+937 Three darts are thrown at the figure given, each landing in a different square. What is the probability that the squares they land in form a row, either horizontally, vertically, or diagonally? Express your answer as a common fraction
+1252 First of all, how many rows are possible? We know there are 3 vertical, 3 horizantal, and two diagonal and so there are \(3+3+2=8\) total rows.
Now, how many combinations are possible? We know any one of the 9 spaces can be hit the first time. Any one of the 9 spaces can be hit the second time besides the first one, so there are 8 total spaces. Any one of the 9 spaces can be hit the third time besides the first or second one, so there are 7 total spaces. To find the total combinations we calculate \(9\times 8 \times 7=504\).
The probability is \(\frac{8}{504}=\frac{1}{63}\).
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+6251 This isn't quite correct.
You've treated the groups of three as permutations but the rows, columns and diagonals as combinations.
You need to be consistent in how you treat them.
If we went w/permuations we have 8 x 3! =48 different permuations for the rows, columns, diagonals
Then the probability becomes p = (48)/(504) = 2/21
If we use combinations we have 8 rows, columns, diagonals but then we have
9C3 = 84 possible arrangements of 3 darts on the board.
This gets us again a probability of p = 8/84 = 2/21
+129852 Number the squares as
1 2 3
4 5 6
7 8 9
We have the following possible sets of three different numbers [ assuming that each dart lands in a different square ] =
C (9 , 3) = 84
But.....there are only 8 different sets hat will form vertical, horizontal or diagonal patterns
So....the probability is 8 / 84 = 2 / 21
