Question 1139181: Jane has five pieces of candy and will eat one piece each of the next five days. Three pieces are wrapped in blue foil and two pieces are wrapped in yellow foil. What is the probability that she will eat a piece of candy wrapped in yellow foil on the 5th day?
Answer by ikleyn(52776)   (Show Source):
You can put this solution on YOUR website! .
To make my solution more understandable, I will call two "yellow" candies A and B.
It is not important, which one of two is "A" and which is "B".
The only important thing is that they have different names.
Let us place the candy "A" for Friday. Then there are 4! = 1*2*3*4 = 24 permutations for the remaining 4 candies.
OK. Now, let us place the candy "B" for Friday. Then there are 4! = 1*2*3*4 = 24 permutations for the remaining 4 candies.
Thus, there are 24 + 24 = 48 different permutations, having "yellow" candy for Friday.
So, the probability under the question is  =  = 0.4 that Jane will eat "yellow" candy on Friday.
Solved.
-------------
O-o-o-p-s !
In my solution, I forgot to explain you that 120 = 5! = 1*2*3*4*5, but I am sure that you know it even without my explanations.
/\/\/\/\/\/\/\/
On Permutations, see introductory lessons
- Introduction to Permutations
- PROOF of the formula on the number of Permutations
- Problems on Permutations
- OVERVIEW of lessons on Permutations and Combinations
in this site.
Also, you have this free of charge online textbook in ALGEBRA-II in this site
- ALGEBRA-II - YOUR ONLINE TEXTBOOK.
The referred lessons are the part of this online textbook under the topic "Combinatorics: Combinations and permutations".
Save the link to this textbook together with its description
Free of charge online textbook in ALGEBRA-II
https://www.algebra.com/algebra/homework/complex/ALGEBRA-II-YOUR-ONLINE-TEXTBOOK.lesson
into your archive and use when it is needed.
|
|
|
