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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

This question was previously asked in
WBJEE 2026 Physics and Chemistry Question Paper (24-May-2026)
The correct answer is
$\frac{16}{625}$

Unit Digits in Multiplication Basics

The unit digit of a product depends solely on the unit digits of the numbers being multiplied. The possible unit digits for any natural number are 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9.

Product Unit Digits Condition

We require the unit digit of the product of four natural numbers to be 1, 3, 7, or 9. This outcome is achievable only if the unit digit of each of the four selected numbers is also from the set {1, 3, 7, 9}.

  • A unit digit of 0, 2, 4, 6, or 8 in any of the numbers results in an even unit digit for the product.
  • A unit digit of 5 in any of the numbers results in a unit digit of 0 or 5 for the product.
  • Consequently, for the product's unit digit to be 1, 3, 7, or 9, all four numbers must independently have unit digits from {1, 3, 7, 9}.

Probability Calculation Steps

Consider a single randomly selected natural number:

  • There are 10 possible unit digits (0 through 9).
  • There are 4 favorable unit digits (1, 3, 7, 9).
  • The probability of a single number having a unit digit from the set {1, 3, 7, 9} is calculated as:

    $ P(\text{Unit digit} \in \{1, 3, 7, 9\}) = \frac{\text{Number of favorable digits}}{\text{Total possible digits}} = \frac{4}{10} = \frac{2}{5} $

Since the four natural numbers are selected randomly and independently, the probability that all four numbers have unit digits from {1, 3, 7, 9} is the product of their individual probabilities:

$ P(\text{Product unit digit} \in \{1, 3, 7, 9\}) = P(\text{1st has favorable unit digit}) \times P(\text{2nd has favorable unit digit}) \times P(\text{3rd has favorable unit digit}) \times P(\text{4th has favorable unit digit}) $

$ P(\text{Product unit digit} \in \{1, 3, 7, 9\}) = \left( \frac{2}{5} \right) \times \left( \frac{2}{5} \right) \times \left( \frac{2}{5} \right) \times \left( \frac{2}{5} \right) = \left( \frac{2}{5} \right)^4 $

Calculating the final value:

$ \left( \frac{2}{5} \right)^4 = \frac{2^4}{5^4} = \frac{16}{625} $

Final Product Unit Digit Probability

The probability that the unit's digit of the product of four randomly selected natural numbers is 1, 3, 7, or 9 is $\frac{16}{625}$.

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