A fair die is tossed repeatedly until a six is obtained. Let X denote the number of tosses required.

Q: Wait, why do we use $q^2$ for $P(X \ge 3)$? Shouldn't we sum up all the probabilities from $X=3$ to infinity?

You absolutely could! Summing the infinite geometric series $\sum_{k=3}^{\infty} q^{k-1}p$ will indeed give you $q^2$. However, think of it logically: if the first six happens at or after the 3rd toss, it simply means the first two tosses were not sixes. The outcome of the 3rd toss and beyond doesn't matter for the condition to be met. It is much faster to think about what *must* happen (two failures) rather than what *could* happen.

Q: Why is the conditional probability $P(X \ge 6 \mid X > 3)$ equal to $P(X \ge 3)$? Is that a coincidence?

It is not a coincidence at all! This is the famous 'Memoryless Property' of the Geometric Distribution. Because each die toss is independent, the die doesn't 'remember' that you already failed three times. Starting from the 4th toss, the probability of waiting another 3 tosses to get a six is exactly the same as the probability of waiting 3 tosses from the very beginning.

Q: How do I know when to use $P(X \ge k)$ versus $P(X > k)$?

Great question! $P(X \ge k)$ means the first success occurs at $k, k+1, k+2, \dots$, which implies the first $k-1$ tosses were failures. Thus, $P(X \ge k) = q^{k-1}$. Conversely, $P(X > k)$ means the first success occurs at $k+1, k+2, \dots$, which implies the first $k$ tosses were failures. Thus, $P(X > k) = q^k$. Always check if the $k$-th toss is included in your 'failure' count!

Q: Can I use the formula for $P(X=k)$ to solve for $P(X \ge 3)$?

Yes, you can! $P(X \ge 3) = 1 - P(X < 3) = 1 - [P(X=1) + P(X=2)]$. If you calculate $1 - (p + qp)$, you will arrive at $q^2$. Both methods are mathematically sound, but the 'failure' logic is often less prone to arithmetic errors.

Q: Is the die toss always a Geometric Distribution?

It is, provided two conditions are met: the trials are independent, and the probability of success remains constant. Since a fair die doesn't change its face distribution after a roll, these conditions hold perfectly. It is the quintessential example of a discrete memoryless process!

Question

Passage:
A fair die is tossed repeatedly until a six is obtained. Let $X$ denote the number of tosses required.

Question 1:
The probability that $X = 3$ equals
(A) 25/216
(B) 25/36
(C) 5/36
(D) 125/216

Question 2:
The probability that $X \ge 3$ equals
(A) 125/216
(B) 25/36
(C) 5/36
(D) 25/216

Question 3:
The conditional probability that $X \ge 6$ given $X > 3$ equals
(A) 125/216
(B) 25/216
(C) 5/36
(D) 25/36

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are sitting at a table with a single, fair six-sided die. You are tasked with a simple yet profound challenge: keep rolling until that elusive 'six' finally appears.

This isn't just a game; it is a gateway into the elegant world of probability theory. When we repeat independent trials—each with a fixed probability of success—until the first success occurs, we enter the domain of the **Geometric Distribution**.

### The Anatomy of a Trial

Let $X$ be the random variable representing the number of tosses required to see our first six. In every toss, there are only two paths: Success ($S$), where we roll a six, and Failure ($F$), where we roll anything else.

Since the die is fair, the probability of success is $p = \frac{1}{6}$, and the probability of failure is $q = 1 - p = \frac{5}{6}$.

When we ask for the probability that $X = 3$, we are asking for a specific narrative: the first toss is a failure, the second is a failure, and the third is a success. Because each toss is independent, we multiply these probabilities:

$$P(X = 3) = P(F) 	imes P(F) 	imes P(S) = q^2 p$$

Substituting our values, we get:

$$P(X = 3) = \left(\frac{5}{6}ight)^2 	imes \frac{1}{6} = \frac{25}{216}$$

This is the probability of the 'first six' appearing exactly on the third attempt.

### The Power of 'At Least'

Now, let us shift our perspective. What if we want $P(X \ge 3)$? This means the first six occurs on the 3rd, 4th, 5th, or any subsequent toss.

If you think about it, this is equivalent to saying that the first two tosses *must* be failures. It doesn't matter what happens after the second toss; as long as the first two are not sixes, the condition $X \ge 3$ is satisfied.

Mathematically, this simplifies beautifully:

$$P(X \ge 3) = q^2 = \left(\frac{5}{6}ight)^2 = \frac{25}{36}$$

This reveals a powerful general rule: for any integer $k \ge 1$, the probability that we wait at least $k$ tosses is $P(X \ge k) = q^{k-1}$. Similarly, the probability that we wait *more than* $k$ tosses is $P(X > k) = q^k$.

These formulas are the keys to unlocking complex conditional probability problems.

### The Memoryless Mystery

Finally, we arrive at the most fascinating part of our journey: the conditional probability $P(X \ge 6 \mid X > 3)$. We want to know the probability that we need at least 6 tosses, given that we have already failed for the first 3 tosses.

Using the definition of conditional probability, we have:

$$P(X \ge 6 \mid X > 3) = \frac{P(X \ge 6 \cap X > 3)}{P(X > 3)}$$

Since the event $X \ge 6$ is a subset of $X > 3$ (if you need at least 6 tosses, you have certainly already exceeded 3), the intersection is simply $P(X \ge 6)$. Thus, our ratio becomes:

$$\frac{P(X \ge 6)}{P(X > 3)} = \frac{q^5}{q^3} = q^2 = \frac{25}{36}$$

Notice the magic here: $P(X \ge 6 \mid X > 3) = P(X \ge 3)$. The die has no memory!

The fact that you have already failed three times does not change the probability of your future success. The process 'resets' at every step. This is the **Memoryless Property**, a cornerstone of stochastic processes.

Whether you are waiting for a bus, a radioactive decay, or a six on a die, the future remains independent of the past. Keep this intuition in your heart, and no probability problem will ever feel insurmountable again.

Animated Solution for Mathematics - Probability: Comprehension Passage

The probability that $X = 3$ equals

Select Answer:

The probability that $X \geq 3$ equals

Select Answer:

The conditional probability that $X \geq 6$ given $X > 3$ equals

Select Answer:

Visualized Solution

Defining the Random Variable $X$

Success and Failure Probabilities

Finding $P (X = 3)$

Calculating $P (X = 3)$

Finding $P (X \geq 3)$

Calculating $P (X \geq 3)$

General Formula for $P (X \geq k)$

Conditional Probability $P (X \geq 6 ∣ X > 3)$

Calculating $P (X \geq 6 ∣ X > 3)$

The Memoryless Property

The Sigma Insight: Random Variables and Probability Distributions

Analyzing the Setup

The Anatomy of a Trial

The Power of 'At Least'

The Memoryless Mystery

Similar Questions

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 :