Problem 23 - Algebra 2 Course in Mathematics for the IIT-JEE and Other Engineering Exams

Question

In a bag there are three tickets numbered 1,2 , 3\. A ticket is drawn at random and put back. This is done four times. The probability that the sum of the numbers is even is (a) \(\frac{41}{81}\) (b) \(\frac{39}{81}\) (c) \(\frac{40}{81}\) (d) None of these (a) The total number of ways of selecting 4 tickets \(=3^{4}=81\) The favourable number of ways \(=\) sum of coefficients of \(x^{2}, x^{4}, \ldots\) in \(\begin{aligned} &\left(x+x^{2}+x^{3}\right)^{4} \\\=& \text { sum of coefficients of } x^{2}, x^{4}, \ldots \text { in } \\ & x^{4}\left(1+x+x^{2}\right)^{4} \end{aligned}\) and \(1=1-a_{1}+a_{2}-a_{3}+\ldots+a_{8}\) (On putting \(x=-1\) ) \(\therefore 3^{4}+1=2\left(1+a_{2}+a_{4}+a_{6}+a_{8}\right)\) \(\Rightarrow a_{2}+a_{4}+a_{6}+a_{8}=41\) Thus sum of the coefficients of \(x^{2}, x^{4}, \ldots=41\) Hence, required probability \(=41 / 81\).

Step-by-Step Solution

Answer

The probability that the sum is even is \(\frac{41}{81}\), answer (a).

Step 1: Understand the Problem

We need to find the probability that the sum of numbers drawn from a bag, where each draw is replaced, is even. The draw is performed four times.

Step 2: Calculate Total Outcomes

Since there are three tickets numbered 1, 2, and 3, and we draw and replace a ticket four times, the total number of outcomes is calculated as \(3^4 = 81\). This reflects all possible sequences of drawing four tickets.

Step 3: Determine Favorable Outcomes

We need the sum of the drawn numbers to be even. Let each draw be represented by the terms \(x, x^2, x^3\) respectively for tickets numbered 1, 2, and 3. The generating function to capture all combinations is \((x + x^2 + x^3)^4\). We seek the sum of coefficients of even powers, namely \(x^2, x^4, \ldots, x^8\) in \(x^4(1 + x + x^2)^4\).

Step 4: Use Roots of Unity to Find Coefficients

Applying roots of unity, by substituting \(x = -1\), the polynomial \(1 + x + x^2\) evaluated at \(x = -1\) becomes zero, simplifying calculations. Calculate \(3^4 + 1 = 2(1 + a_2 + a_4 + a_6 + a_8)\) which gives \(a_2 + a_4 + a_6 + a_8 = 41\). This is the sum of coefficients of even powers.

Step 5: Calculate Probability

The probability that the sum is even is given by the favorable outcomes over total outcomes, thus \(\frac{41}{81}\).

Step 6: Identify Correct Answer

Comparing with the given options, the correct answer is (a) \(\frac{41}{81}\).

Key Concepts

Generating Functions

Generating functions are a powerful tool used in combinatorics to count sequences or find probabilities. In this problem, generating functions help determine the probability of drawing tickets and achieving a specific sum through polynomial expressions. Here’s how they work in simple terms.
When we express the options of drawing a ticket (numbered 1, 2, and 3) using algebra, each option is given a weight: 1 can be expressed as \(x\), 2 as \(x^2\), and 3 as \(x^3\). Thus, our generating function for one draw is \(x + x^2 + x^3\). This function accumulates all possibilities of drawing one ticket at a time.
Since we are drawing four tickets in total, and each draw is independent, we raise this function to the power four, resulting in \((x + x^2 + x^3)^4\). This expression can be expanded to capture all possible results of drawing four tickets in sequence. By analyzing the coefficients of resulting terms, we can determine how often each sum occurs.
In our case, we are interested in terms with even exponents (like \(x^2, x^4\)), because these represent even sums. The coefficients of these terms tell us the number of favorable outcomes where the sum of numbers drawn is even.

Roots of Unity

Roots of unity provide a clever mathematical shortcut in solving polynomial problems like our exercise above. They stem from complex numbers, specifically ones where powers cycle back to the starting number.
In the context of this problem, one powerful usage is when substituting values into our generating functions to simplify evaluation. By inputting \(x = -1\), we are leveraging roots of unity to simplify the calculation of the sum of coefficients of the even powers.
Why does this work? By choosing \(-1\), we directly target and simplify terms involving even powers. This substitution contributes to summing coefficients that align with our goal (favorable sequences leading to even sums). It allows the complex calculation of coefficients in the polynomial \((1 + x + x^2)^4\) to be reduced to a form we can easily assess:

• \((x = -1)\): Coefficients addressed in even degrees.

Utilizing roots of unity in such problems reduces complexity, offering efficient paths to solution.

Even Sum

An even sum occurs when the result of adding several numbers results in an even number. Even numbers are integers divisible by 2 without a remainder (e.g., 2, 4, 6, etc.).
In our context, for the sum of the numbers from four ticket draws to be even, the product of their respective polynomial terms must maintain an even power of \(x\) throughout the expansion.
Consider this:

• Summing even numbers or pairing two odd numbers results in an even number.
• A practical example: Drawing tickets showing \(1, 1, 2, 2\) (i.e., sum is 6 = even), illustrates this principle within the chosen polynomial setup.

By focusing on even exponents in our generating function expansion, we align with situations where the aggregated draws yield an even result, capturing all favorable outcomes.

Combinatorial Probability

Combinatorial probability helps in finding out how likely a certain combination of events occurs within a set of possibilities. This approach calculates the likeliness based on all possible outcomes and the number of ways favorable outcomes can happen.
In our problem, with 3 potential ticket outcomes appearing per draw and four independent draws, there are \(3^4 = 81\) total possible outcomes. Combinatorial methods aid in narrowing down how many sequences add to an even sum, compared to the total.
To compute probability:

• Total outcomes: All permutations from four draws.
• Favorable outcomes: Sequence yields an even number.

Using the generating function and concepts like roots of unity, we find the number of favorable sequences (41) to calculate the probability. Therefore, probability is the ratio of these favorable outcomes to the total possibilities, calculated as \(\frac{41}{81}\), providing us with the required probability measure of their occurrence.