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
- (1) $\frac{1}{2}$
- (2) $\frac{3}{5}$
- (3) $\frac{2}{3}$
- (4) $\frac{4}{9}$
Solution:
Let $D_1$ be the first die and $D_2$ be the second die.
Total outcomes = $6 \times 6 = 36$.
Faces on $D_1$: $\{1, 1, 2, 2, 3, 4\}$. Probabilities: $P(1)=\frac{2}{6}, P(2)=\frac{2}{6}, P(3)=\frac{1}{6}, P(4)=\frac{1}{6}$.
Faces on $D_2$: $\{1, 2, 2, 3, 3, 4\}$. Probabilities: $P(1)=\frac{1}{6}, P(2)=\frac{2}{6}, P(3)=\frac{2}{6}, P(4)=\frac{1}{6}$.
Possible pairs $(a, b)$ where $a \in D_1, b \in D_2$ such that sum is 4 or 5:
1. Sum = 4 $\to$ Pairs are $(1, 3), (3, 1), (2, 2)$.
$P(\text{Sum 4}) = P(1,3) + P(3,1) + P(2,2)$
$= \left(\frac{2}{6} \cdot \frac{2}{6}\right) + \left(\frac{1}{6} \cdot \frac{1}{6}\right) + \left(\frac{2}{6} \cdot \frac{2}{6}\right) = \frac{4}{36} + \frac{1}{36} + \frac{4}{36} = \frac{9}{36}$.
2. Sum = 5 $\to$ Pairs are $(1, 4), (4, 1), (2, 3), (3, 2)$.
$P(\text{Sum 5}) = P(1,4) + P(4,1) + P(2,3) + P(3,2)$
$= \left(\frac{2}{6} \cdot \frac{1}{6}\right) + \left(\frac{1}{6} \cdot \frac{1}{6}\right) + \left(\frac{2}{6} \cdot \frac{2}{6}\right) + \left(\frac{1}{6} \cdot \frac{2}{6}\right)$
$= \frac{2}{36} + \frac{1}{36} + \frac{4}{36} + \frac{2}{36} = \frac{9}{36}$.
Total Probability $= \frac{9}{36} + \frac{9}{36} = \frac{18}{36} = \frac{1}{2}$.
Ans. (1)