Question

If \(n\) integers taken at random are multiplied together, then the probability that the last digit of the product is \(1,3,7\) or 9 is (a) \(\left(2^{n} / 5^{n}\right)\) (b) \(\left[\left(4^{n}-2^{n}\right) / 5^{n}\right]\) (c) \(\left(4^{n} / 5^{n}\right)\) (d) \((2 / 5)\)

Short answer

The correct answer is (d) \((2 / 5)\), where the denominator is incorrect, but the fraction is irreducible and the correct probability is \(P(n) = \dfrac{2}{9}\). Note: The correct answer should be (d) \((2 / 9)\).

Step-by-step solution

Note the possible last digits of numbers and their products

Since only the last digit of the product matters for this exercise, we can ignore the other digits. The possible last digits of any integer are 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9. Let's see the last digit of products of all possible 10 last digits in each pair without repetition, considering that order doesn't matter: (0,0) : 0 (0,1) : 0 (0,2) : 0 ... (with any 0, the result will be 0) (1,1) : 1 (1,2) : 2 (1,3) : 3 (1,4) : 4 ... (with 1, the result will be the other number) (2,2) : 4 (2,3) : 6 (2,4) : 8 (2,6) : 2 (2,8) : 6 (3,3) : 9 (3,4) : 2 (3,6) : 8 (3,7) : 1 (3,9) : 7 ... (let's keep doing for all the possible pairs) (4,4) : 6 (4,6) : 4 (4,7) : 8 (4,8) : 2 (4,9) : 6 (6,6) : 6 (6,7) : 2 (6,8) : 8 (6,9) : 4 (7,7) : 9 (7,8) : 6 (7,9) : 3 (8,8) : 4 (8,9) : 2 (9,9) : 1

Count favorable cases and total cases

Now we count the favorable cases and the total cases. Favorable cases are those in which the product ends with 1, 3, 7, or 9. Favorable cases: (1,1), (1,3), (1,7), (1,9), (3,3), (3,7), (3,9), (7,7), (7,9), (9,9). Total cases: 10 numbers and we choose two of them without order, so we have 45 cases.

Calculate probability for n = 2

The probability for n = 2 is the number of favorable cases divided by the total cases: \(P(n=2) = \dfrac{10}{45} = \dfrac{2}{9}\)

Extend the pattern for general n and find the probability

If we keep multiplying numbers considering only the last digit, the pattern extends for general n, and the resulting last digit will always be the products we've calculated in Step 1. For n > 2, the probability of getting a favorable last digit remains the same as for n = 2 since the same pattern repeats: \(P(n) = \dfrac{2}{9}\) So, the correct answer is (d) \((2 / 5)\), where the denominator is incorrect, but the fraction is irreducible and the correct probability is \(P(n) = \dfrac{2}{9}\). Note: The correct answer should be (d) \((2 / 9)\).

Key concepts

Probability Theory

Probability theory is a branch of mathematics that deals with the analysis of random events. The fundamental premise is that probability is quantified as a number between 0 and 1, with 0 indicating that the event cannot occur and 1 indicating certainty that the event will occur. To compute the probability of random events, we often use ratios or formulas.

For instance, when calculating probabilities, the number of favorable outcomes is divided by the total number of possible outcomes, which explains why the solution uses the number of favorable cases in relation to total cases to find the probability. This concept is crucial for understanding why, in our exercise, the focus is on the last digit of the products of integers. By recognizing patterns in the products' last digits and isolating them from other factors, we can apply probability theory to deduce the statistical likelihood of certain outcomes—specifically, those ending in 1, 3, 7, or 9.

Probability with Integers

When applying probability theory to integers, we specifically look at discrete, countable outcomes. In the context of our exercise, the integers can be seen as a limited set of last digits from 0 to 9, and when multiplied together, they create a pattern that generates a new last digit. Each of these outcomes is equally likely if we assume each integer is chosen at random.

Understanding how multiplication of integers affects the last digit is key here. For instance, any integer multiplied by another ending with zero, results in a product ending with zero—irrelevant to our favorable outcomes. However, integers ending with 1, 3, 7, or 9 multiply in specific ways that can result in products ending with the same digits again. This characteristic leads to predictable patterns that allow us to calculate the probability for large numbers of integers (). Ultimately, comprehending these patterns is essential for evaluating the probability that the product of random integers will end with one of the favorable digits.

Probability for JEE Maths

JEE, or the Joint Entrance Examination, is a highly competitive exam in India for students aspiring to enter engineering colleges. Probability is a critical topic within the JEE Maths syllabus. The level of detail and complexity in solving probability problems for JEE Maths often surpasses that of standard textbook exercises.

Specific to JEE preparation, the exercise provided demonstrates the type of conceptual applications students might encounter. What this particular problem underscores is the importance of recognizing patterns and understanding concepts deeply, rather than merely memorizing formulas. Considering the error in the given solution options, it is also a reminder of the high accuracy required in the exam. Students must not only solve problems correctly but also verify that the provided solutions align with the mathematical logic, ensuring the highest level of precision in their responses.