One die has two faces marked $1$, two faces marked $2$, one face marked $3$ and one face marked $4$. Another die has one face marked $1$, two faces marked $2$, two faces marked $3$ and one face marked $4$. The probability of getting the sum of numbers to be $4$ or $5$, when both the dice are thrown together, is:
Step-by-Step Solution
Key Concept: Each face is still equally likely. Probability of face value $v$ on Die $j$ = $\dfrac{\text{count of faces marked }v}{6}.$ Sum probabilities over all $(a,b)$ with $a+b\in\{4,5\}.$
Pairs $(a,b)$ with $a+b=4$ or $5$ ($a$ on Die 1, $b$ on Die 2):
$(1,3),(1,4),(2,2),(2,3),(3,1),(3,2),(4,1).$
Probabilities (count$_{1}$/6 $\times$ count$_{2}$/6):
$(1,3):\,\tfrac{2}{6}\cdot\tfrac{2}{6}=\tfrac{4}{36};\ (1,4):\,\tfrac{2}{6}\cdot\tfrac{1}{6}=\tfrac{2}{36};\ (2,2):\,\tfrac{2}{6}\cdot\tfrac{2}{6}=\tfrac{4}{36};$
$(2,3):\,\tfrac{2}{6}\cdot\tfrac{2}{6}=\tfrac{4}{36};\ (3,1):\,\tfrac{1}{6}\cdot\tfrac{1}{6}=\tfrac{1}{36};\ (3,2):\,\tfrac{1}{6}\cdot\tfrac{2}{6}=\tfrac{2}{36};\ (4,1):\,\tfrac{1}{6}\cdot\tfrac{1}{6}=\tfrac{1}{36}.$
Sum: $\dfrac{4+2+4+4+1+2+1}{36}=\dfrac{18}{36}=\dfrac{1}{2}.$
Correct Answer: 2