Competition problem — Combinatorics

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Combinatorics Difficulty 3.0 Junior Find the answer

There are two red balls and one white ball of the same size in a pocket. Balls are drawn with replacement, and a sequence $\{a\_n\}$ is defined as follows: $a\_n=\begin{cases} -1, \text{if a red ball is drawn on the nth draw}, \\ 1, \text{if a white ball is drawn on the nth draw}. \end{cases}$ If $S\_n$ is the sum of the first $n$ terms in the sequence $\{a\_n\}$ , what is the probability that $S\_7=3$ ?

A) $\frac{26}{3^{6}}$
B) $\frac{28}{3^{6}}$
C) $\frac{29}{3^{6}}$
D) $\frac{31}{3^{6}}$

Your answer

Multiple choice: answer with the letter of the option you want.

Solution

This problem tests our understanding of the probability multiplication rule for independent events. To obtain $S\_7=3$ , we must have drawn exactly two red balls in seven draws. Since each draw is independent, we can use the probability multiplication rule.

Step 1: Determine the number of ways to draw exactly two red balls in seven draws. This is given by the binomial coefficient $C\_7^2 = \frac{7!}{2!(7-2)!} = 21$ ways.

Step 2: Calculate the probability of drawing exactly two red balls and five white balls. For each favorable outcome, the probability is $(\frac{2}{3})^2 \cdot (\frac{1}{3})^5$ .

Step 3: Multiply the number of favorable outcomes by the probability of each outcome: $21 \cdot (\frac{2}{3})^2 \cdot (\frac{1}{3})^5 = \frac{28}{3^{6}}$ .

Thus, the probability that $S\_7=3$ is $\boxed{\frac{28}{3^{6}}}$ , and the correct answer is (B).

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.