Problem 20 - Algebra 2 Course in Mathematics for the IIT-JEE and Other Engineering Exams

Question

A die is thrown three times. Getting a 3 or a 6 is considered a success. Then the probability of at least 2 successes is (a) \(2 / 9\) (b) \(7 / 27\) (c) \(1 / 27\) (d) None of these

Step-by-Step Solution

Answer

The probability is \( \frac{7}{27} \), which corresponds to option (b).

Step 1: Define the concept of success

A success is when the die shows a 3 or 6. This means to have a success on a single throw, the results should be a 3 or 6.

Step 2: Calculate the probability of success on one throw

The die has six faces (1 through 6). The numbers 3 and 6 represent successful outcomes. Thus, the probability of success on any single throw is \( \frac{2}{6} = \frac{1}{3} \).

Step 3: Calculate the probability of failure on one throw

A failure occurs when the die shows any number other than 3 or 6, which happens with probability \( \frac{4}{6} = \frac{2}{3} \).

Step 4: Use binomial probability formula

The number of successes in 3 throws follows a binomial distribution, where the probability of success is \( p = \frac{1}{3} \) and the number of trials is \( n = 3 \). We are looking for the probability of at least 2 successes.

Step 5: Calculate the probability of exactly 0 successes

\[ P(X = 0) = \binom{3}{0} \left(\frac{1}{3}\right)^0 \left(\frac{2}{3}\right)^3 = 1 \cdot 1 \cdot \frac{8}{27} = \frac{8}{27} \]

Step 6: Calculate the probability of exactly 1 success

\[ P(X = 1) = \binom{3}{1} \left(\frac{1}{3}\right)^1 \left(\frac{2}{3}\right)^2 = 3 \cdot \frac{1}{3} \cdot \frac{4}{9} = \frac{12}{27} \]

Step 7: Calculate probability of at least 2 successes

Since we need the probability of at least 2 successes, we can do this by taking the complement of the combined probabilities of having 0 or 1 success: \[ P(X \geq 2) = 1 - P(X = 0) - P(X = 1) = 1 - \frac{8}{27} - \frac{12}{27} = \frac{7}{27} \]

Step 8: Interpret the result

The probability of getting at least 2 successes in 3 throws of the die is \( \frac{7}{27} \).

Key Concepts

Probability of Success

When we talk about the probability of success, we mean the likelihood that a specific event will happen. In our original exercise, "success" is defined as rolling a 3 or a 6 on a die. The probability of success on a single throw is calculated by dividing the number of successful outcomes by the total number of possible outcomes. In this case, there are 2 successful outcomes (rolling a 3 or a 6) out of 6 possible outcomes (since a die has 6 faces): \[ P(\text{success}) = \frac{2}{6} = \frac{1}{3} \] This fraction means that for every attempt, there's a one-in-three chance of rolling a 3 or a 6. Remember that probability is always a number between 0 and 1, where 0 means the event is impossible, and 1 means it is certain.Because each roll of the die is an independent event, the outcome of one roll does not affect the next. This is a crucial detail in probability that ensures our calculations remain consistent across multiple trials.

Binomial Distribution

The binomial distribution is a probability distribution that summarizes the likelihood that a value will take one of two independent states across several trials or experiments. In the context of the die-throwing exercise, the binomial distribution helps us determine the probability of achieving a given number of successes (rolling a 3 or 6), in a series of dice throws.Key features of the binomial distribution include:

• The number of trials, \( n \), must be fixed. Here, \( n = 3 \) because the die is thrown three times.
• Each trial must have two possible outcomes: success or failure.
• The probability of success, \( p \), must remain constant across all trials.

The formula for the binomial distribution is: \[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \]Where:

• \( n \) is the number of trials
• \( k \) is the number of successes
• \( p \) is the probability of success on an individual trial
• \( (1-p) \) is the probability of failure

Understanding binomial distribution makes it easier to calculate probabilities for specific outcomes over multiple trials, much like gambling scenarios or quality control processes.

Complement Rule

The complement rule in probability is a useful technique used to find the probability of the event "at least" one outcome by calculating the "not" case first. When tasked with finding the probability of at least 2 successes in the exercise, we can apply the complement rule:1. Calculate the probability of the complement event (0 or 1 success), which is usually simpler.2. Subtract the probability of the complement event from 1 to find the desired probability.Applying this to our problem:

• Calculate \( P(X = 0) \), the probability of no successes.
• Calculate \( P(X = 1) \), the probability of exactly one success.
• Use the complement rule: \( P(X \geq 2) = 1 - P(X = 0) - P(X = 1) \)

This method leverages the idea that the sum of probabilities of all possible outcomes of an event must equal 1. Therefore, the probability of an event occurring plus the probability of it not occurring will always add up to 1. It often simplifies calculations, especially in "at least" scenarios.

Probability Calculation

Calculating probability systematically often involves using a combination of mathematical principles. In our exercise, we calculated the probability for different outcomes for a series of die throws.First, calculate the probability for the least likely events (0 and 1 successes), then use these probabilities to find at least 2 successes:

• Calculate for no successes: \( P(X = 0) \)
• Calculate for exactly one success: \( P(X = 1) \)
• Apply the complement rule to find the probability of at least 2 successes: \( P(X \geq 2) = 1 - P(X = 0) - P(X = 1) \)

For each scenario, use the binomial formula:\[ P(X = k) = \binom{n}{k} \left(\frac{1}{3}\right)^k \left(\frac{2}{3}\right)^{n-k} \] Where you substitute values for \( n \), \( k \), \( p \).By structuring the problem this way—using combinations, probabilities, and complement rules—we can effectively evaluate the probabilities of complex events. This structured approach is particularly useful in statistics and can be applied to a wide range of real-life situations where we assess the likelihood of various outcomes.