Respuesta :
Answer:
Step-by-step explanation:
To find the number of balls of each color using the given probabilities, we can use the formula for inverse probability or the formula for combining.
Inverse probability formula:
Number of objects with a specific characteristic = (total number of objects) x (probability of that characteristic) (assuming all characteristics are independent) In our case, let's assume that each ball has an equal chance of being picked, so all probabilities are independent. Therefore, we can use this formula to find the number of balls of each color as follows: Red balls: Number of red balls = (total number of balls) x (probability of picking a red ball) Red balls = 10 x 0.3 = 3 red balls
Green balls: Number of green balls = (total number of balls) x (probability of picking a green ball) Green balls = 10 x 0.2 = 2 green balls
Purple balls: Number of purple balls = (total number of balls) x (probability of picking a purple ball) Purple balls = 10 x 0.2 = 2 purple balls
Blue balls: Number of blue balls = (total number of balls) x (probability of picking a blue ball) Blue balls = 10 x 0.1 = 1 blue ball
Orange balls: Number of orange balls = (total number of balls) x (probability of picking an orange ball) Orange balls = 10 x 0.1 = 1 orange ball
Formula for combining:
Number of ways to choose r objects from a set of n distinct objects is given by: nCr = n! / [r! (n-r)!]
Let's say we want to find the number of ways to choose 3 balls from the bag, with 1 red, 1 green, and 1 purple ball. Using the formula for combining, we get:
Number of ways to choose 3 balls with 1 red, 1 green, and 1 purple ball = 10Cr3 = 10! / [3! (10-3)!] = 120 ways
So, there are 120 ways to choose 3 balls with the given color distribution.
1E: The formula for finding the sum of the first n even numbers is: sum = n(2a) = 2n(a)
1F: The formula for finding the sum of the first n odd numbers is: sum = n(2a+1) = 2n(a+1/2)
In both cases, a is the first term of the sequence.
I hope this helps! Let me know if you have any further questions.