A box contains 6 red balls, 4 blue balls, and 2 green balls. Three balls are randomly drawn (without replacement) from the box. What is the probability that the three balls selected are all the same color?
im lost
A bag contains 5 yellow balls, 5 orange balls, 6 white balls and 6 red balls. Five balls are drawn and NOT replaced each time
To find the probability that the three balls selected are all the same color, we first need to find the total number of ways to select three balls from the box. Then, we need to find the number of ways to select three balls of the same color.
The total number of ways to select three balls from the box can be calculated using combinations. Since we are selecting without replacement, the total number of ways to select three balls is given by the combination formula:
nCr = n! / (r! * (n-r)!)
where n is the total number of balls in the box, and r is the number of balls we want to select (in our case, 3).
In this case, n = 6 (number of red balls) + 4 (number of blue balls) + 2 (number of green balls) = 12. So, n = 12.
The number of ways to select three balls of the same color can be found by adding the combinations for selecting three red balls, three blue balls, and three green balls.
Number of ways to select three red balls (nCr):
6C3 = 6! / (3! * (6-3)!) = 6! / (3! * 3!) = (6 * 5 * 4) / (3 * 2 * 1) = 20
Similarly, the number of ways to select three blue balls (nCr) is:
4C3 = 4! / (3! * (4-3)!) = 4! / (3! * 1!) = (4 * 3 * 2) / (3 * 2 * 1) = 4
The number of ways to select three green balls (nCr) is:
2C3 = 2! / (3! * (2-3)!) = 2! / (3! * (-1)!) = 0
Adding the combinations for each color gives us:
Number of ways to select three balls of the same color = 20 + 4 + 0 = 24
Therefore, the probability of selecting three balls that are all the same color is:
Probability = Number of ways to select three balls of the same color / Total number of ways to select three balls
Probability = 24 / nCr(12, 3)
Now, we can calculate the probability.
follow the basic methodology that Drwls showed you in the previous post. Assume each color ball can be distinguished from each other. (e.g., name the green balls g1 and g2) First, the denominator. How many different ways can 3 balls from 12 be chosen. 12-choose-3 is 12!/3!(12-3)!
Now the numerator. How many different ways can 3 red balls be chosen from 6? Plus how many different ways can 3 blue balls be chosen from 4.
Take it from here.