Question

A fair die is tossed repeatedly until a six is obtained. Let X denote the number of tosses required and let $a=P(X=3), b=P(X≥3)$, and $( c = P(X \geq 6|X>3).$ Then $\frac{b+c}{a}$ is equals to ____.

Answer 1

Solution:

Step 1. Calculate $a = P(X = 3)$:
$a = \frac{5}{6} \cdot \frac{5}{6} \cdot \frac{1}{6} = \frac{25}{216}$

Step 2. Calculate $b = P(X \geq 3)$:
$b = \frac{5}{6} + \frac{5}{6} \cdot \frac{5}{6} + \frac{5}{6} \cdot \frac{5}{6} \cdot \frac{1}{6} + \dots = \frac{25}{36}$

Step 3. Calculate $c = P(X \geq 6 \mid X \geq 3)$:
$c = \left(\frac{5}{6}\right)^3 \cdot \frac{1}{6} + \dots = \frac{25}{36}$

Step 4. Compute $\frac{b + c}{a}$:
$\frac{b + c}{a} = 12$