Problem 25 - Algebra 2 Course in Mathematics for the IIT-JEE and Other Engineering Exams

Question

Four tickets marked \(00,01,10,11\), respectively, are placed in a bag. A ticket is drawn at random five times, being replaced each time. The probability that the sum of the numbers on tickets thus drawn is 23 will be [DCE-99] (a) \(25 / 256\) (b) \(100 / 256\) (c) \(231 / 256\) (d) None of these (a) Total number of ways in which 4 tickets can be drawn 5 times \(=4^{5} .\) Favourable cases of getting a sum of 23 \(=\) Coefficients of \(x^{23}\) in \((1+x)^{5}\left(1+x^{10}\right)^{5}\) \(=\) Coefficients of \(x^{23}\) in \(\left(1+5 x+10 x^{2}+10 x^{3}\right.\) \(\left.+5 x^{4}+x^{5}\right)\left(1+5 x^{10}+10 x^{20}+10 x^{30}+\ldots\right)\) \(=100\) \(\therefore\) Required probability \(=\frac{100}{4^{5}}=\frac{100}{1024}=\frac{25}{256}\)

Step-by-Step Solution

Answer

The probability is \( \frac{25}{256} \).

Step 1: Understanding the Problem

We have four tickets with values 00, 01, 10, and 11. Each ticket can be represented by numbers: 0, 1, 10, and 11. We need to figure out the probability that the sum of five drawn tickets equals 23.

Step 2: Total Number of Outcomes

Since each ticket is replaced back into the bag, this is a scenario of independent events. Thus, the total number of ways to draw a ticket five times is computed by raising the number of ticket options (4) to the power of the number of draws (5): \[ 4^5 = 1024 \]

Step 3: Representing the Desired Sum

The favorable cases involve determining how many ways we can end up with a sum of 23 among the numbers: 0, 1, 10, and 11. This can be done using generating functions. Let’s interpret these numbers through powers of x: 0 as \(x^0\), 1 as \(x^1\), 10 as \(x^{10}\), 11 as \(x^{11}\).

Step 4: Setting up the Generating Function

We must consider five draws, generating the function: \[ (1 + x + x^{10} + x^{11})^5 \] to account for all draws. Since \(x^9\) is missing, adjust the function to: \[ (1 + x)^5 (1 + x^{10})^5 \] because powers of 10 and above (representing big-ticket numbers) can't be in all draws.

Step 5: Finding Coefficient of Specific Power

Determine which term has an overall power of 23. For this, we expand: \[ (1 + x)^5 \] gives \(1 + 5x + 10x^2 + 10x^3 + 5x^4 + x^5\) and \[ (1 + x^{10})^5 \] gives \(1 + 5x^{10} + 10x^{20} + 10x^{30} + \ldots\) Mix terms from both polynomials such that exponents total 23. The suitable coefficient calculation results in a sum of favorable cases as 100.

Step 6: Calculating the Probability

Finally, divide the number of favorable cases by the total number of possible cases:\[ \frac{100}{1024} = \frac{25}{256} \], thus arriving at the required probability.

Key Concepts

Generating Functions

Generating functions are a powerful tool in probability and combinatorics for simplifying complex problems. They help in representing sequences and calculating probabilities by using algebraic structures. In the context of our ticket-drawing problem, generating functions are used to find the number of ways we can achieve a specific sum of drawn ticket values.

Understanding generating functions starts with recognizing the numbers on the tickets as powers of a variable, typically denoted by \(x\). For example, ticket values 0, 1, 10, and 11 are represented as \(x^0, x^1, x^{10},\) and \(x^{11}\), respectively. This transformation allows us to construct a polynomial where each term represents a possible outcome when drawing tickets.

To address the problem of finding the sum of 23 in five draws, we create the generating function:

• \((1 + x + x^{10} + x^{11})^5\)

This function models drawing five tickets. However, since certain powers aren't feasible with repeated numbers between 0 and 1, the function can be expanded as:

• \((1 + x)^5(1 + x^{10})^5\)

This approach simplifies the calculation of probabilities by converting a summation problem into finding a specific coefficient, which corresponds to the number of favorable outcomes.

Independent Events

In probability theory, events are called independent if the outcome of one event does not affect the probability of another event. This concept is crucial in calculating probabilities for situations where actions repeat under the same conditions, like drawing and replacing tickets.

In our exercise, each draw from the bag is an independent event because each ticket is replaced before the next draw. This means:

• Each draw is unaffected by the previous draw's outcome.
• The probability of drawing any given ticket remains constant with every draw.

Thus, the total number of possible outcomes when a ticket is drawn five times can be calculated by raising the number of ticket options to the power of the number of draws. For four tickets, this results in:

• \(4^5 = 1024\)

Understanding independent events makes calculating probabilities straightforward, as you can multiply the probabilities of individual events when sequences or combinations are involved.

Favorable Outcomes

Favorable outcomes are those specific results of an event that satisfy the condition we're interested in, such as achieving a particular sum when drawing tickets. The likelihood of the event is determined by comparing these favorable outcomes to the total possible outcomes.

In the context of our problem, a favorable outcome would be any sequence of five ticket values that sum up to 23. Using the generating function approach helps us pin down exactly how many such combinations exist. By equating the sum to the power of 23, we determine the coefficient representing these cases:

When expanding the generating function \((1 + x)^5(1 + x^{10})^5\), we identify the term \(x^{23}\) and its coefficient gives us the number of favorable outcomes. In this case, the coefficient is:

• 100, meaning there are 100 ways to attain the desired sum.

The probability is then the ratio of favorable outcomes to total outcomes. Understanding how to calculate favorable outcomes using generating functions or other methods is vital in solving complex probability questions. The probability hence is:

• \(\frac{100}{1024} = \frac{25}{256}\)

which helps us arrive at our answer efficiently.