Question

two children are playing a code - breaking game. one child makes a sequence of three colors from red, yellow, blue, and purple. the other child must guess the sequence of colors in the correct order. once one color is used, it cannot be repeated in the sequence. what is the probability that the sequence is guessed on the first try?
\\(\frac{1}{24}\\)
\\(\frac{1}{8}\\)
\\(\frac{1}{4}\\)
\\(\frac{1}{3}\\)

Explanation:

Step1: Calculate total permutations

We need to find the number of permutations of 4 colors taken 3 at a time. The formula for permutations is \( P(n, r)=\frac{n!}{(n - r)!} \), where \( n = 4 \) (total colors: red, yellow, blue, purple) and \( r = 3 \) (sequence length).

$$ P(4, 3)=\frac{4!}{(4 - 3)!}=\frac{4!}{1!}=\frac{4\times3\times2\times1}{1}=24 $$

Step2: Determine favorable outcomes

There is only 1 correct sequence (the one made by the first child), so the number of favorable outcomes is 1.

Step3: Calculate probability

Probability is the number of favorable outcomes divided by the number of total outcomes. So the probability \( P=\frac{\text{favorable}}{\text{total}}=\frac{1}{24} \).

Answer:

\(\frac{1}{24}\) (corresponding to the first option)