Probability of Prime Numbers on Dice | Class 10 Maths

Quick answer: When a fair die is thrown once, the prime numbers are $2, 3, 5$, so $P(\text{prime}) = \frac{3}{6} = \frac{1}{2}$. The composite numbers are $4, 6$, so $P(\text{composite}) = \frac{1}{3}$. The number $1$ is neither prime nor composite, so $P(\text{neither}) = \frac{1}{6}$. With two dice, $P(\text{product is prime}) = \frac{6}{36} = \frac{1}{6}$ and $P(\text{sum is prime}) = \frac{15}{36} = \frac{5}{12}$.

Throwing a fair six-sided die is one of the simplest probability experiments: the faces $1, 2, 3, 4, 5, 6$ are equally likely, so each has probability $\frac{1}{6}$. Board questions often make it trickier by asking about types of numbers:

• Find the probability of getting a prime number.
• Find the probability of getting a composite number.
• Find the probability of getting a number that is neither prime nor composite.
• Two dice are thrown. Find the probability that the product of the numbers is a prime number.

The most common mistake is how students treat the number 1.

In CBSE Class 10 Mathematics, Chapter 14 (Probability), die questions based on prime and composite numbers are frequently asked in board exams.

What You Will Learn

• The sample space for one die, $\{1, 2, 3, 4, 5, 6\}$, with $6$ outcomes
• Why $1$ is neither prime nor composite
• Probabilities for one die: prime ($\frac{1}{2}$), composite ($\frac{1}{3}$), neither ($\frac{1}{6}$) and odd prime ($\frac{1}{3}$)
• The sample space for two dice, with $36$ outcomes
• The probability that the product of two dice is prime
• The probability that the sum of two dice is prime

1. Classifying the Numbers on a Die

When a fair die is thrown once, the possible outcomes are $1, 2, 3, 4, 5, 6$, so the total number of outcomes is $6$.

The number 1 trap: A prime number has exactly two factors, $1$ and itself. A composite number has more than two factors. The number $1$ has only one factor, so it is neither prime nor composite. Never include $1$ when counting primes or composites.

2. Solved Problem: One Die

Question

A die is thrown once. Find the probability of getting:

(i) a prime number

(ii) a composite number

(iii) a number that is neither prime nor composite

(iv) an odd prime number

(v) a number lying between $2$ and $6$

Solution

Total number of outcomes $= 6$.

(i) Prime number. Favourable outcomes: $2, 3, 5$, which is $3$ outcomes.

$P(\text{prime}) = \frac{3}{6} = \frac{1}{2}$

(ii) Composite number. Favourable outcomes: $4, 6$, which is $2$ outcomes.

$P(\text{composite}) = \frac{2}{6} = \frac{1}{3}$

(iii) Neither prime nor composite. Favourable outcome: $1$ only.

$P(\text{neither prime nor composite}) = \frac{1}{6}$

(iv) Odd prime number. The primes are $2, 3, 5$. Leaving out $2$ (it is even), the odd primes are $3, 5$, which is $2$ outcomes.

$P(\text{odd prime}) = \frac{2}{6} = \frac{1}{3}$

(v) Number between 2 and 6. "Between" does not include $2$ and $6$, so the favourable outcomes are $3, 4, 5$, which is $3$ outcomes.

$P(\text{between } 2 \text{ and } 6) = \frac{3}{6} = \frac{1}{2}$

Check: $P(\text{prime}) + P(\text{composite}) + P(\text{neither}) = \frac{3}{6} + \frac{2}{6} + \frac{1}{6} = 1$. ✓

3. Solved Problem: Two Dice

Question

Two dice are thrown at the same time. Find the probability that:

(i) the product of the numbers on the two dice is a prime number

(ii) the sum of the numbers on the two dice is a prime number

Solution

Each die has $6$ outcomes, so the total number of outcomes is $6 \times 6 = 36$.

Part (i): Product Is a Prime Number

For a product of two numbers from $1$ to $6$ to be prime, one number must be $1$ and the other must be prime ($2$, $3$ or $5$). If neither number is $1$, the product has extra factors and is composite (for example, $2 \times 2 = 4$ and $2 \times 3 = 6$).

Favourable outcomes:

• First die shows $1$: $(1, 2), (1, 3), (1, 5)$
• Second die shows $1$: $(2, 1), (3, 1), (5, 1)$

That is $6$ favourable outcomes.

$P(\text{product is prime}) = \frac{6}{36} = \frac{1}{6}$

Answer (i): $\frac{1}{6}$

Part (ii): Sum Is a Prime Number

The sum can be any number from $2$ to $12$. The prime sums are $2, 3, 5, 7$ and $11$.

$P(\text{sum is prime}) = \frac{15}{36} = \frac{5}{12}$

Answer (ii): $\frac{5}{12}$

4. Summary Table

5. Exam Tips and Common Mistakes

Exam Tip: Always list the favourable outcomes before writing the probability. It makes your working clear and earns step marks.

Common Mistake: Counting $1$ as a prime number. Writing $\{1, 2, 3, 5\}$ and getting $\frac{4}{6}$ is wrong. The correct answer is $\frac{3}{6} = \frac{1}{2}$.

Common Mistake: Forgetting that $2$ is even. For an odd prime, leave out $2$, so only $3$ and $5$ count.

Common Mistake: With two dice, $(1, 2)$ and $(2, 1)$ are different outcomes. Count both.

Frequently Asked Questions

What is the probability of getting a prime number on a die?

The primes on a die are $2$, $3$ and $5$, so the probability is $\frac{3}{6} = \frac{1}{2}$.

Is 1 a prime number?

No. A prime number has exactly two factors. The number $1$ has only one factor, so it is neither prime nor composite.

What is the probability of getting a composite number on a die?

The composite numbers on a die are $4$ and $6$, so the probability is $\frac{2}{6} = \frac{1}{3}$.

What is the probability that the product of two dice is prime?

There are $6$ favourable outcomes out of $36$, so the probability is $\frac{6}{36} = \frac{1}{6}$.

What is the probability that the sum of two dice is prime?

There are $15$ favourable outcomes out of $36$, so the probability is $\frac{15}{36} = \frac{5}{12}$.