100 balls, numbered from 1 to 100. If three balls are selected at random and with replacement from the bo

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sambati wrote: ↑Tue Jul 21, 2020 10:56 amA 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. $$\frac{1}{4}$$

B. $$\frac{3}{8}$$

C. $$\frac{1}{2}$$

D. $$\frac{5}{8}$$

E. $$\frac{3}{4}$$

**Solution:**

To get the sum of the three numbers to be odd, EITHER all three numbers are odd OR two numbers are even and the third number is odd.

The probability of three odd numbers is ½ x ½ x ½ = 1/8. The probability of two even numbers and one odd number, in any order, (see note below) is (½ x ½ x ½) x 3 = 3/8. Therefore, the overall probability of the sum of the three numbers is odd is ⅛ + ⅜ = 4/8 = ½.

(Note: Let O be odd and E be even, then two even numbers and one odd number can be arranged in 3 ways: OEE, EOE, EEO.)

**Answer: C**

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