SOLUTION: A pair of 6 sided dice are tossed. What is the probability that at least one of the dice has a value greater than or equal to 5?
Algebra ->
Permutations
-> SOLUTION: A pair of 6 sided dice are tossed. What is the probability that at least one of the dice has a value greater than or equal to 5?
Log On
Question 1197565: A pair of 6 sided dice are tossed. What is the probability that at least one of the dice has a value greater than or equal to 5? Found 2 solutions by math_tutor2020, ikleyn:Answer by math_tutor2020(3816) (Show Source):
Let's call the dice A and B
Let's have die A roll to 4 or smaller. Same goes for die B.
There are 4*4 = 16 outcomes where both dice are 4 or smaller.
That must mean 36-16 = 20 outcomes are when either
A = 5 or 6
OR
B = 5 or 6
or both scenarios happen
We can confirm this with a chart
1
2
3
4
5
6
1
X
X
2
X
X
3
X
X
4
X
X
5
X
X
X
X
X
X
6
X
X
X
X
X
X
The 'X's represent the situation when either dice is 5 or larger (i.e. 5 or 6)
There are 6 X's in row 5 and 6 X's in row 6
That's 6+6 = 12 X's total so far.
We have 6 X's in column 5 and 6 X's in column 6
That's another 6+6 = 12 X's
We're up to 12+12 = 24
But in the bottom right hand corner, we're double counting those four X's since they are in the overlapped region of the four columns mentioned.
We need to subtract off 4 to correct for this erroneous double counting.
24-4 = 20
Or note that the blank portion of the table is 4 rows tall and 4 columns across.
This represents the 4*4 = 16 cases where both dice are 4 or smaller, as mentioned earlier.
This then gives us the 36-16 = 20 copies of X in total.
The opposite event is when both dice values are from 1 to 4.
There are 4*4 = 16 such "unlucky" cases.
The rest 36-16 = 20 are "lucky" cases.
The probability to have a lucky case is = . ANSWER
(