Two distinct natural numbers from 1 to 9 are picked at random. What is the probability that their product has 1 in its unit place?
Calculating Probability of Product Unit Digit 1
The question asks for the probability that the product of two distinct natural numbers picked from 1 to 9 has 1 in its unit place. Let's break this down into calculating the total possible outcomes and the number of favorable outcomes.
Total Possible Outcomes When Picking Two Distinct Numbers
We are picking two distinct natural numbers from the set {1, 2, 3, 4, 5, 6, 7, 8, 9}. The order in which the numbers are picked matters when considering permutations. There are 9 choices for the first number and, since the numbers must be distinct, 8 choices remaining for the second number.
The total number of ways to pick two distinct numbers in order is the number of permutations of 9 items taken 2 at a time, denoted as \(P(9, 2)\).
\[P(9, 2) = \frac{9!}{(9-2)!} = \frac{9!}{7!} = 9 \times 8 = 72\]
So, there are 72 possible ordered pairs of distinct numbers that can be picked from the set {1, 2, ..., 9}.
Identifying Favorable Outcomes
We are looking for pairs of distinct numbers \((a, b)\) from {1, 2, 3, 4, 5, 6, 7, 8, 9} such that the unit digit of their product \((a \times b)\) is 1.
The unit digit of a product depends only on the unit digits of the numbers being multiplied. For the unit digit of a product to be 1, the unit digits of the two numbers must multiply to a number whose unit digit is 1.
Let's consider the unit digits of numbers from 1 to 9 and their possible products:
- 1 x 1 = 1
- 3 x 7 = 21 (Unit digit 1)
- 7 x 3 = 21 (Unit digit 1)
- 9 x 9 = 81 (Unit digit 1)
Now we need to find pairs of distinct numbers from {1, 2, ..., 9} that correspond to these unit digit combinations.
- For unit digits 1 and 1: This would require picking the number 1 twice, but the numbers must be distinct. So, (1, 1) is not a valid pair.
- For unit digits 3 and 7: This corresponds to picking the numbers 3 and 7. These are distinct numbers from the set. The product is \(3 \times 7 = 21\), which has 1 in the unit place. The ordered pairs are (3, 7) and (7, 3).
- For unit digits 9 and 9: This would require picking the number 9 twice, but the numbers must be distinct. So, (9, 9) is not a valid pair.
The only pairs of distinct numbers from 1 to 9 whose product has a unit digit of 1 are (3, 7) and (7, 3).
Thus, there are 2 favorable outcomes.
Calculating the Probability
The probability of an event is calculated as:
\[\text{Probability} = \frac{\text{Number of Favorable Outcomes}}{\text{Total Number of Possible Outcomes}}\]
Number of Favorable Outcomes = 2
Total Number of Possible Outcomes = 72
\[\text{Probability} = \frac{2}{72} = \frac{1}{36}\]
Therefore, the probability that the product of two distinct natural numbers picked at random from 1 to 9 has 1 in its unit place is \( \frac{1}{36} \).
| Event | Description | Count |
|---|---|---|
| Total Outcomes | Picking 2 distinct numbers from 9 (ordered) | \(P(9, 2) = 72\) |
| Favorable Outcomes | Picking 2 distinct numbers whose product's unit digit is 1 | (3, 7), (7, 3) - 2 outcomes |
Revision Table: Probability Concepts
| Concept | Description | Formula/Example |
|---|---|---|
| Sample Space | The set of all possible outcomes of a random experiment. | For rolling a single die, S = {1, 2, 3, 4, 5, 6} |
| Event | A subset of the sample space. | Getting an even number when rolling a die, E = {2, 4, 6} |
| Probability | The likelihood of an event occurring. | \(P(E) = \frac{\text{Number of outcomes in E}}{\text{Total number of outcomes in S}}\) |
| Permutation | The number of ways to arrange a set of items where order matters. | \(P(n, k) = \frac{n!}{(n-k)!}\) |
| Combination | The number of ways to choose a set of items where order does not matter. | \(C(n, k) = \frac{n!}{k!(n-k)!}\) |
Additional Information: Unit Digits of Products
The unit digit of a product is determined solely by the unit digits of the numbers being multiplied. This is a useful property for solving problems involving the unit digit of large numbers or products.
To find the unit digit of \(a \times b\), you only need to find the unit digit of (Unit digit of \(a\)) \(\times\) (Unit digit of \(b\)).
For example, to find the unit digit of \(23 \times 47\), you look at the unit digits 3 and 7. Their product is \(3 \times 7 = 21\). The unit digit of 21 is 1. So, the unit digit of \(23 \times 47\) is 1. (Indeed, \(23 \times 47 = 1081\)).
In our problem, we needed the unit digit of the product to be 1. We listed all pairs of unit digits from 1-9 whose product ends in 1: (1,1), (3,7), (7,3), (9,9). We then checked which pairs of distinct numbers from 1 to 9 correspond to these unit digits.
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Important Questions from Probability
- If one event is unaffected by the outcome of another event, the two events are said to be:
1. Dependent
2. Independent
3. Mutually exclusive
4. Mutually inclusive - Let $M$ be a randomly chosen non-empty subset of $S = \{1, 2, 3, \dots, 2026\}$.
Which of the following is the probability that the product of all the elements of $M$ is even? - A fair coin is tossed 6 times. The probability of getting at least 5 heads is _________________. (rounded off to two decimal places)
- Consider a random experiment of rolling a fair die. The probability of getting an odd number or a number less than 4 is ______ (Rounded off to two decimal places).
- In a null hypothesis significance test, the significance level $\alpha$ was set to 0.01. For which one or more of the following $p$ values would the null hypothesis be rejected?
