Problem 31 - Algebra 2 Course in Mathematics for the IIT-JEE and Other Engineering Exams

Question

A die is thrown four times. The probability of getting at most two 6 is (a) \(0.984\) (b) \(0.802\) (c) \(0.621\) (d) \(0.721\)

Step-by-Step Solution

Answer

The probability is approximately 0.984, so the answer is (a).

Step 1: Understand the Scenario

We have a die being thrown four times. We need to calculate the probability of getting at most two 6s. This means we are interested in events where the number of times 6 appears is 0, 1, or 2.

Step 2: Define the Probability of Rolling a 6

The probability of rolling a 6 on a single die throw is \( \frac{1}{6} \). The probability of not rolling a 6 is \( 1 - \frac{1}{6} = \frac{5}{6} \).

Step 3: Use the Binomial Probability Formula

The scenario of rolling four dice and counting the successes (rolling a 6) follows a binomial distribution with \( n = 4 \) trials (throws), success probability \( p = \frac{1}{6} \) for rolling a 6. The binomial probability formula is \[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \] where \( \binom{n}{k} \) represents the binomial coefficient.

Step 4: Calculate Probability for 0, 1, and 2 sixes

We calculate the binomial probability for \( X = 0 \), \( X = 1 \), and \( X = 2 \):- \( P(X = 0) = \binom{4}{0} \left( \frac{1}{6} \right)^0 \left( \frac{5}{6} \right)^4 = 1 \times 1 \times \left( \frac{5}{6} \right)^4 \)- \( P(X = 1) = \binom{4}{1} \left( \frac{1}{6} \right)^1 \left( \frac{5}{6} \right)^3 \)- \( P(X = 2) = \binom{4}{2} \left( \frac{1}{6} \right)^2 \left( \frac{5}{6} \right)^2 \)

Step 5: Compute Each Probability

Calculate each term:- \( P(X = 0) = \frac{625}{1296} \)- \( P(X = 1) = 4 \times \frac{1}{6} \times \frac{125}{216} = \frac{500}{1296} \)- \( P(X = 2) = 6 \times \frac{1}{36} \times \frac{25}{36} = \frac{150}{1296} \)

Step 6: Sum the Probabilities

Sum the probabilities for 0, 1, and 2 sixes: \[ P(X \leq 2) = \frac{625}{1296} + \frac{500}{1296} + \frac{150}{1296} = \frac{1275}{1296} \approx 0.9838 \]

Step 7: Select the Closest Answer

The calculated probability \(0.9838\) is closest to option (a) \(0.984\).

Key Concepts

Binomial Distribution

The Binomial Distribution is a fundamental concept in probability theory. It is used to describe the number of successful outcomes in a fixed number of independent trials, which each have the same probability of success. In our die-throwing problem, each throw of the die is a trial. We are particularly interested in how many times we can roll a 6, which is considered a successful outcome.

Thought of in terms of a real-life scenario:

• Trials: Rolling the die four times.
• Success: Getting a six on the die.
• Probability of success in one trial: \( \frac{1}{6} \)

The binomial distribution is appropriate here because each throw is independent of the others. Each throw basically resets the chance of rolling a 6 to \( \frac{1}{6} \). The number of trials is finite (four throws), making this a perfect setup for using the binomial formula. We calculate the probability for a certain number of successes, \( k \), among the \( n \) trials. Each has a success probability \( p \) with the formula:\[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \]Where \( \binom{n}{k} \) is the binomial coefficient, denoting how many ways we can choose \( k \) successes out of \( n \) trials. In this scenario, the terms \( (1-p) \) and \( p \) represent the probabilities of failure and success respectively.

Probability of Rolling a Die

Understanding the probability of rolling a die is essential for solving problems like the one at hand. A standard six-sided die has six possible outcomes: 1, 2, 3, 4, 5, and 6.

Each side has an equal chance of landing face up when the die is rolled. Therefore, the probability of any specific number, such as rolling a 6, is \( \frac{1}{6} \). This value comes from dividing one successful outcome (rolling a 6) by the six possible outcomes (1 through 6).

The probability of not rolling a 6, which includes any of the outcomes 1, 2, 3, 4, or 5, is calculated by taking the complement of rolling a 6, which comes out to \( 1 - \frac{1}{6} = \frac{5}{6} \).

Being consistent in understanding both probabilities (rolling a 6 and not rolling a 6) is crucial for breaking down the problem into manageable parts. It helps us confidently apply the binomial formula and evaluate cases like having 0, 1, or 2 rolls of 6 in four attempts.

Mathematical Problem Solving

Mathematical problem solving is a skill that builds on methodical approaches and logical reasoning. Let’s break down our problem-solving process for this exercise so it becomes more intuitive.

First, clearly define what the problem is asking. Here, we need to find the probability of getting at most two 6s in four throws. This requires us to think about the various potential outcomes and categorize them into zero, one, or two 6s.

Second, identify known probabilities and necessary formulas. Knowing that the event follows a binomial distribution allows us to apply the binomial probability formula. This step involves recognizing:

• The number of trials \( n = 4 \)
• The probability of success \( p = \frac{1}{6} \)
• The different cases to evaluate: \( X = 0, 1, \) and \( 2 \)

For each case, calculate the probability using the binomial formula,
then sum these probabilities to get the total probability.

Finally, after computation, compare your result to the provided options. By following this structured solving method, you not only solve the problem effectively but enhance your understanding and confidence in tackling similar problems in the future.