Sum at least 9 with the first die even

Problem

Two dice are rolled. What is the probability that the sum is $\ge 9$ given that the first die showed an even number?

Solution

We condition on the event «first die even», that is on the values $\{2,4,6\}$: these are $3\cdot 6 = 18$ equally likely outcomes. For each value of the first die, we count the outcomes of the second that bring the sum to at least $9$:

• first $=2$: we need the second $\ge 7$, impossible → $0$ cases;
• first $=4$: we need the second $\ge 5$, that is $\{5,6\}$ → $2$ cases;
• first $=6$: we need the second $\ge 3$, that is $\{3,4,5,6\}$ → $4$ cases.

In all $0+2+4 = 6$ favourable cases out of $18$: $P(\text{somma}\ge 9 \mid \text{primo pari}) = \frac{6}{18} = \boxed{\dfrac{1}{3}}$

Links

Topics: Probabilita
Concepts: Probabilita condizionata
Methods: Prob condizionata
Skills: Calcolo probabilita · Ragionare per casi
Exercise type: Problema probabilita