Question

What is the probability of drawing $2$ yellow marbles, if the first one is NOT placed back into the bag before the second draw? There is $10$ marbles total, $2$ yellow, $3$ pink, and $5$ blue.

Answer 1

Find the probability of drawing $2$ yellow marbles if the first one is NOT placed back into the bag before the second draw:

Given that bag has total $10$ marbles and $2$ yellow marbles.

Apply the formula $\mathrm{P}(\mathrm{yellow}\mathrm{marble}\mathrm{on}\mathrm{first}\mathrm{draw})=\frac{\mathrm{number}\mathrm{of}\mathrm{yellow}\mathrm{marble}}{\mathrm{total}\mathrm{number}\mathrm{of}\mathrm{marble}}$ ;

$$\begin{gathered} \mathrm{P}\left(\mathrm{yellow}\mathrm{marble}\mathrm{on}\mathrm{first}\mathrm{draw}\right)=\frac{2}{10} \\ =\frac{1}{5} \end{gathered}$$

Chances of drawing yellow marble on first draw is $\frac{1}{5}$.

The yellow marble is not replaced after first draw so remaining yellow marble is $1$and total number of remaining marbles are 9.

Hence $\mathrm{P}(\mathrm{yellow}\mathrm{marble}\mathrm{on}\mathrm{second}\mathrm{draw})=\frac{1}{9}$ .

Therefore $\mathrm{P}(\mathrm{yellow}\mathrm{marble})=\mathrm{P}\left(\mathrm{yellow}\mathrm{marble}\mathrm{at}\mathrm{first}\mathrm{draw}\right)\times \mathrm{P}\left(\mathrm{yellow}\mathrm{marble}\mathrm{at}\mathrm{second}\mathrm{draw}\right)$.

Substitute the known values;

$$\begin{gathered} \mathrm{P}(\mathrm{yellow}\mathrm{marble})=\frac{1}{5}\times \frac{1}{9} \\ \mathrm{P}(\mathrm{yellow}\mathrm{marble})=\frac{1}{45} \end{gathered}$$

The probability of drawing $2$ yellow marbles if the first one is NOT placed back into the bag before the second draw is $\frac{1}{45}$.