Answer:
3/11
Explanation:
From the above question, we have the following information
Total number of balls = 12
Number of white balls = 4
Number of blue balls = 3
Number of red balls = 5
We solve this question using combination formula
C(n, r) = nCr = n!/r!(n - r)!
We are told that 3 balls are drawn out at random.
The chance/probability of drawing out 3 balls = 12C3 = 12!/3! × (12 - 3)! = 12!/3! × 9!
= 12 × 11 × 10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1/(3 × 2 × 1) × (9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1)
= 220 ways
The chance of selecting 3 balls at random = 220
To find the chance that all the three balls are selected,
= [Chance of selecting (white ball) × Chance of selecting(blue ball) × Chance of selecting(red balls)]/ The chance/probability of drawing out 3 balls
Chance of selecting (white ball)= 4C1
Chance of selecting(blue ball) = 3C1 Chance of selecting(red balls) = 5C1
Hence,
= [4C1 × 3C1 × 5C1]/ 220
= 60/220
= 6/22
= 3/11
The chance that all three are selected is = 3/11