Sammy and Sally each carry a bag containing a banana, a chocolate bar, and a licorice stick. Simultaneously, they take out a single food item and consume it. The possible pairs of food items that Sally and Sammy consumed are as follows.

chocolate bar - chocolate bar
licorice stick - chocolate bar
banana - banana
chocolate bar - licorice stick
licorice stick - licorice stick
chocolate bar – banana
banana - licorice stick
licorice stick - banana
banana - chocolate bar
Find the probability that no chocolate bar was eaten.

You have 9 possibilities. Divide that into the number of possibilities without chocolate.

I came up with 4/9 and it was wrong. A little more help please!

To find the probability that no chocolate bar was eaten, we need to calculate the number of favorable outcomes (pairs of food items where no chocolate bar was eaten) and divide it by the total number of possible outcomes.

First, let's count the number of favorable outcomes. From the given pairs, the favorable outcomes are:

1. licorice stick - licorice stick
2. banana - banana
3. licorice stick - banana
4. banana - licorice stick

So, there are 4 favorable outcomes.

Now, let's count the total number of possible outcomes. Each person has 3 options (banana, chocolate bar, licorice stick), so there are 3 options for Sammy and 3 options for Sally, resulting in a total of 3 x 3 = 9 possible outcomes.

Therefore, the probability that no chocolate bar was eaten is 4/9.

To find the probability, you can divide the number of favorable outcomes by the total number of possible outcomes. So in this case, the probability would be 4/9.