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

The task is to calculate \(\frac{b+c}{a}\) where \(a = P(X=3)\), \(b = P(X \geq 3)\), and \(c = P(X \geq 6 | X > 3)\). The problem involves the geometric distribution, where each die roll is an independent event with a probability \(p = \frac{1}{6}\) for rolling a six, and \(q = \frac{5}{6}\) for not rolling a six.

1. Calculate \(a = P(X=3)\): 
The probability that the first two rolls are not six, and the third roll is six:
\(a = q^2 \times p = \left(\frac{5}{6}\right)^2 \times \frac{1}{6} = \frac{25}{216}\)

2. Calculate \(b = P(X \geq 3)\):
This is the probability that the first two rolls are not six:
\(b = q^2 = \left(\frac{5}{6}\right)^2 = \frac{25}{36}\)

3. Calculate \(c = P(X \geq 6 | X > 3)\):
Using the geometric distribution property,
\(c = q^{2} = \left(\frac{5}{6}\right)^2 = \frac{25}{36}\)

4. Calculate \( \frac{b+c}{a} \):
Plug in the values:
\(\frac{b+c}{a} = \frac{\frac{25}{36} + \frac{25}{36}}{\frac{25}{216}} = \frac{\frac{50}{36}}{\frac{25}{216}} = \frac{50}{36} \times \frac{216}{25} = \frac{50 \times 216}{36 \times 25} = \frac{10800}{900} = 12\)

The computed value is 12, which fits within the expected range of 12 to 12.

Answer 2

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