Answer:
7/12.
Step-by-step explanation:
Let's go over the details.
Box A contains 3, 5, and 6, and 9. Box B contains 1, 4, and 7.
Only 1 card is picked from each box.
For the sake of simplicity, let's take a step back. Let's assume that the actual values of the cards only indicate whether they are odd or even. So, we can replace the values with O for odd and E for even.
Box A now contains O(3), O(5), E(6), O(9), and Box B now contains O(1), E(4), O(7).
From logic, we can deduce that O + O = E, E + E = E, and E + O = O.
We're only looking at the combinations that result in an even sum. So we're specifically looking for OO and EE combinations.
Now that we know what we're looking for, we can go back to original problem.
There are a total of 12 possibilities if you use the values of the cards. Therefore, there is a [tex]\frac{1}{12}[/tex] chance that you will get a specific combination.
To find the probability of getting a specific combination, you need to multiply the number of cards from Box A with the number of cards in Box B. So the possible combinations would be:
(3,1), (5,1), (6,1), (9,1), (3,4), (5,4), (6,4), (9,4), (3,7), (5,7), (6,7), (9,7)
But since we're looking only for (O,O) and (E,E) combinations, we can ignore all of the (O,E) and (E,O) combinations.
(3,1), (5,1), (9,1), (6,4), (3,7), (5,7), (9,7)
These account for 7 of the 12 possible combinations.
So the probability of getting two numbers whose sum is even, or (to put it in simpler terms) the probability of getting an OO or EE combination, is [tex]\frac{7}{12}[/tex].
