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

Correct Answer: 12

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 \( \frac{b + c}{a} \):  
\( \frac{b + c}{a} = 12 \)