Q.An unbiased die is thrown again and again until three sixes are obtained. Find the probability of obtaining a third six in the sixth throw of the die.

Geometric Distribution

Imagine you're rolling a fair die, waiting for a 6 to show up. You roll once — not a 6. Roll again — still not a 6. You keep going. The question is: how many rolls until you finally see that first 6?

That "number of trials until the first success" is exactly what the geometric distribution describes.

The Intuition

The geometric distribution models a very simple experiment: you repeat an independent trial (like rolling a die, flipping a coin, testing a light bulb) where each trial has the same probability $p$ of "success." You stop the moment the first success occurs. The random variable $X$ is the number of trials needed to get that first success.

Key features:

• Trials are independent — the outcome of one trial doesn't affect the next.
• Each trial has exactly two outcomes: success (probability $p$) or failure (probability $q = 1-p$).
• You keep going until success happens, no matter how long it takes.

So if $X = 5$, it means you failed 4 times in a row and then succeeded on the 5th trial.

The Probability Formula

What's the chance that the first success occurs on the $k$-th trial? You need $k-1$ failures followed by one success. Since trials are independent, multiply the probabilities:

That's the geometric distribution.

Why It's Called "Geometric"

The probabilities form a geometric sequence: each term is $(1-p)$ times the previous one.

The ratio between consecutive probabilities is constant: $\frac{P(X=k+1)}{P(X=k)} = 1-p$.

Expected Value (Mean)

How many trials do you expect to wait on average? Intuition says: if success probability is $p$, you'd expect about $1/p$ trials. For a fair die ($p = 1/6$), you expect about 6 rolls. That's exactly right:

Variance

The smaller $p$ is, the larger the variance — which makes sense: rare successes mean you could get lucky early or wait a very long time.

A Concrete Example

Suppose a student has a 20% chance of passing a driving test on any given attempt ($p = 0.2$). What's the probability they pass on their third attempt?

So about a 12.8% chance. And the expected number of attempts is $1/0.2 = 5$. …