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 ...
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 grape or a banana?
Answer Details
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.