There are 8balls numbered 1 through 8 placed in a bucket. What is the probability of reaching into the bucket and randomly drawing four balls numbered 1, 4, 8, and 6 without replacement, in that order? Express your answer as a fraction in lowest terms or a decimal rounded to the nearest millionth.

Answer:Probability of drawing four balls numbered 1, 4, 8, and 6 from a 8 ball bucket = 0.041667Step-by-step explanation:Probability is  of the ratio of number of favorable outcomes to the total number of outcomes.Here outcome is drawing 4 balls from a 8 ball bucket.Favorable outcome is randomly drawing four balls numbered 1, 4, 8, and 6.Total number of ways in which 4 balls from a 8 ball bucket can be drawn is given by 8 x 7 x 6 x 5 = 1680.Total number of ways in which drawing four balls numbered 1, 4, 8, and 6 from a 8 ball bucket is given by 8C₄     [tex]8C_4=\frac{8\times 7\times 6\times 5}{1\times 2\times 3\times 4} =70[/tex]Probability of drawing four balls numbered 1, 4, 8, and 6 from a 8 ball bucket is given by        [tex]P(n)=\frac{70}{1680} =\frac{1}{24} =0.041667[/tex]Probability of drawing four balls numbered 1, 4, 8, and 6 from a 8 ball bucket = 0.041667