When two fair 6-sided dice are tossed, the numbers a and b are obtained. What is the probability that the two digit number ab (where a and b are digits) and a and b are all divisible by 3?
Im not sure if 1/9 is right...
Can someone give me an answer and explain?
THANK YOU IF YOU CAN!
Make a 6 x 6 grid. In each square put the possible combination. Count the ones that satisfy the conditions.
It'll look something like the one below
1,1 1,2 1,3 1,4 1,5 1,6
2,1 2,2 2,3 2,4 2,5 2,6
3,1 3,2 3,3 3,4 3,5 3,6
4,1 4,2 4,3 4,4 4,5 4,6
5,1 5,2 5,3 5,4 5,5 5,6
6,1 6,2 6,3 6,4 6,5 6,6
You see in the first two rows that none of the pairs satisfies the requirement that a be divisible by 3
In the third row, 3,3 and 3,6 work. In the sixth row, 6,3 and 6,6 work.
That's four of them. So, 4 out of 36 is 4/36 and reduces to 1/9. You were right. Good work!
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