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Is there direct correspondence between the rule of multiplication of combinations and probability multiplication in case of independent events?

For combinations, whichever cup I choose, I can chose any spoon, which makes me cups x spoon choices. If spoon choice would be affected by the cup, the relationship would be more complex. I guess that dependencies work the same in case of probability computations and you have the multiplication in the simplest, independent case. Is it right? How does it come out that multiplication comes out in both cases, despite, unlike combination multiplication, probability multiplication reduces the number of ways.

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Yes, there is a cleare connection in elementary probability. It is easy to see in small basic examples. You throw two dice, one red and one blue. What is the probability of getting an even number eyes on both dice?

We can use the usual $\frac{\text{# favourable}}{\text{# possible}}$. Let $R$ be the number eyes on the red dice and $B$ the number eyes on the blue dice. Number of possible eyes on each dice is 6, and number possible is 3. Since possibilities for each dice can be combined freely, the total number of possible outcomes is $6 \times 6=36$ and total number of favourable is $3 \times 3 = 9$. So by our basic definition of probability, the probability of even number on both dice is $\frac{9}{36}=\frac{3}{6}\cdot\frac{3}{6}$, which is what we also get from using multiplication rule for independent events.

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