Question

A bag contains $5$ black balls and $3$ white balls.

Two balls are drawn at random, one after another from the bag without replacement.

What the probability that both of the balls have the same color?

Answer 1

Step 1. Find the probability of both the balls and the balls having the same color or not:

Given, that the bag contains $5$ black balls and $3$ white balls.

Two balls are drawn at random, one after another from the bag without replacement.

Step 2. Find the probability of choosing Black:

There is a $\frac{5}{8}$ chance that a black marble will be chosen.

Without hesitation, black marble is selected.

A black marble has a $\frac{4}{7}$ chance of being chosen since there are now $4$ black marbles in a bag of $7$.

Now, let's add the two fractions to obtain our probability:

$\begin{array}{rcl}\left(\frac{5}{8}\right)x\left(\frac{4}{7}\right)& =& \frac{20}{56}\\ & =& \frac{5}{14}\end{array}$

The probability of selecting black color is $\frac{5}{14}$

Step 3. Find the probability of choosing White:

White marble will be selected with a $\frac{3}{8}$ chance.

The chosen marble is not changed.

A white marble now has a $\frac{2}{7}$ chance of being chosen since there are now $2$ white marbles in a bag of $7$.

As we did previously, let's combine the two fractions:

$\begin{array}{rcl}\left(\frac{3}{8}\right)x\left(\frac{2}{7}\right)& =& \frac{6}{56}\\ & =& \frac{3}{28}\end{array}$

The probability of choosing white color is $\frac{3}{28}$.

Hence, the probability that both of the balls have the same color depends on which color is chosen twice.