. 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 picks a fruit a random. What is the probability that Jameson gets a banana and Brittany gets a pear?
91
First, we need to find the probability of Jameson picking a banana. There are a total of 14 fruits in the basket, so the probability is:
P(Jameson gets a banana) = 2/14 = 1/7
Since Jameson does not replace the fruit, there are now 13 fruits left in the basket. We need to find the probability of Brittany picking a pear out of the remaining fruits. There are only 2 pears left, so the probability is:
P(Brittany gets a pear) = 2/13
To find the probability of both events happening, we multiply the individual probabilities:
P(Jameson gets a banana and Brittany gets a pear) = (1/7) x (2/13) = 2/91
Therefore, the probability of Jameson getting a banana and Brittany getting a pear is 2/91.
To find the probability that Jameson gets a banana and Brittany gets a pear, we need to consider the total number of possible outcomes and the number of favorable outcomes.
Total number of fruits in the basket: 3 apples + 2 oranges + 2 bananas + 2 pears + 5 peaches = 14 fruits
To calculate the probability for Jameson, we need to determine the total number of options he has to choose from and the number of bananas in the basket.
Total number of options for Jameson: 14 fruits
Number of bananas in the basket: 2 bananas
Therefore, the probability that Jameson gets a banana is 2/14.
After Jameson picks a fruit, there are now 13 fruits left in the basket (since Jameson does not replace the fruit). To calculate the probability for Brittany, we need to determine the total number of options she has to choose from and the number of pears remaining in the basket.
Total number of options for Brittany: 13 fruits
Number of pears in the basket: 2 pears
Therefore, the probability that Brittany gets a pear is 2/13.
To find the probability of both events happening (Jameson getting a banana and Brittany getting a pear), we need to multiply their individual probabilities:
Probability of Jameson getting a banana * Probability of Brittany getting a pear = (2/14) * (2/13) = 4/182 = 2/91
So, the probability that Jameson gets a banana and Brittany gets a pear is 2/91.
To find the probability that Jameson gets a banana and Brittany gets a pear, we need to first determine the total number of possible outcomes and the number of favorable outcomes.
Total number of possible outcomes:
Since Jameson picks a fruit at random and does not replace it, after Jameson makes his selection, there will be one less fruit in the basket. Therefore, the total number of possible outcomes for Brittany's selection will be the number of remaining fruits in the basket.
Total number of favorable outcomes:
Jameson needs to select one out of the 2 bananas in the basket, so the number of possible outcomes favorable for Jameson is 2.
After Jameson's selection, Brittany needs to select one out of the 2 remaining pears in the basket, so the number of possible outcomes favorable for Brittany is 2.
So, the total number of favorable outcomes is 2 * 2 = 4.
To calculate the probability, you divide the number of favorable outcomes by the total number of possible outcomes:
Probability = favorable outcomes / total outcomes
= 4 / (total number of remaining fruits in the basket)
Now, let's determine the total number of remaining fruits in the basket. Initially, there were 3 apples, 2 oranges, 2 bananas, 2 pears, and 5 peaches. Since Jameson selected a banana, there is one less banana in the basket. Therefore, the remaining fruits in the basket are: 3 apples, 2 oranges, 1 banana, 2 pears, and 5 peaches.
The total number of remaining fruits in the basket is 3 + 2 + 1 + 2 + 5 = 13.
Now we can calculate the probability:
Probability = 4 / 13 ≈ 0.3077
So, the probability that Jameson gets a banana and Brittany gets a pear is approximately 0.3077 or 30.77%.