Q99 / 155 · WBJEE 2026 · Mathematics
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
- Correct answerA$\dfrac{16}{625}$
- B$\dfrac{18}{625}$
- C$\dfrac{4}{625}$
- D$\dfrac{5}{625}$
Solution
The unit digit of the product of four natural numbers will be $1, 3, 7$ or $9$ if and only if the product is an odd number and not a multiple of $5$.
This implies that none of the four numbers can be even or have $5$ as their unit digit.
Therefore, the unit digit of each of the four numbers must be chosen from the set $\{1, 3, 7, 9\}$.
The probability of a natural number having its unit digit as $1, 3, 7$ or $9$ is $\dfrac{4}{10} = \dfrac{2}{5}$.
Since the four numbers are selected independently, the required probability is $\left(\dfrac{2}{5}\right)^4 = \dfrac{16}{625}$.
Answer: $\dfrac{16}{625}$
This implies that none of the four numbers can be even or have $5$ as their unit digit.
Therefore, the unit digit of each of the four numbers must be chosen from the set $\{1, 3, 7, 9\}$.
The probability of a natural number having its unit digit as $1, 3, 7$ or $9$ is $\dfrac{4}{10} = \dfrac{2}{5}$.
Since the four numbers are selected independently, the required probability is $\left(\dfrac{2}{5}\right)^4 = \dfrac{16}{625}$.
Answer: $\dfrac{16}{625}$
Hence, the correct answer is option: 'A'.