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 :

Let's solve the problem step by step.

1. Determining Total Number of Sequences:
- The child can choose from 4 different colors: red, yellow, blue, and purple.
- The sequence is made up of 3 positions.
- Since once a color is used, it cannot be repeated, we are dealing with permutations of 4 colors taken 3 at a time.
- The number of possible sequences is calculated as [tex]\(P(4,3)\)[/tex], which is the permutations of 4 items taken 3 at a time.
- The formula for permutations is given by [tex]\[ P(n, r) = \frac{n!}{(n - r)!} \][/tex]
- Plugging in the values: [tex]\( n = 4 \)[/tex] and [tex]\( r = 3 \)[/tex], we get:
[tex]\[ P(4, 3) = \frac{4!}{(4 - 3)!} = \frac{4!}{1!} = \frac{24}{1} = 24 \][/tex]

2. Calculating the Probability of Guessing the Correct Sequence on the First Try:
- There is only 1 correct sequence out of these 24 possible sequences.
- The probability [tex]\( P \)[/tex] of guessing the correct sequence on the first try is therefore:
[tex]\[ P = \frac{1}{24} \][/tex]

Thus, the probability that the sequence is guessed on the first try is [tex]\( \frac{1}{24} \)[/tex].

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