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 that the sequence is guessed correctly on the first try, let's break down the problem step by step.

1. Identify the number of available colors:
There are 4 different colors available: red, yellow, blue, and purple.

2. Determine the length of the sequence:
The sequence to be guessed consists of 3 colors.

3. Understand the constraints:
Each color can only be used once in the sequence, meaning there are no repetitions in a given sequence.

4. Calculate the total number of possible sequences:
To compute this, we need to consider the permutations of 4 colors taken 3 at a time. The formula for permutations of [tex]\( n \)[/tex] items taken [tex]\( r \)[/tex] at a time is given by:
[tex]\[ P(n, r) = \frac{n!}{(n-r)!} \][/tex]
Plugging in our values [tex]\( n = 4 \)[/tex] and [tex]\( r = 3 \)[/tex]:
[tex]\[ P(4, 3) = \frac{4!}{(4-3)!} = \frac{4!}{1!} = 4 \cdot 3 \cdot 2 \cdot 1 = 24 \][/tex]
Therefore, there are 24 different possible sequences of three colors.

5. Calculate the probability of guessing the correct sequence on the first try:
The probability of guessing the correct sequence is the reciprocal of the total number of possible sequences since there is only one correct sequence:
[tex]\[ \text{Probability} = \frac{1}{24} \][/tex]

Therefore, the probability that the sequence is guessed correctly on the first try is:

[tex]\(\boxed{\frac{1}{24}}\)[/tex]