A bag contains 3 red balls and 5 white balls. Two balls are drawn at random at the same time. What is the probability that both are red?

Step-by-step solution

1. 1

   Decide whether order matters. The two balls are drawn together, so a draw is an unordered pair of balls. Counting unordered pairs with combinations keeps the numerator and denominator consistent; you would get the same answer counting ordered pairs, as long as you do it on both sides of the fraction.

2. 2

   Count all equally likely outcomes. The bag holds $3+5=8$ balls, and every pair of distinct balls is equally likely, so the sample space has

   $\binom{8}{2}=\frac{8\cdot 7}{2}=28$

   elements.

3. 3

   Count the favourable outcomes. Both balls red means choosing $2$ of the $3$ red balls:

   $\binom{3}{2}=3$

   The white balls play no role beyond inflating the denominator.

4. 4

   Divide to get the probability.

   $P(\text{both red})=\frac{\binom{3}{2}}{\binom{8}{2}}=\frac{3}{28}\approx 0.107$

   $\boxed{\dfrac{3}{28}}$

5. 5

   Confirm with the sequential (multiplication rule) method. Draw one ball at a time. The first is red with probability $\tfrac{3}{8}$; given that, only $2$ red balls remain among $7$, so the second is red with probability $\tfrac{2}{7}$:

   $\frac{3}{8}\cdot\frac{2}{7}=\frac{6}{56}=\frac{3}{28}$

   The agreement is a good check that the drawing was treated as being without replacement. With replacement the answer would instead be $\left(\tfrac{3}{8}\right)^2=\tfrac{9}{64}$, noticeably larger.