Question

Please help me!!
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. 1/24
B. 1/8
C. 1/4
D. 1/3

Answer 1

I think it would be 1/3 since there are 3 colors and you cannot use another one. That is just what I think.

Answer 2

how many permutations can you make with 3 distinct colors ?

Answer 3

What do you mean @phi?

Answer 4

make that 4 distinct colors

Answer 5

I assume you are studying permutations

you have 4 choices for the first color
3 for the 2nd
2 for the 3rd
1 for the last

4*3*2*1 (or 4!) different permutations
and you have 1/(4!) chance of picking the correct one

Answer 6

Yes I am

Answer 7

So it would be 1/4?

Answer 8

1 over 4 factorial
4! is 4*3*2*1

Answer 9

Okay, so for that i wouldn't have to multiply right?

Answer 10

I would first figure out what 4! is
then write 1/that number

Answer 11

So it will be 1/24

Answer 12

yes.
there are 24 permutations you could make
and the chance of picking the correct one is 1/24

Answer 13

Thank you so much :)

Answer 14

ok after going through all of that I re-read the question:

One child makes a sequence of three colors from red, yellow, blue, and purple.

so the number of permutations is 4*3*2 = 24
and 1/24 is the correct answer

Answer 15

Thank you I got it :)

Answer 16

I appreciate for explaining step by step and taking your time :)