Question

An unbiased die is tossed until 5 appears. If \( X \) denotes the number of tosses required, \( \frac{25}{P(X=5)} \) is equal to

• A. \( \frac{25}{36} \)
• B. \( \frac{125}{216} \)
• C. \( \frac{216}{125} \)
• D. \( \frac{36}{25} \)
• E. \( \frac{216}{25} \)

Hint:

For repeated trials until success, always use geometric distribution formula.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
This problem describes a geometric distribution. A geometric distribution models the number of trials needed to get the first success in a series of independent Bernoulli trials. The random variable X is the number of tosses required to get the first '5'.
Step 2: Key Formula or Approach:
The probability mass function (PMF) for a geometric distribution is given by: \[ P(X = k) = (1-p)^{k-1}p \] where `p` is the probability of success on a single trial, and `k` is the number of trials until the first success. In this problem:
A 'success' is rolling a 5. The probability of success is \(p = \frac{1}{6}\).
The probability of 'failure' (not rolling a 5) is \(1-p = \frac{5}{6}\).
Step 3: Detailed Explanation:
1. Calculate P(X=2). This is the probability that the first toss is a failure and the second toss is a success. \[ P(X=2) = (1-p)^{2-1}p = \left(\frac{5}{6}\right)^1 \left(\frac{1}{6}\right) = \frac{5}{36} \] 2. Calculate P(X=5). This is the probability that the first four tosses are failures and the fifth toss is a success. \[ P(X=5) = (1-p)^{5-1}p = \left(\frac{5}{6}\right)^4 \left(\frac{1}{6}\right) \] 3. Calculate the ratio. \[ \frac{P(X=2)}{P(X=5)} = \frac{\left(\frac{5}{6}\right)^1 \left(\frac{1}{6}\right)}{\left(\frac{5}{6}\right)^4 \left(\frac{1}{6}\right)} \] The \(\left(\frac{1}{6}\right)\) term cancels out. \[ = \frac{\left(\frac{5}{6}\right)^1}{\left(\frac{5}{6}\right)^4} = \left(\frac{5}{6}\right)^{1-4} = \left(\frac{5}{6}\right)^{-3} \] \[ = \left(\frac{6}{5}\right)^3 = \frac{6^3}{5^3} = \frac{216}{125} \] Step 4: Final Answer:
The value of the ratio is \(\frac{216}{125}\).