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Is the answer 1/9?
“If each number is odd”
So the total number of possibilities to get odd on both the dice is 3 (1,3, and 5 on first dice) x3 (1,3, and 5 on second dice)=9
“Sum is 10”
Only one possibility (5 and 5 on both)
So \frac{events-in-favor}{total-number-of-events} = \frac{1}{9}
This is what I got. Is the answer 1/36?
The given answer is 1/36
he solved as if the “if” is replaced by when.
Aah I did the same.
Maybe what they’re mentioning is, out of all the possibilities that give a result of 10, consider only the ones in which both are odd. But the general event is still getting any number (even ir odd) when rolling the dice.