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?

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): You can put this solution on YOUR website!

There are 6 sides on each dice, giving 6*6 = 36 outcomes.

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 summary so far is that there are 20 ways to have at least one of the dice be 5 or 6.

This is out of 36 ways to roll two dice
20/36 = (4*5)/(4*9) = 5/9

Answer: 5/9

Answer by ikleyn(52778) (Show Source): You can put this solution on YOUR website!
.

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

Solved.

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