Question:
Four natural numbers selected at random are multiplied together, then the probability that the digit in the unit's place in the product be 1, 3, 7 or 9 is:
Four natural numbers selected at random are multiplied together, then the probability that the digit in the unit's place in the product be 1, 3, 7 or 9 is:
Show Hint
This problem highlights a very neat shortcut condition: the set of digits $\{1, 3, 7, 9\}$ forms a closed group under multiplication (meaning multiplying any of these numbers together will always produce another number ending in 1, 3, 7, or 9). This guarantees that as long as individual numbers stay inside this group, the final product is guaranteed to stay inside it too!
Updated On: May 28, 2026
- $\frac{16}{625}$
- $\frac{18}{625}$
- $\frac{4}{625}$
- $\frac{5}{625}$
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The Correct Option is A
Solution and Explanation
Concept:
The unit's digit of a product of integers is entirely determined by the product of the unit's digits of the individual numbers. For a product of numbers to end in an odd digit that is not 5 (meaning ending in 1, 3, 7, or 9), none of the individual numbers can be even, and none of them can end in the digit 5.
Step 1: Identify the valid unit digits for an individual number selection.
A natural number can end in any of the 10 basic digits from 0 to 9: \[ \text{Total digit options} = \{0, 1, 2, 3, 4, 5, 6, 7, 8, 9\} \implies 10 \text{ choices} \] For the final overall multiplied product to end specifically in the set $\{1, 3, 7, 9\}$, let us look at what digits are allowed for each individual number:
• If any individual number ends in an even digit ($0, 2, 4, 6, 8$), the entire final product will automatically become even, making it impossible to end in an odd number.
• If any individual number ends in the digit 5, and is multiplied exclusively by other odd numbers, the final product will inevitably end in 5, which is excluded from our target set. Therefore, every single individual number chosen must have a unit digit belonging exclusively to the set: \[ \text{Valid individual digits} = \{1, 3, 7, 9\} \implies 4 \text{ choices} \]
Step 2: Calculate the probability of a single number being valid.
Since each digit choice has an equal chance of appearing at the end of a random natural number, the probability $P_1$ that any single chosen number ends in a valid digit is: \[ P_1 = \frac{\text{Valid Choices}}{\text{Total Choices}} = \frac{4}{10} = \frac{2}{5} \]
Step 3: Extend the probability calculation across all four independent selections.
We are selecting four natural numbers completely at random. Since the selection of each number represents an independent event, we find the total combined probability by multiplying the individual probabilities together: \[ \text{Total Probability } (P) = P_1 \times P_2 \times P_3 \times P_4 = \left(\frac{2}{5}\right)^4 \]
Step 4: Evaluate the final fractional probability value.
Expand the exponential powers for both the numerator and the denominator components: \[ P = \frac{2^4}{5^4} = \frac{16}{625} \] This matches option (A) perfectly.
A natural number can end in any of the 10 basic digits from 0 to 9: \[ \text{Total digit options} = \{0, 1, 2, 3, 4, 5, 6, 7, 8, 9\} \implies 10 \text{ choices} \] For the final overall multiplied product to end specifically in the set $\{1, 3, 7, 9\}$, let us look at what digits are allowed for each individual number:
• If any individual number ends in an even digit ($0, 2, 4, 6, 8$), the entire final product will automatically become even, making it impossible to end in an odd number.
• If any individual number ends in the digit 5, and is multiplied exclusively by other odd numbers, the final product will inevitably end in 5, which is excluded from our target set. Therefore, every single individual number chosen must have a unit digit belonging exclusively to the set: \[ \text{Valid individual digits} = \{1, 3, 7, 9\} \implies 4 \text{ choices} \]
Step 2: Calculate the probability of a single number being valid.
Since each digit choice has an equal chance of appearing at the end of a random natural number, the probability $P_1$ that any single chosen number ends in a valid digit is: \[ P_1 = \frac{\text{Valid Choices}}{\text{Total Choices}} = \frac{4}{10} = \frac{2}{5} \]
Step 3: Extend the probability calculation across all four independent selections.
We are selecting four natural numbers completely at random. Since the selection of each number represents an independent event, we find the total combined probability by multiplying the individual probabilities together: \[ \text{Total Probability } (P) = P_1 \times P_2 \times P_3 \times P_4 = \left(\frac{2}{5}\right)^4 \]
Step 4: Evaluate the final fractional probability value.
Expand the exponential powers for both the numerator and the denominator components: \[ P = \frac{2^4}{5^4} = \frac{16}{625} \] This matches option (A) perfectly.
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