Probability That Die Tossed Until Six Requires At Least Three Tosses

To find the probability that $X \geq 3$, we analyze the random variable $X$, which denotes the number of tosses required until a six is obtained.

For a fair six-sided die:

• The probability of rolling a six on any given toss is $p = \frac{1}{6}$.
• The probability of not rolling a six on any given toss is $q = 1 - p = \frac{5}{6}$.

Since the tosses are independent, the random variable $X$ follows a geometric distribution with parameter $p = \frac{1}{6}$, where:
$P(X = k) = q^{k-1}p \quad \text{for } k = 1, 2, 3, \dots$

The event $X \geq 3$ means that the first six does not appear on the first or second toss. In other words, the first two tosses must both be failures (not a six):
$P(X \geq 3) = P(\text{no six on the 1st toss and no six on the 2nd toss})$

Using the complementary probability:
$P(X \geq 3) = 1 - P(X < 3) = 1 - [P(X = 1) + P(X = 2)]$

Substituting the probabilities:
$P(X = 1) = p = \frac{1}{6}$
$P(X = 2) = q \cdot p = \left(\frac{5}{6}\right)\left(\frac{1}{6}\right) = \frac{5}{36}$

Thus:
$P(X \geq 3) = 1 - \left(\frac{1}{6} + \frac{5}{36}\right) = 1 - \frac{11}{36} = \frac{25}{36}$

Equivalently, this is simply the probability that the first two tosses are not six:
$P(X \geq 3) = q^2 = \left(\frac{5}{6}\right)^2 = \frac{25}{36}$

Therefore, the correct option is B.