Two red balls without replacement

Problem

An urn contains $5$ red balls and $3$ green balls. Two balls are drawn without replacement. What is the probability that both are red?

Solution

On the first draw there are $5$ red balls out of $8$; on the second, having removed one red, there remain $4$ out of $7$. The two draws combine by multiplying the probabilities: $P(\text{RR}) = \frac{5}{8}\cdot\frac{4}{7} = \frac{20}{56} = \boxed{\dfrac{5}{14}}$

Links

Topics: Probabilita
Concepts: Casi favorevoli casi possibili · Probabilita
Skills: Calcolo probabilita
Exercise type: Problema probabilita