Question

3 numbers are randomly drawn from the set {1, 2, 3, ....., 300}. Find the probability that sum of these 3 numbers is divisible by 3

Answer 1

\frac{14851}{44551}

Answer 2

To find the probability that the sum of three randomly drawn numbers from the set $S = \{1, 2, 3, \ldots, 300\}$ is divisible by 3, we first categorize the numbers based on their remainder when divided by 3.

Let $N_0$ be the count of numbers in $S$ that are divisible by 3. Let $N_1$ be the count of numbers in $S$ that leave a remainder of 1 when divided by 3. Let $N_2$ be the count of numbers in $S$ that leave a remainder of 2 when divided by 3.

The set $S$ contains 300 numbers. Since 300 is perfectly divisible by 3, each remainder category will have an equal number of elements: $N_0 = \{3, 6, \ldots, 300\}$, so $N_0 = \frac{300}{3} = 100$. $N_1 = \{1, 4, \ldots, 298\}$, so $N_1 = \frac{298-1}{3} + 1 = \frac{297}{3} + 1 = 99 + 1 = 100$. $N_2 = \{2, 5, \ldots, 299\}$, so $N_2 = \frac{299-2}{3} + 1 = \frac{297}{3} + 1 = 99 + 1 = 100$. So, we have $N_0 = N_1 = N_2 = 100$. Let $k=100$.

The total number of ways to choose 3 numbers from 300 is given by $\binom{300}{3}$.

$$\text{Total outcomes} = \binom{300}{3} = \frac{300 \times 299 \times 298}{3 \times 2 \times 1} = 50 \times 299 \times 298 = 4455100$$

For the sum of three numbers ($x_1 + x_2 + x_3$) to be divisible by 3, the sum of their remainders when divided by 3 must be divisible by 3. Let $r_1, r_2, r_3$ be the remainders of $x_1, x_2, x_3$ respectively. We need $r_1 + r_2 + r_3 \equiv 0 \pmod{3}$. The possible combinations of remainders are:

1. All three numbers have remainder 0: $(0, 0, 0)$. Number of ways = $\binom{N_0}{3} = \binom{100}{3}$.
2. All three numbers have remainder 1: $(1, 1, 1)$. (Sum of remainders = 3) Number of ways = $\binom{N_1}{3} = \binom{100}{3}$.
3. All three numbers have remainder 2: $(2, 2, 2)$. (Sum of remainders = 6) Number of ways = $\binom{N_2}{3} = \binom{100}{3}$.
4. One number from each remainder category: $(0, 1, 2)$. (Sum of remainders = 3) Number of ways = $N_0 \times N_1 \times N_2 = 100 \times 100 \times 100 = 100^3$.

Now, let's calculate the number of favorable outcomes:

$$\binom{100}{3} = \frac{100 \times 99 \times 98}{3 \times 2 \times 1} = 100 \times 33 \times 49 = 161700$$ $$100^3 = 1000000$$

Total favorable outcomes = $3 \times \binom{100}{3} + 100^3$

$$= 3 \times 161700 + 1000000 = 485100 + 1000000 = 1485100$$

The probability is the ratio of favorable outcomes to total outcomes:

$$P(\text{sum is divisible by 3}) = \frac{1485100}{4455100}$$

We can simplify this fraction by dividing the numerator and denominator by 100:

$$P(\text{sum is divisible by 3}) = \frac{14851}{44551}$$

Alternatively, for the general case where $N = 3k$ and $N_0=N_1=N_2=k$, the probability is given by:

$$P = \frac{3 \binom{k}{3} + k^3}{\binom{3k}{3}} = \frac{3 \frac{k(k-1)(k-2)}{6} + k^3}{\frac{3k(3k-1)(3k-2)}{6}}$$

Multiply numerator and denominator by 6:

$$P = \frac{k(k-1)(k-2) + 2k^3}{k(3k-1)(3k-2)} = \frac{k(k^2-3k+2) + 2k^3}{k(9k^2-9k+2)}$$ $$P = \frac{k^3-3k^2+2k + 2k^3}{9k^3-9k^2+2k} = \frac{3k^3-3k^2+2k}{9k^3-9k^2+2k}$$

Factor out $k$ from numerator and denominator:

$$P = \frac{k(3k^2-3k+2)}{k(9k^2-9k+2)} = \frac{3k^2-3k+2}{9k^2-9k+2}$$

Substitute $k=100$: Numerator = $3(100)^2 - 3(100) + 2 = 30000 - 300 + 2 = 29702$. Denominator = $9(100)^2 - 9(100) + 2 = 90000 - 900 + 2 = 89102$. So, $P = \frac{29702}{89102}$. Dividing both by 2 gives $\frac{14851}{44551}$.

This fraction is in its simplest form because the greatest common divisor of 14851 and 44551 is 1. (Any common factor must divide $44551 - 3 \times 14851 = 148$. Since 14851 is odd and not divisible by 37, there are no common factors).

The final answer is $\frac{14851}{44551}$.

Explanation of the solution:

• Categorize the 300 numbers into three groups based on their remainder when divided by 3: $N_0$ (remainder 0), $N_1$ (remainder 1), $N_2$ (remainder 2). Each group contains 100 numbers.
• Calculate the total number of ways to choose 3 numbers from 300 using combinations: $\binom{300}{3}$.
• Determine the conditions for the sum of three numbers to be divisible by 3:
  • All three numbers from $N_0$.
  • All three numbers from $N_1$.
  • All three numbers from $N_2$.
  • One number from each group ($N_0, N_1, N_2$).
• Calculate the number of ways for each condition and sum them to find the total favorable outcomes.
• Divide the total favorable outcomes by the total possible outcomes to get the probability.