Problem 151 - A Complete Resource Book in Mathematics for JEE Main

Question

A fair die is tossed eight times. The probability that a third six is observed on the eight throw, is: [2002] (A) \(\frac{{ }^{7} C_{2} \times 5^{5}}{6^{7}}\) (B) \(\frac{{ }^{7} C_{2} \times 5^{5}}{6^{8}}\) (C) \(\frac{{ }^{7} C_{2} \times 5^{5}}{6^{6}}\) (D) none of these

Step-by-Step Solution

Answer

Option (B), \(\frac{ ^{7}C_{2} \times 5^5 }{ 6^8 }\).

Step 1: Understanding the Problem

We are asked to find the probability that the third occurrence of a '6' happens exactly on the 8th toss of a fair die. This means there are exactly two '6's in the first seven tosses and the 8th toss must be a '6'.

Step 2: Choosing the Probability Formula

To find this probability, we use the formula for finding the exact term in a negative binomial distribution, which is suitable for finding the probability of getting the k-th success on the n-th trial. Here, the probability of rolling a '6' (success) is \(\frac{1}{6}\), and not rolling a '6' (failure) is \(\frac{5}{6}\).

Step 3: Setting Up the Terms

For the negative binomial probability, we need to calculate the number of ways to choose the positions of the first two '6's from the first seven tosses. This is given by \( ^{7}C_{2} \). Then, the remaining five positions must be non-six outcomes. Finally, the probability of '6' on the 8th toss is \(\frac{1}{6}\).

Step 4: Calculating the Probability

The probability that exactly two '6's occur in the first seven tosses is\( ^{7}C_{2} \left(\frac{1}{6}\right)^2 \left(\frac{5}{6}\right)^5 \). Then multiply this by the probability of a '6' on the 8th toss, \(\frac{1}{6}\). So the total probability is \( ^{7}C_{2} \left(\frac{1}{6}\right)^3 \left(\frac{5}{6}\right)^5 \). Simplifying gives: \(\frac{ ^{7}C_{2} \times 5^5 }{ 6^8 }\).

Step 5: Choosing the Correct Answer

The calculated probability \(\frac{ ^{7}C_{2} \times 5^5 }{ 6^8 }\) matches option (B).

Key Concepts

Probability Theory

Probability theory is the mathematical framework used to measure uncertainty and predict the likelihood of events. When a die is tossed, each outcome has a chance or probability of occurring which informs the overall predictions we make.

In our exercise, we need the probability of getting the third six on the eighth toss of a die. This falls under the category of probability theory as we're quantifying the likelihood of this specific outcome. The event itself is random, meaning there is an inherent chance associated with it.

• Each toss of the die is called a trial.
• A 'success' here is rolling a six.
• The specific condition is observing the third success on the eighth trial.

Understanding probability theory allows us to use the right models, like the negative binomial distribution in this situation, to calculate such probabilities successfully. Using this theory, we derive that the probability of early trials meeting certain conditions drastically influences the likelihood of subsequent events.

Combinatorics

Combinatorics is a branch of mathematics that deals with counting, organizing, and arranging elements within a set. It is crucial for determining the number of ways events can occur, particularly when certain conditions apply.

In our problem, combinatorics helps us calculate the number of ways to position the first two successes (rolling a six) within seven dice tosses. We use the combination formula, which is expressed as \[ ^{7}C_{2} = \frac{7!}{2!(7-2)!}\]Here, the '!' denotes a factorial, which means the product of all positive integers up to that number. This calculation tells us how many different sequences can produce two sixes in seven rolls.

• The combination is read as "seven choose two," indicating the different ways of selecting two specific trials to result in success.
• It's critical for laying out scenarios meeting the problem's conditions, like having exactly two wins before the last trial.

Through such combinatorial methods, we systematically understand how to organize and solve problems regarding complex event sequences.

Dice Probability

Dice probability refers to calculating the chance of one or more specific outcomes when rolling a die. Since a standard die has six sides, each face (numbers 1-6) has an equal probability of appearing: \[ P( ext{rolling a six}) = \frac{1}{6}\]And correspondingly, the probability of not rolling a six is: \[ P( ext{not a six}) = \frac{5}{6}\]

Our exercise focuses on finding when exactly rolling a six becomes the third occurrence at the eighth toss. This means:

• The eighth roll is a success \(\left(\frac{1}{6}\right)\).
• The five non-six outcomes must occur within the first seven tosses.

Dice probability is essential here to calculate the likelihood of sequences within the first seven rolls and then the final success on the eighth trial, laying the action framework through probability.