Question

There are 10 balls numbered 1 through 10 placed in a bucket. What is the probability of reaching into the bucket and randomly drawing four balls numbered 8, 5, 10, and 7 without replacement, in that order? Express your answer as a fraction in lowest terms or a decimal rounded to the nearest millionth.

Answer 1

Given that there are 10 balls in the bucket, you know that these are numbered from 1 through 10.

You can determine that the probability of drawing a ball numbered 8 is:

`P_8=(1)/(10)`

Since there is no replacement, the probability of drawing a ball numbered 5 in the second selection is:

`P_5=(1)/(9)`

Therefore:

`\begin{gathered} P_(10)=(1)/(8) \\ \\ P_7=(1)/(7) \end{gathered}`

Hence, the probability of drawing four balls numbered 8, 5, 10, and 7 without replacement is;

`P=(1)/(10)\cdot(1)/(9)\cdot(1)/(8)\cdot(1)/(7)`
`P=(1)/(5040)`

Therefore, the answer is:

`P=(1)/(5040)`