A fair die is tossed until six is obtained on it. Let X be the number of required tosses, then the c

Q: Wait, why is $X \ge 5$ the same as $X > 4$?

That is a fantastic observation! Since $X$ represents the number of tosses, it must be a positive integer. If $X$ is at least 5, the smallest possible value is 5. If $X$ is strictly greater than 4, the smallest possible integer value is also 5. Thus, for discrete variables like this, $X \ge 5$ and $X > 4$ are mathematically identical.

Q: Why can't we just calculate the individual probabilities for $X=5, 6, 7...$ and sum them up?

You absolutely can! That would involve summing an infinite geometric series: $\sum_{k=5}^{\infty} p q^{k-1}$. However, using the property $P(X > n) = q^n$ is a much more elegant and faster way to reach the answer, and it helps you avoid the common pitfalls of infinite series calculations.

Q: What is this 'memoryless property' mentioned in the solution?

The memoryless property is a unique feature of the geometric distribution. It means that the probability of needing more tosses does not depend on how many tosses you have already made. If you have already failed twice, the probability of needing at least three more tosses is exactly the same as the probability of needing at least three tosses from the very beginning. It is as if the die 'forgets' its past failures!

Question

A fair die is tossed until six is obtained on it. Let $X$ be the number of required tosses, then the conditional probability $P(X \ge 5 | X > 2)$ is :

Options:

(A) $\frac{125}{216}$

(B) $\frac{11}{36}$

(C) $\frac{5}{6}$

(D) $\frac{25}{36}$

Sigma Percentile · Accepted Answer

### The Waiting Game: Unlocking the Memoryless Property

Imagine you are sitting at a table with a single, fair six-sided die. Your goal is simple: roll a six. But there is a catch—you must keep rolling until that six appears.

The number of rolls it takes is our random variable, $X$. This is the classic setup for a **Geometric Distribution**, a beautiful mathematical model that describes the waiting time for the first success in a series of independent trials.

### The Geometric Nature of the Experiment

In this experiment, the probability of success (rolling a six) is $p = \frac{1}{6}$. Consequently, the probability of failure (rolling anything else) is $q = 1 - p = \frac{5}{6}$.

When we ask about the probability that the first success occurs after $n$ trials, we are essentially asking for the probability of $n$ consecutive failures. Since each roll is independent, we simply multiply the probabilities:

$$P(X > n) = q^n$$

This formula is your best friend in these types of problems.

### Navigating the Condition

The problem asks for the conditional probability $P(X \ge 5 \mid X > 2)$. This notation can look intimidating, but let's break it down.

We are given that $X > 2$, meaning we already know the first two rolls were not sixes. We are now standing at the start of the third roll, looking forward. We want to know the probability that we will need at least five total rolls, given that we have already failed twice.

Using the conditional probability formula, we have:

$$P(X \ge 5 \mid X > 2) = \frac{P(X \ge 5 \cap X > 2)}{P(X > 2)}$$

### The Intersection and Simplification

Look at the numerator: $P(X \ge 5 \cap X > 2)$. If $X$ is greater than or equal to 5, it is automatically greater than 2.

Therefore, the intersection of these two events is simply $P(X \ge 5)$. As we discussed, since $X$ must be an integer, $X \ge 5$ is equivalent to $X > 4$. Our expression now becomes:

$$\frac{P(X > 4)}{P(X > 2)}$$

Using our handy formula $P(X > n) = q^n$, we substitute $n=4$ for the numerator and $n=2$ for the denominator:

$$\frac{q^4}{q^2} = q^2$$

### The Beauty of Memorylessness

Notice what happened? The result is $q^2$. This is the **memoryless property** in action!

The probability of needing at least two more rolls (given we have already failed twice) is exactly the same as the probability of needing at least two rolls from the very start. The past two failures have vanished from the equation.

We are left with a simple calculation:

$$q^2 = \left(\frac{5}{6}\right)^2 = \frac{25}{36}$$

And there you have it! The probability is **$\frac{25}{36}$**. By understanding the geometric nature of the experiment and the memoryless property, we transformed a potentially complex conditional probability problem into a simple algebraic one.

Animated Solution for Mathematics - Probability: A fair die is tossed until six is obtained on it. Let $X$ be the number of required tosses, then the conditional probability $P (X \geq 5∣ X > 2)$ is :

Select Answer:

Visualized Solution

The Experiment

Geometric Distribution

Probability of $X > n$

The Given Condition

The Target Event

Conditional Probability Formula

Applying the Formula

Finding the Intersection

Simplifying the Numerator

Substituting the $q^{n}$ Formula

Algebraic Simplification

Final Substitution

Final Answer

The Sigma Insight: Random Variables and Probability Distributions

The Waiting Game

The Geometric Nature of the Experiment

Navigating the Condition

The Intersection and Simplification

The Beauty of Memorylessness

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