Final answer:
To calculate the probabilities for the two conditions, combinations formula is used to count the possible ways balls can be drawn from the bag. Different colors are calculated by pairing the quantities of each color, while at least one blue involves using complement rule.
Step-by-step explanation:
Let's solve each part of the question step by step:
Part I: Probability of picking two balls of different colors
The total number of ways to pick two balls from the bag is calculated by the combination formula C(n, k) = n! / (k!(n - k)!), where n is the total number of balls, and k is the number of balls we want to choose. With 25 balls in total (10 red + 8 green + 7 blue), the number of ways to pick any two balls is C(25, 2).
To find the combinations where the balls are of different colors, we calculate the sum of all the possible combinations of picking a red and a green, a red and a blue, and a green and a blue:
- Picking a red and green: C(10, 1) * C(8, 1)
- Picking a red and blue: C(10, 1) * C(7, 1)
- Picking a green and blue: C(8, 1) * C(7, 1)
The probability is then the sum of these combinations divided by the total number of ways to pick any two balls.
Part II: Probability of picking at least a blue ball
Picking at least one blue ball includes scenarios where either the first or second ball is blue, or both are blue. To find this probability, we can use the complement rule: P(at least one blue) = 1 - P(no blue).
We can calculate P(no blue) by considering the combinations to pick two balls that are not blue, which would be from the red and green balls:
- Picking both balls that are not blue: C(18, 2) where 18 is the sum of red and green balls (10 + 8).
Subtracting this from 1 gives us the required probability for at least one blue ball.