a basket contains the following pieces of fruit: 3 apples, 2 oranges, 2 bananas, 2 pears, and 5 peaches. Jameson picks a fruit at random and does not replace it. Then Brittany pricks a fruit at random. What is the probability that Jameson gets a banana and Brittany gets a pear?

A. 4/27
B. 1/49
C. 2/91
D. 27/91

I think A?

Wait what?

14 total fruits, so

P = 2/14 * 2/13

2/14x2/13=4/182=2/91

2 2

14 13
1 2
7

and how does this help?

Are you right?

I cannot verify if the answers provided are completely accurate. It is always best to double-check your work and answers to ensure accuracy.

To determine the probability that Jameson picks a banana and Brittany picks a pear, we need to calculate the individual probabilities and then multiply them together.

Step 1: Calculate the probability of Jameson picking a banana:
- There are a total of 3 + 2 + 2 + 2 + 5 = 14 fruits in the basket.
- Jameson has 2 bananas to choose from out of these 14 fruits.
- Therefore, the probability of Jameson picking a banana is 2/14.

Step 2: Calculate the probability of Brittany picking a pear:
- After Jameson picks a fruit, there are now 13 fruits remaining in the basket.
- Out of these 13 fruits, there are 2 pears left.
- Therefore, the probability of Brittany picking a pear is 2/13.

Step 3: Multiply the probabilities together:
- The probability of Jameson picking a banana is 2/14.
- The probability of Brittany picking a pear, given that Jameson has already picked a fruit, is 2/13.
- So, the probability of both events happening together is (2/14) * (2/13).

Now, let's simplify this product:
(2/14) * (2/13) = 4/182 = 2/91

Therefore, the correct answer is option C. The probability that Jameson gets a banana and Brittany gets a pear is 2/91.

1. A

2. C
3. B
4. B
5. C
6. B
7. B
8. C
9. D
10. B
11. C
12. C
13. D
14. A
15. A
16. C
17. A
18. C
i got 12/18 searching the answers one by one.
these should get you 100%