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?

A. [tex]\frac{1}{24}[/tex]
B. [tex]\frac{1}{8}[/tex]
C. [tex]\frac{1}{4}[/tex]
D. [tex]\frac{1}{3}[/tex]



Answer :

To determine the probability of guessing the correct sequence on the first try, let's go through the problem step by step.

1. Identify the total number of colors and sequence length:
- The available colors are: red, yellow, blue, and purple.
- The sequence length is 3.

2. Calculate the total number of possible sequences:
Since each color can only be used once and the sequence matters (the order is crucial), this is a problem of permutations.

The number of permutations of [tex]\( n \)[/tex] items taken [tex]\( k \)[/tex] at a time is given by:
[tex]\[ P(n, k) = \frac{n!}{(n - k)!} \][/tex]

Here, [tex]\( n = 4 \)[/tex] (the number of colors) and [tex]\( k = 3 \)[/tex] (the length of the sequence).

We need to find [tex]\( P(4, 3) \)[/tex]:
[tex]\[ P(4, 3) = \frac{4!}{(4 - 3)!} = \frac{4!}{1!} = \frac{4 \times 3 \times 2 \times 1}{1} = 4 \times 3 \times 2 = 24 \][/tex]

3. Determine the probability:
The probability of guessing the sequence correctly on the first try is the reciprocal of the number of possible sequences.

[tex]\[ \text{Probability} = \frac{1}{P(4, 3)} = \frac{1}{24} \][/tex]

Thus, the probability that the sequence is guessed correctly on the first try is:
[tex]\[ \boxed{\frac{1}{24}} \][/tex]

This corresponds to the first choice provided in the given options.