Question

There is an urn containing $10$ marbles numbered $1$ to $10$.

We sample $4$ marbles without replacement.

What is the probability that our sample contains the marbles labeled $1$ and $2$.

Answer 1

Calculate the probability that the sample contains the marbles labeled $1$ and $2$:

Recall that the number of ways to select $r$ objects from $n$ distinct objects is ${}^n{C}_r=\frac{n!}{\left(n-r\right)!r!}$.

Number of ways to select $4$ marbles without replacement from $10$ is ${}^{10}{C}_4$.

This is the total number of possible outcomes in the given experiment.

Next, consider that the $4$ marbles selected marbles must include the marbles labeled $1$ and $2$.

This is the simultaneous occurrence of two tasks.

The first is to choose the two marbles labeled $1$ and $2$, there is only one way to do this.

The second is to choose two more marbles from the rest of the $8$ marbles, there are ${}^8{C}_2$ ways of doing this.

By the Fundamental Principle of counting, both tasks can be done simultaneously in $1\times {}^8{C}_2$ ways.

Thus, the $4$ marbles selected can include the marbles labeled $1$ and $2$ in $1\times {}^8{C}_2$ ways.

This is the number of favorable outcomes in the given experiment,

The probability of the event that the $4$ marbles selected can include the marbles labeled $1$ and $2$ is

$\begin{array}{rcl}\frac{\text{number of favourable outcomes}}{\text{total possible outcomes}}& =& \frac{1\times {}^8{C}_2}{{}^{10}{C}_4}\\ & =& \frac{8!}{\left(8-2\right)!2!}÷\frac{10!}{\left(10-4\right)!4!}\left(\text{Replaced}{}^{n}C_r\text{with}\frac{n!}{\left(n-r\right)!r!}\right)\\ & =& \frac{8!}{6!2!}\times \frac{6!4!}{10!}\\ & =& \frac{8!4!}{2!10!}\\ & =& \frac{4\times 3}{10\times 9}\\ & =& \frac{2}{5\times 3}\\ & =& \frac{2}{15}\end{array}$

Hence, the probability that a sample of $4$ marbles selected out of $10$ contains the marbles labeled $1$ and $2$ is $\frac{\mathbf{2}}{\mathbf{15}}$.