I am a retired person who studies maths as a hobby.
This is the problem I am struggling with:
2 unbiased dice are thrown. Find the probability that the product of the scores is a) odd, b) a multiple of 3, c) a multiple of 12.
I know the total number of possibilities is $6 \times 6 = 36$
As regards a) an odd number x an odd number = an odd number
so the probability is $\frac{1}{2} \times\frac{1}{2} = \frac{1}{4}$
Now for b) I would have thought if one dice rolls a 3 or a 6 then any score on the other dice would give a product which is a multiple of 3. Since there is a $\frac{1}{3}$ chance of this happening on any one dice I would have thought the probability is $\frac{1}{3}$.
But the book says $\frac{5}{9}$
As regards c) I was able to work out this as $\frac{7}{36}$, but only by actually counting the possible outcomes. I don't know the method.
Best Answer
For second, the favorable outcomes are as follows -
If you take $3$ on first dice, $6$ favorable outcomes
If you take $6$ on first dice, again $6$ favorable outcomes
If you take $3$ on second dice, only $4$ additional outcomes as two were counted with first being $3$ or $6$.
Similarly for $6$ on second dice.
So total favorable outcomes $ = 6 + 6 + 4 + 4 = 20$.
But there is an easier way to do this. As we know, $3$ or $6$ on any dice would give product as multiple of $3$. So for product not to be multiple of $3$, we have to get from $(1, 2, 4, 5)$ on both dice. That probability is $\frac{4}{6} \times \frac{4}{6} = \frac{4}{9}$.
So probability that the product is multiple of $3 = 1 - \frac{4}{9} = \frac{5}{9}$.
For third as you rightly said, we count the favorable outcomes as below.
$12 = 3 \times 4$ or $2 \times 6$ so $4$ favorable outcomes - $(3,4), (4,3),(2,6),(6,2)$
$24 = 4 \times 6$ so $2$ favorable outcomes $(4,6), (6,4)$.
$36 = 6 \times 6$ so $1$ favorable outcome $(6,6)$.
That leads to $\frac{7}{36}$.