There are 4 rows and 3 columns in a table, and each slot is painted with black or white with equal p

Jorge Lawson

Jorge Lawson

Answered question

2022-05-28

There are 4 rows and 3 columns in a table, and each slot is painted with black or white with equal probability, and I wish to find the probability that one, and only one of the rows is painted black. What I did was:
P = 4 ( 1 / 2 ) 3 ( 7 / 8 ) 3 = 343 1024
and it indeed seems to be the correct answer. I thought I should also try solving it using combinatorics, but that is something I fail to do. My try is:
4 1 ( 2 9 7 ) 2 12 = 505 1024 343 1024
what have I done wrong?

Answer & Explanation

Carter Escobar

Carter Escobar

Beginner2022-05-29Added 9 answers

Please note that ( 2 9 7 )   does not count where each of the remaining three rows cannot be painted all black.
It should rather be ( 2 3 1 ) 3   permutations for the remaining three rows. Each row can be painted in 2 3 ways but one of those is where all columns are black and we subtract that.
So the answer should be,
  4 1 ( 2 3 1 ) 3 2 12  

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