A3. Ronald throws with three dice. They look just like ordinary dice, only different numbers are on the six faces.

On the first die, the numbers are: $1,1,2,2,3$ and 3.

On the second die, the numbers are: $2,2,4,4,6$ and 6.

And on the third die, the numbers are: 1, 1, 3, 3, 5 and 5.

He adds up the three numbers he throws.

What is the probability that the result is an odd number?
(A) $\frac{1}{4}$
(B) $\frac{1}{3}$
(C) $\frac{1}{2}$
(D) $\frac{2}{3}$
(E) $\frac{3}{4}$
A3. Ronald throws with three dice. They look just like ordinary dice, only different numbers are on the six faces. On the first die, the numbers are: 1, 1, 2, 2, 3, and 3. On the second die, the numbers are: 2, 2, 4, 4, 6, and 6. And on the third die, the numbers are: 1, 1, 3, 3, 5, and 5. He adds up the three numbers he throws. What is the probability that the result is an odd number? (B) $\frac{1}{3}$

Solution The second die always gives an even number and the third die always gives an odd number. The question is thus actually how great the probability is that the first die lands on an even number. This happens in 2 out of the 6 cases, so with a probability of $\frac{2}{6}=\frac{1}{3}$. $\square$