SOLUTION: Two employees will be chosen at random from a group of five to go to a conference. What is the probability that employee 3 and employee 5 will be chosen? 5C2=5*4*3*2*!/(3*2*1)(

SOLUTION: Two employees will be chosen at random from a group of five to go to a conference. What is the probability that employee 3 and employee 5 will be chosen? 5C2=5432!/(321)(

Question 1147982: Two employees will be chosen at random from a group of five to go to a conference. What is the probability that employee 3 and employee 5 will be chosen?
5C2=5*4*3*2*!/(3*2*1)(2*1)=20 So 20 total combinations but then I didn't know what to do. I tried various ways but could not come up with one of the multiple choice answers

Answer by ikleyn(52814) (Show Source): You can put this solution on YOUR website!
.


In all, there are   =  = 10 possible combinations (= unordered pairs) of 5 employees taken 2 at a time.


    (It is your space of events in this problem).


Of them, only one unordered pair (3,5) is favorable.


Thus the probability under the question is  .    ANSWER

Solved, answered and explained.

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To develop your knowledge on the subject, look into the relevant lessons on Combinations
    - Introduction to Combinations
    - PROOF of the formula on the number of Combinations
    - Problems on Combinations

and into the relevant lesson on Probability
    - Elementary Probability problems related to 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 topics
"Combinatorics: Combinations and permutations" and "Solved problems on Probability".

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.

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