Question

There are 7 balls numbered I through 7 placed in a bucket What is the probability of reaching into the bucket and randomly drawing two balls numbered 6 and 3 without replacement, in that order? Express your answer as a fraction in lowest terms or a decimal rounded to the nearest millionth.

Answer 1

We have:

- Numbers of balls from 1 to 7 = 7

- Number of balls with number 6 = 1

- Number of ball with number 3 = 1

Then, the probability of ramdomly choosing a 6 is

`p(6)=(1)/(7)`

Once we chose a ball, there are 6 balls into the bucket. Then the probability of ramdomly choosing a 3 is

`P(3)=(1)/(6)`

Then, the probability of randomly choosing a 6 and 3 in that order, is

`\begin{gathered} P(6\text{and}3)=P(6)\cdot P(3)=(1)/(7)\cdot(1)/(6) \\ P(6\text{ and 3)=}(1)/(7\cdot6) \\ P(6\text{ and 3)=}(1)/(42) \end{gathered}`

that is, the probability is 1/42 = 0.023809