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two children are playing a code - breaking game. one child makes a sequ…

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 use permutations as order matters and no repetition. The formula for permutations is \( P(n, r)=\frac{n!}{(n - r)!} \), where \( n = 4 \) (colors: red, yellow, blue, purple) and \( r=3 \) (sequence length).
\( P(4,3)=\frac{4!}{(4 - 3)!}=\frac{4!}{1!}=4\times3\times2 = 24 \).

Step2: Determine favorable outcomes

There's 1 correct sequence. Probability is \( \frac{\text{favorable}}{\text{total}}=\frac{1}{24} \).

Answer:

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