Two numbers x and y are chosen at random from a set of the first 10 natural numbers. What is the probability that (x + y) is divisible by 4 ?
Calculating Probability: Sum of Two Numbers Divisible by 4
The problem asks for the probability that the sum of two numbers, x and y, chosen at random from the first 10 natural numbers, is divisible by 4. The set of the first 10 natural numbers is S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}.
When "two numbers x and y are chosen at random", it typically implies selecting two distinct numbers without replacement, and the order of selection matters (i.e., (x, y) is different from (y, x)). Let's proceed with this interpretation.
Step 1: Determine the Total Possible Outcomes
We are choosing two distinct numbers x and y from a set of 10 numbers, and the order matters. The total number of ordered pairs (x, y) with x ≠ y is the number of permutations of 10 items taken 2 at a time, denoted as &(\(\text{P}(10, 2)\)&).
&(\(\text{Total outcomes} = \text{P}(10, 2) = 10 \times 9 = 90\))
Step 2: Determine the Favourable Outcomes
We need to find the number of pairs (x, y) such that x ≠ y, x, y &(\(\in\)) S, and (x + y) is divisible by 4.
Let's find the pairs (x, y) where x ≠ y and x + y is a multiple of 4. The minimum possible sum is 1 + 2 = 3, and the maximum possible sum is 10 + 9 = 19. The multiples of 4 within this range are 4, 8, 12, and 16.
We can systematically list the pairs (x, y) with x ≠ y from S for each target sum:
- Sum = 4: Possible pairs are (1, 3) and (3, 1). (2, 2) is not allowed as x ≠ y. There are 2 favourable pairs.
- Sum = 8: Possible pairs are (1, 7), (7, 1), (2, 6), (6, 2), (3, 5), (5, 3). (4, 4) is not allowed. There are 6 favourable pairs.
- Sum = 12: Possible pairs are (2, 10), (10, 2), (3, 9), (9, 3), (4, 8), (8, 4), (5, 7), (7, 5). (6, 6) is not allowed. There are 8 favourable pairs.
- Sum = 16: Possible pairs are (6, 10), (10, 6), (7, 9), (9, 7). (8, 8) is not allowed. There are 4 favourable pairs.
The total number of favourable outcomes is the sum of the counts for each target sum:
&(\(\text{Total favourable outcomes} = 2 + 6 + 8 + 4 = 20\))
Step 3: Calculate the Probability
The probability of the event is the ratio of the total favourable outcomes to the total possible outcomes.
&(\(\text{Probability} = \frac{\text{Total favourable outcomes}}{\text{Total possible outcomes}} = \frac{20}{90}\))
Simplify the fraction:
&(\(\text{Probability} = \frac{20}{90} = \frac{2}{9}\))
Alternatively, we can use the concept of remainders modulo 4. The numbers in the set S = {1, 2, ..., 10} can be classified by their remainder when divided by 4:
- Remainder 0: R0 = {4, 8} (2 numbers)
- Remainder 1: R1 = {1, 5, 9} (3 numbers)
- Remainder 2: R2 = {2, 6, 10} (3 numbers)
- Remainder 3: R3 = {3, 7} (2 numbers)
The sum x + y is divisible by 4 if and only if the sum of their remainders modulo 4 is congruent to 0 modulo 4. Since x ≠ y, the possible combinations of remainder classes &(\(([x]_4, [y]_4)\) &) for (x + y) &(\(\equiv\)) 0 (mod 4) are:
- &(\(([x]_4, [y]_4) = (0, 0)\)&): x, y &(\(\in\)) R0, x ≠ y. Number of pairs = &(\(\text{P}(2, 2) = 2 \times 1 = 2\)). Pairs: (4, 8), (8, 4).
- &(\(([x]_4, [y]_4) = (1, 3)\)&): x &(\(\in\)) R1, y &(\(\in\)) R3. Number of pairs = &(\(|R_1| \times |R_3| = 3 \times 2 = 6\)).
- &(\(([x]_4, [y]_4) = (3, 1)\)&): x &(\(\in\)) R3, y &(\(\in\)) R1. Number of pairs = &(\(|R_3| \times |R_1| = 2 \times 3 = 6\)).
- &(\(([x]_4, [y]_4) = (2, 2)\)&): x, y &(\(\in\)) R2, x ≠ y. Number of pairs = &(\(\text{P}(3, 2) = 3 \times 2 = 6\)).
Total number of favourable pairs = 2 + 6 + 6 + 6 = 20.
Total possible ordered pairs (x, y) with x ≠ y = 90.
&(\(\text{Probability} = \frac{20}{90} = \frac{2}{9}\))
Both methods yield the same probability, &(\(\frac{2}{9}\)&).
| Target Sum (Multiple of 4) | Favourable Pairs (x, y) | Count |
|---|---|---|
| 4 | (1, 3), (3, 1) | 2 |
| 8 | (1, 7), (7, 1), (2, 6), (6, 2), (3, 5), (5, 3) | 6 |
| 12 | (2, 10), (10, 2), (3, 9), (9, 3), (4, 8), (8, 4), (5, 7), (7, 5) | 8 |
| 16 | (6, 10), (10, 6), (7, 9), (9, 7) | 4 |
| Total Favourable Pairs | 20 | |
Revision Table: Key Concepts
| Concept | Description |
|---|---|
| Sample Space | The set of all possible outcomes of an experiment. In this case, ordered pairs (x, y) chosen from {1, ..., 10} with x ≠ y. |
| Event Space | The subset of the sample space containing only the outcomes where the desired event occurs (x + y is divisible by 4). |
| Probability | The ratio of the number of favourable outcomes to the total number of possible outcomes. |
| Permutation | An arrangement of items where the order matters. Used here for selecting two distinct numbers (x, y). |
| Divisibility Rule for 4 | A number is divisible by 4 if its remainder when divided by 4 is 0. For sums, this can be checked using remainders of individual numbers. |
Additional Information: Probability with Replacement
If the question implied choosing two numbers with replacement (x and y can be the same), the total number of outcomes would be &(\(10 \times 10 = 100\)).
The favourable outcomes (x+y divisible by 4) would include pairs like (2, 2), (4, 4), (6, 6), (8, 8), (10, 10), plus other pairs where x ≠ y that sum to a multiple of 4.
- Sum = 4: (1, 3), (2, 2), (3, 1) - 3 pairs
- Sum = 8: (1, 7), (2, 6), (3, 5), (4, 4), (5, 3), (6, 2), (7, 1) - 7 pairs
- Sum = 12: (2, 10), (3, 9), (4, 8), (5, 7), (6, 6), (7, 5), (8, 4), (9, 3), (10, 2) - 9 pairs
- Sum = 16: (6, 10), (7, 9), (8, 8), (9, 7), (10, 6) - 5 pairs
- Sum = 20: (10, 10) - 1 pair
Total favourable outcomes with replacement = 3 + 7 + 9 + 5 + 1 = 25.
Probability with replacement = &(\(\frac{25}{100} = \frac{1}{4}\)&).
Since &(\(\frac{1}{4}\)&) is not among the options and &(\(\frac{2}{9}\)&) is, the interpretation of choosing two distinct numbers without replacement (ordered pairs) is the intended one for this problem.
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