Probabilidad Condicional con Canicas

Problem

A bag contains $5$ red marbles and $3$ blue marbles, and two marbles are drawn without replacement; given that at least one marble is red, find the probability that both marbles are red.

Step 1: Find $P(\text{both red})$

The two red-draw path is $RR$, so

$$P(RR)=\dfrac{5}{8}\cdot\dfrac{4}{7}=\dfrac{20}{56}=\dfrac{10}{28}.$$

Step 2: Find $P(\text{at least one red})$

Use the complement event "no red at all," which is $BB$:

$$P(BB)=\dfrac{3}{8}\cdot\dfrac{2}{7}=\dfrac{6}{56}=\dfrac{3}{28}.$$

So the probability of at least one red is

$$P(\text{at least one red})=1-\dfrac{3}{28}=\dfrac{25}{28}.$$

Step 3: Apply conditional probability

Using $P(A\mid B)=\dfrac{P(A\cap B)}{P(B)}$ with $A=$ "both red" and $B=$ "at least one red",

$$P(\text{both red}\mid \text{at least one red})=\dfrac{10/28}{25/28}=\dfrac{10}{25}=\dfrac{2}{5}.$$

Answer

The probability is $\dfrac{2}{5}$.