In a basket of fruits, there are 6 grapes, 11 bananas, and 13 oranges. If one fruit is chosen at random, what is the probability that the fruit is either a ...

Probability is calculated by dividing the number of favorable outcomes by the total number of possible outcomes.

First, calculate the total number of fruits in the basket:

• Number of grapes = 6
• Number of bananas = 11
• Number of oranges = 13

Total fruits = 6 + 11 + 13 = 30

Next, find the number of ways to choose a fruit that is either a grape or a banana:

• Number of grapes = 6
• Number of bananas = 11

So, there are 6 + 11 = 17 fruits that meet the condition (either grape or banana).

The probability of choosing either a grape or a banana is:

\[ \text{Probability} = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{17}{30} \]

This is correct because:

• There is no overlap between the kinds (a fruit cannot be both a grape and a banana!), so we add the numbers directly.
• We count only the grapes and bananas, since the question asks for "either a grape or a banana".

In summary, the probability that the fruit is either a grape or a banana is \(\frac{17}{30}\) because 17 out of the 30 fruits are either grapes or bananas.