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

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

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

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

4. Calculation of \( \frac{b+c}{a} \):
Substituting the computed 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 calculated result is 12, which falls within the expected range of 12 to 12.