Problem 34 - Algebra 2 Course in Mathematics for the IIT-JEE and Other Engineering Exams

Question

A pair of fair dice is thrown independently, 4 times. The probability of getting a sum of exactly 7 twice is: (a) \(\frac{5}{81}\) (b) \(\frac{25}{243}\) (c) \(\frac{25}{216}\) (d) \(\frac{125}{648}\)

Step-by-Step Solution

Answer

The probability is \( \frac{25}{216} \), answer (c).

Step 1: Calculate the probability of getting a sum of 7 with a pair of dice

To find the probability of rolling a sum of 7 with two dice, count the favorable outcomes. There are 6 outcomes: (1,6), (2,5), (3,4), (4,3), (5,2), and (6,1). The total number of possible outcomes with two dice is 36 (6*6). Thus, the probability of rolling a sum of 7 in one roll is \( \frac{6}{36} = \frac{1}{6} \).

Step 2: Determine the probability of not getting a sum of 7

The probability of not rolling a sum of 7 is the complement of the probability of rolling a sum of 7. Hence, this probability is \( 1 - \frac{1}{6} = \frac{5}{6} \).

Step 3: Use Binomial probability formula

We need the probability of rolling a sum of 7 exactly twice in 4 throws. Use the binomial probability formula: \( P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \), where \( n = 4 \), \( k = 2 \), and \( p = \frac{1}{6} \).

Step 4: Calculate the required probability

Using the binomial formula, substitute the values: \( P(X = 2) = \binom{4}{2} \left(\frac{1}{6}\right)^2 \left(\frac{5}{6}\right)^2 \). Calculate each part: \( \binom{4}{2} = 6 \), \( \left(\frac{1}{6}\right)^2 = \frac{1}{36} \), and \( \left(\frac{5}{6}\right)^2 = \frac{25}{36} \). So, \( P(X = 2) = 6 \cdot \frac{1}{36} \cdot \frac{25}{36} = \frac{150}{1296} = \frac{25}{216} \).

Key Concepts

Binomial Probability

Binomial probability is used to determine the likelihood of a specific number of successes in a fixed number of independent trials. Each trial must have only two possible outcomes: success or failure. In the context of rolling dice, getting a sum of 7 is a success.
First, we must decide what defines a "success" and how often we expect it to occur. With dice, a success is rolling a sum of 7. This occurs with a probability of \(\frac{1}{6}\).

• In this exercise, we throw the dice 4 times. The number of successes we want is rolling a sum of 7 exactly twice.
• The formula to solve this is called the Binomial Probability Formula, which is: \[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \]
• Here, \(n\) is the total number of trials (4), \(k\) is the number of desired successes (2), and \(p\) is the probability of a single success (\(\frac{1}{6}\)).

This calculation results in the probability of two successful sum-of-7 rolls in four attempts.

Combinatorics

Combinatorics is a core part of probability that involves counting the number of possible ways an event can occur. The concept helps break down how arrangements and choices can be put together.
In our dice example:

• We're interested in counting how many ways we can achieve exactly two successful rolls (sums of 7) in four dice tosses.
• This is determined using combinations, which calculate the number of possible groupings without regard to order. For 4 trials and 2 successes, it is represented by \(\binom{4}{2}\).
• The combination formula is: \[ \binom{n}{k} = \frac{n!}{k!(n-k)!} \]
• Calculating \(\binom{4}{2}\), we find there are 6 ways to choose 2 successes out of 4 trials.

This shows that combinatorics aids in finding the number of success patterns possible within the constraints of a problem.

Complementary Probability

Complementary probability is all about finding the likelihood of the opposite of a desired event. It simplifies calculations by subtracting the probability of an event from 1.
For instance, if the chance of rolling a 7 on two dice is \(\frac{1}{6}\), then the probability of not rolling a 7 is the complement, which is \(1 - \frac{1}{6} = \frac{5}{6}\).

• Knowing one probability instantly allows us to calculate the other, using the complement rule.
• In our exercise, understanding complementary probability helps configure the equation, where \((1-p)^{n-k}\) accounts for the failures.
• The complement simplifies complex outcomes in problems with multiple steps, ensuring no possible scenario remains unconsidered.

This understanding makes solving probability problems more intuitive and accurate.