
Kenneth S. answered 06/21/18
Tutor
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Expert Help in Algebra/Trig/(Pre)calculus to Guarantee Success in 2018
A B C D Let's tabulate possible distinct 'distributions'
5 0 0 0 ... one possibility
4 1 0 0 ... three possibilities
0 1 0
0 0 1
3 0 0 2 ... six possibilities
0 2 0
2 0 0
1 1 0
1 0 1
0 1 1
2 3 0 0 ... ten possibilities
0 3 0
0 0 3
2 0 1
2 1 0
1 0 2
1 2 0
1 1 1
0 1 2
0 2 1
I will let YOU tabulate all the possibilities where 1 goes into the A column. I predict there will be 15 of those!
Why? Because it appears that these counts are the TRIANGULAR numbers.
Do not fail to take note that I am leaving to you the final steps of tabulating/demonstrating all the possibilities that begin with 1 and 0 in the A position. Or better yet, discover a shortcut to computing that.
Ultimately, you'll have P(of the one event of interest) = 1/n where n is the total number of distinct distribution possibilities.