A bag contains z+2 black balls, z+5 red balls and z+8 white balls. The probability of getting a black ball is ¼, what is the probability of getting three balls of different colour, when three balls are drawn?
Question
A bag contains z+2 black balls, z+5 red balls and z+8 white balls. The probability of getting a black ball is ¼, what is the probability of getting three balls of different colour, when three balls are drawn?
Solution
To solve this problem, we first need to find the total number of balls in the bag.
Given that the probability of drawing a black ball is 1/4, we can set up the following equation:
(z + 2) / (total number of balls) = 1/4
The total number of balls is the sum of the black, red, and white balls, which is (z + 2) + (z + 5) + (z + 8) = 3z + 15.
Substituting this into our equation gives us:
(z + 2) / (3z + 15) = 1/4
Solving this equation for z gives us z = 2.
Now that we know z, we can find the number of each color of ball:
Black balls = z + 2 = 2 + 2 = 4 Red balls = z + 5 = 2 + 5 = 7 White balls = z + 8 = 2 + 8 = 10
The total number of balls is 4 + 7 + 10 = 21.
The probability of drawing three balls of different colors is the product of the probabilities of drawing one ball of each color.
This is calculated as follows:
P(Black) * P(Red) * P(White) = (4/21) * (7/20) * (10/19) = 0.035
Therefore, the probability of drawing three balls of different colors is 0.035 or 3.5%.
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