1984 AMC 12 Problem 19

Question

A box contains

11

balls, numbered

1,

2,

3,

\ldots,

11.

If

6

balls are drawn simultaneously at random, what is the probability that the sum of the numbers on the balls drawn is odd?

\frac{100}{231}

\frac{115}{231}

\frac12

\frac{118}{231}

\frac6{11}

There are

6

odd-numbered balls and

5

even-numbered balls

Count selections containing

1,

3,

or

5

odd balls

An odd sum requires an odd number of the six selected balls to be odd. There are

6

odd and

5

even balls, so the favorable count is

\begin{aligned} &\binom61\binom55+\binom63\binom53\\ &\quad+\binom65\binom51\\ &=6+200+30=236. \end{aligned}

Of the

\binom{11}{6}=462

selections, the desired probability is

\frac{236}{462}=\frac{118}{231}.

Therefore, the correct answer is D.

Problem 19 in other years