Problem 160 - A Complete Resource Book in Mathematics for JEE Main

Question

A pair of fair dice is thrown independently three times. The probability of getting a score of exactly 9 twice is [2007] (A) \(1 / 729\) (B) \(8 / 9\) (C) \(8 / 729\) (D) \(8 / 243\)

Step-by-Step Solution

Answer

The probability is (D) \( \frac{8}{243} \).

Step 1: Calculate Probability for One Roll

First, we find the probability of getting a score of exactly 9 in a single roll of two dice. To obtain a score of 9, the dice combinations are (3,6), (4,5), (5,4), and (6,3). There are 4 favorable outcomes. Since there are 36 possible outcomes when rolling two dice (6 sides on the first die and 6 sides on the second), the probability for one roll is \( \frac{4}{36} = \frac{1}{9} \).

Step 2: Define Binomial Distribution

Since the dice are rolled three times, we want exactly two of those rolls to result in a score of 9. This is a binomial probability distribution where \( n = 3 \) (rolls), \( k = 2 \) (successful rolls with score of 9), and the probability of success (getting a 9) for each roll \( p = \frac{1}{9} \).

Step 3: Use Binomial Probability Formula

The binomial probability formula is: \[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \]Substitute \( n = 3, \ k = 2, \ p = \frac{1}{9} \):- \( \binom{3}{2} = 3 \) (as there are 3 different successful results: \{ success, success, failure; success, failure, success; failure, success, success \})- \( \left( \frac{1}{9} \right)^2 = \frac{1}{81} \)- \( \left( \frac{8}{9} \right)^1 = \frac{8}{9} \) (probability of not scoring 9 in one roll) Therefore, the probability is: \[ P(\text{exactly 2 nines in 3 rolls}) = 3 \times \frac{1}{81} \times \frac{8}{9} \]

Step 4: Solve the Expression

Continue the calculation from the above step:\[ 3 \times \frac{1}{81} \times \frac{8}{9} = \frac{24}{729} \]This fraction \( \frac{24}{729} \) simplifies to \( \frac{8}{243} \).

Step 5: Select the Correct Answer

The solution for the probability of getting exactly a score of 9 twice is \( \frac{8}{243} \), which matches option (D).

Key Concepts

Binomial Distribution

The binomial distribution is a statistical method that models the probability of a specific number of successes in a fixed number of independent trials. Each trial has only two possible outcomes: success or failure. When it comes to rolling dice multiple times, each roll is an independent event.
For example, when rolling a pair of dice, obtaining a specific sum, like 9, is a success. If you roll these dice three times, you're conducting three independent trials. In this context:

• The number of trials, denoted by \( n \), is 3.
• The probability of success, denoted by \( p \), is the probability of rolling a 9 in one trial.
• The number of successful rolls you're interested in, denoted by \( k \), could be one or more, varying with your exercise.

Understanding binomial distribution helps quantify and analyze the likelihood of different outcomes, such as rolling a sum of 9 twice in three attempts.

Fair Dice

Fair dice are essential when discussing probabilities because they ensure each side has an equal chance of landing face up. This assumption of fairness is crucial in our calculations.

• A fair six-sided die will land on any number between 1 and 6 with equal probability, which is \( \frac{1}{6} \).
• When two fair dice are rolled together, the total outcomes increase: combining each outcome of one die with each outcome of another die results in 36 possible combinations.

This assumption of fairness allows us to use simple fractions to express probabilities when dealing with dice outcomes. For example, with two fair dice, the probability of any specific combination, such as (3,6), is \( \frac{1}{36} \). The reliability of fair dice simplifies complex probability models, especially in computing outcomes and patterns, such as the sum of numbers.

Probability of Sum on Dice

Calculating the probability of a sum, such as 9, when rolling two dice includes finding all possible favorable combinations. The sum is the total value from both dice. For example, to get a sum of 9:

• Possible outcomes are (3,6), (4,5), (5,4), and (6,3).
• These are the only pairs of numbers from two dice that add up to 9.
• There are 4 favorable outcomes.

Since there are 36 possible outcomes total (since each die has 6 sides), the probability of rolling a sum of 9 in one shot is \( \frac{4}{36} = \frac{1}{9} \).
If you want to find the probability of getting this sum in a series of rolls, you use these outcomes as the basis, using methods like the binomial distribution discussed earlier. This approach highlights the structured pathway to understanding probability with fair dice.