a bag of fruit contains 3 apples, 2 oranges, 1 banana, and 4 pears. Gerald will randomly select 2 fruit at a time from the bag and not put them back. what is the probability that the first piece of fruit jared selects will be a banana and the second piece of fruit will be of fruit be a pear
CAN SOMEONE PLEASE GUIDE ME THROUGH THIS IK THE ANSWER BUT IDK THE STEPS TO TAKE I NEED HELP ASAP PLEASE! THANK YOU!
The answer is 2/45. :)
Well, Jared seems to be quite the fruit enthusiast! Let's calculate the probability for him.
Step 1: Find the total number of fruit combinations.
Jared will be selecting 2 fruit at a time, so we need to calculate the total number of 2-fruit combinations from the given bag. We can use the formula for combinations:
nCr = n! / (r!(n-r)!)
In this case, n (the total number of fruits in the bag) is 10, and we are selecting r (the number of fruits Jared selects at a time) which is 2.
So, the total number of 2-fruit combinations is:
10C2 = 10! / (2!(10-2)!), which simplifies to 10*9 / (2*1) = 45.
Step 2: Find the number of combinations that have a banana as the first fruit.
Since there is only 1 banana in the bag, we need to select 1 banana and 1 other fruit from the remaining 9 fruits. So, the number of combinations with a banana as the first fruit is:
1C1 * 9C1 = 1 * 9 = 9.
Step 3: Find the number of combinations with a pear as the second fruit.
After selecting the banana as the first fruit, there are 4 pears remaining in the bag. We need to select 1 of those pears along with the remaining fruit (total remaining fruits = 8). So, the number of combinations with a pear as the second fruit is:
4C1 * 8C1 = 4 * 8 = 32.
Step 4: Find the probability.
The probability is calculated by dividing the favorable outcomes (number of combinations with a banana as the first fruit and a pear as the second fruit) by the total number of outcomes (total number of 2-fruit combinations).
Probability = Favorable outcomes / Total outcomes
= (9 * 32) / 45
= 288 / 45
≈ 6.4
So, the probability that Jared selects a banana as his first fruit and a pear as his second fruit is approximately 6.4%.
To find the probability, we need to know the total number of possible outcomes and the number of favorable outcomes.
Step 1: Total number of possible outcomes
When Gerald selects the first fruit, there are 3 apples, 2 oranges, 1 banana, and 4 pears in the bag. So, the total number of fruits is 3 + 2 + 1 + 4 = 10.
Step 2: Number of favorable outcomes
Since we want the first fruit to be a banana, and there is only 1 banana in the bag, there is one favorable outcome for the first fruit.
After selecting the banana in the first round, there will be 2 apples, 2 oranges, and 4 pears left in the bag. So, the total number of fruits for the second round is 2 + 2 + 4 = 8. Since we want the second fruit to be a pear, and there are 4 pears in the bag, there will be 4 favorable outcomes for the second fruit.
Step 3: Multiply the probabilities
To find the probability of both events happening, we multiply the probability of the first event (selecting a banana) by the probability of the second event (selecting a pear).
Probability of selecting a banana: 1/10
Probability of selecting a pear: 4/8 (after selecting the banana)
Final Step: Calculate the overall probability
To find the overall probability, we multiply the probabilities of the two events together:
Probability = (1/10) * (4/8)
Simplifying further:
Probability = 1/10 * 1/2
Probability = 1/20
So, the probability that Jared selects a banana first and a pear second is 1/20.
There are 10 fruits.
P(banana) = 1/10
since there is only one banana. After the 1st pick, there are only 9 fruits left.
Now it should be clear what P(pear) is.
P(both) = P(banana) * P(pear)