**Problem:**
A bag of fruits contains 3 mangoes, 5 bananas, 2 oranges, and 6 guavas. A fruit is chosen at random from the bag, and then replaced. A second fruit is chosen. What is the probability of choosing a banana and a guava?
**Solution:**
The probability of choosing a banana and a guava, with replacement, is the product of the probability of choosing a banana and the probability of choosing a guava.
The probability of choosing a banana is 5 / 16, since there are 5 bananas in the bag and 16 total fruits.
The probability of choosing a guava is 6 / 16, since there are 6 guavas in the bag and 16 total fruits.
Therefore, the probability of choosing a banana and a guava, with replacement, is (5 / 16) * (6 / 16) = **15 / 128**.
In other words, there is a 15 / 128 chance of choosing a banana and a guava, with replacement, if you randomly choose two fruits from the bag.
**Explanation:**
The probability of event A happening, then event B, is the probability of event A happening times the probability of event B happening given that event A already happened. In this case, event A is choosing a banana and event B is choosing a guava.
The probability of choosing a banana is 5 / 16, since there are 5 bananas in the bag and 16 total fruits. The probability of choosing a guava is also 6 / 16, since there are 6 guavas in the bag and 16 total fruits.
However, since the fruit is replaced after it is chosen, the probability of choosing a guava is not affected by the fact that a banana was chosen first. Therefore, the probability of choosing a guava given that a banana was already chosen is still 6 / 16.
So, the probability of choosing a banana and a guava, with replacement, is (5 / 16) * (6 / 16) = **15 / 128**.
I hope this helps! Let me know if you have any other questions.