A box contains 8 blue balls and 2 red balls. Three balls are selected from the box at random without replacement. Find the probability that two balls are blue and one ball is red.
7/15
To find the probability, we need to determine the number of favorable outcomes (selecting two blue balls and one red ball) and the total number of possible outcomes (selecting any three balls from the box).
First, let's calculate the number of favorable outcomes:
We need to choose two blue balls from the eight available blue balls, which can be represented as C(8, 2) or "8 choose 2". This is calculated using the combination formula: C(n, r) = n! / (r! * (n-r)!).
C(8, 2) = 8! / (2! * (8-2)!) = 8! / (2! * 6!) = (8 * 7) / (2 * 1) = 28.
Similarly, we need to choose one red ball from the two available red balls: C(2, 1) = 2! / (1! * (2-1)!) = (2 * 1) / (1 * 1) = 2.
Since we need to select two blue balls and one red ball, we multiply the number of favorable outcomes:
Number of favorable outcomes = C(8, 2) * C(2, 1) = 28 * 2 = 56.
Now, let's calculate the number of total possible outcomes:
To select any three balls from the ten balls in the box, we need to calculate C(10, 3):
C(10, 3) = 10! / (3! * (10-3)!) = 10! / (3! * 7!) = (10 * 9 * 8) / (3 * 2 * 1) = 120.
Finally, we can calculate the probability by dividing the number of favorable outcomes by the number of total possible outcomes:
Probability = Number of favorable outcomes / Number of total possible outcomes = 56 / 120 = 7 / 15.
Therefore, the probability that two balls are blue and one ball is red is 7/15 or approximately 0.467.