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)(
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-> 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)(
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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
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
The referred lessons are the part of this online textbook under the topics
"Combinatorics: Combinations and permutations" and "Solved problems on Probability".
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https://www.algebra.com/algebra/homework/complex/ALGEBRA-II-YOUR-ONLINE-TEXTBOOK.lesson
