Conditional Probability That Number of Tosses Is At Least Six Given Greater Than Three

To find the conditional probability $P(X \geq 6 \mid X > 3)$, we first identify the probability distribution of the random variable $X$.

Let success be defined as rolling a six on a fair die.
The probability of success on any single toss is:
$p = \frac{1}{6}$
The probability of failure (not rolling a six) is:
$q = 1 - p = \frac{5}{6}$

The random variable $X$ represents the number of tosses until the first six appears, which follows a geometric distribution:
$P(X = k) = q^{k-1}p \quad \text{for } k = 1, 2, 3, \dots$

The probability that more than $n$ tosses are required is the probability that the first $n$ tosses are all failures:
$P(X > n) = q^n$

Since $X$ is an integer-valued random variable, we have:

• For $X > 3$, this corresponds to $n = 3$:
  $P(X > 3) = q^3 = \left(\frac{5}{6}\right)^3$

• For $X \geq 6$, this is equivalent to $X > 5$ (i.e., the first $5$ tosses are failures):
  $P(X \geq 6) = P(X > 5) = q^5 = \left(\frac{5}{6}\right)^5$

Since the event $\{X \geq 6\}$ is a subset of $\{X > 3\}$, their intersection is:
$\{X \geq 6\} \cap \{X > 3\} = \{X \geq 6\}$

Now, applying the definition of conditional probability:
$P(X \geq 6 \mid X > 3) = \frac{P(X \geq 6 \cap X > 3)}{P(X > 3)} = \frac{P(X \geq 6)}{P(X > 3)}$

Substituting the values:
$P(X \geq 6 \mid X > 3) = \frac{q^5}{q^3} = q^2 = \left(\frac{5}{6}\right)^2 = \frac{25}{36}$

Thus, the required conditional probability is $\frac{25}{36}$, which corresponds to option D.