Single and Two Dice Probability Experiments for CBSE Class 10

From ancient board games played in the Indus Valley civilization to modern tabletop games like Monopoly and Ludo, the six-sided cube known as a die (plural: dice) is one of humanity's oldest generators of random chance. Marked with small dots called pips numbering from $1$ to $6$, an unbiased die provides an ideal physical model for studying classical discrete probability.

In CBSE Class 10 Mathematics, Chapter 14 (Probability), single and two-dice experiments represent a cornerstone of board exam questions. Mastering the $36$-outcome grid, understanding prime and composite numbers, analyzing doublets, and predicting the probability distribution of the sum of two dice is essential for complete exam mastery.

What You Will Learn

• Sample space and probabilities for throwing a single die ($n = 6$)
• Prime numbers, composite numbers, and common classification traps on dice
• The complete $6 \times 6$ outcome grid for throwing two dice simultaneously ($n = 36$)
• What are Doublets?
• The probability distribution of the Sum of Two Numbers (ranging from $2$ to $12$)
• Solving the famous NCERT problem: "5 will not come up either time"
• Board exam tips, tabular methods, and common student errors

1. Throwing a Single Die ($n = 6$)

A standard fair die has six faces numbered $1, 2, 3, 4, 5, 6$.

• Sample Space: $S = \{1, 2, 3, 4, 5, 6\}$
• Total number of possible outcomes: $n(S) = \mathbf{6}$.

Common Event Classifications on a Single Die:

                            The 6 Faces of a Die
                      1       2       3       4       5       6
                      |       |       |       |       |       |
              Neither |   +---+---+---+   +---+---+   |       |
              Prime   |   |   PRIMES  |   | COMPOSITES|       |
              nor     |   | (2, 3, 5) |   |  (4, 6)   |       |
              Composite   +-----------+   +-----------+       |

1. Getting a Prime Number:
   • The prime numbers on a die are $2, 3,$ and $5$ (Total $= 3$).
   • <u>Remember: $1$ is NEITHER prime nor composite!</u>
   • $P(\text{Prime}) = \frac{3}{6} = \mathbf{\frac{1}{2}}$
2. Getting an Odd Number:
   • Odd numbers: $\{1, 3, 5\} \implies P(\text{Odd}) = \frac{3}{6} = \mathbf{\frac{1}{2}}$.
3. Getting an Even Number:
   • Even numbers: $\{2, 4, 6\} \implies P(\text{Even}) = \frac{3}{6} = \mathbf{\frac{1}{2}}$.
4. Getting a Number Lying Between 2 and 6:
   • Numbers strictly between 2 and 6 are $3, 4,$ and $5$ (excludes 2 and 6!):
   • $P(2 < x < 6) = \frac{3}{6} = \mathbf{\frac{1}{2}}$

2. Throwing Two Dice Simultaneously ($n = 36$)

Whether two dice (say, one blue and one grey) are rolled together, or a single die is rolled twice in succession, the total number of outcomes is: $\mathbf{n(S) = 6 \times 6 = 36}$

The Complete $6 imes 6$ Sample Space Grid:

Every outcome is written as an ordered pair $(x, y)$, where $x$ is the score on the first die and $y$ is the score on the second die:

3. What are Doublets?

A doublet is an outcome in which both dice show the exact same number.

The diagonal of the $6 \times 6$ grid contains all doublets: $\text{Doublets} = \{(1, 1), \; (2, 2), \; (3, 3), \; (4, 4), \; (5, 5), \; (6, 6)\}$

• Total favourable outcomes: $n(\text{Doublet}) = 6$.
• Probability of rolling a doublet: $\mathbf{P(\text{Doublet}) = \frac{6}{36} = \frac{1}{6}}$

4. Probability Distribution of the Sum of Two Numbers

When two dice are thrown, the sum of the numbers can range from a minimum of $2$ ($(1, 1)$) to a maximum of $12$ ($(6, 6)$):

Important: <u>A sum of 7 is the MOST PROBABLE outcome when rolling two dice (probability = 1/6), because there are 6 distinct combinations that sum to 7!</u>

5. Solved CBSE Board Examination Problems

Solved Example 1: Sum of Dice Problems (NCERT Classic)

Problem: Two dice are thrown at the same time. Determine the probability of getting: (i) sum of the two numbers is $8$
(ii) sum is at least $10$
(iii) sum is less than or equal to $12$

Solution: Total possible outcomes $n(S) = 36$.

1. (i) Sum is 8: Favourable outcomes: $\{(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)\} \implies n(E_1) = 5$. $P(\text{Sum is 8}) = \mathbf{\frac{5}{36}}$

2. (ii) Sum is at least 10: Means sum can be $10, 11,$ or $12$:

   • Sum 10: $(4, 6), (5, 5), (6, 4)$ (3 outcomes)
   • Sum 11: $(5, 6), (6, 5)$ (2 outcomes)
   • Sum 12: $(6, 6)$ (1 outcome) Total favourable outcomes $= 3 + 2 + 1 = 6$. $P(\text{Sum } \ge 10) = \frac{6}{36} = \mathbf{\frac{1}{6}}$

3. (iii) Sum is less than or equal to 12: Since the maximum possible sum is $6 + 6 = 12$, ALL 36 outcomes have a sum $\le 12$! This is a Sure Event: $P(\text{Sum } \le 12) = \frac{36}{36} = \mathbf{1}$

Solved Example 2: "5 Will Not Come Up Either Time" (CBSE Classic)

Problem: A die is thrown twice. What is the probability that: (i) $5$ will not come up either time?
(ii) $5$ will come up at least once?

Solution:

1. Let $E$ be the event that "$5$ comes up at least once".
2. Identify all outcomes where $5$ appears on the first die or the second die:
   • $5$ on the first die: $(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)$ (6 outcomes)
   • $5$ on the second die: $(1, 5), (2, 5), (3, 5), (4, 5), (6, 5)$ (5 outcomes, excluding $(5, 5)$ already counted)
   • Total outcomes where $5$ appears at least once: $n(E) = 6 + 5 = \mathbf{11}$
3. (ii) Probability that $5$ will come up at least once: $P(E) = \mathbf{\frac{11}{36}}$
4. (i) Probability that $5$ will not come up either time: This is the complementary event of $5$ coming up at least once: $P(\text{not } E) = 1 - P(E) = 1 - \frac{11}{36} = \frac{36 - 11}{36} = \mathbf{\frac{25}{36}}$
5. Therefore:
   • <u>(i) Probability that 5 will not come up either time is $\frac{25}{36}$</u>.
   • <u>(ii) Probability that 5 will come up at least once is $\frac{11}{36}$</u>.

6. Summary and Examination Tips

Exam Tip: In two-dice problems, always check that you didn't double-count the intersection outcome $(5, 5)$! There are 6 outcomes with 5 on the first die and 6 with 5 on the second die, but $(5, 5)$ is shared, giving $6 + 6 - 1 = 11$ outcomes!

Common Mistake: Writing $(2, 3)$ and $(3, 2)$ as the same outcome. When two dice are thrown (e.g. red and blue), getting 2 on red and 3 on blue is physically different from getting 3 on red and 2 on blue!