Question

A fair die is tossed until six is obtained on it. Let X be the number of required tosses, then the conditional probability P(\(X \ge 5 | X>2\)) is :

• A. \(\frac{11}{36}\)
• B. \(\frac{25}{36}\)
• C. \(\frac{5}{6}\)
• D. \(\frac{125}{216}\)

Hint:

The memoryless property of the geometric distribution is a very powerful shortcut. It essentially says that if you haven't succeeded yet, the probability of future outcomes is the same as if you were starting from scratch. Recognizing this can save significant calculation time.

Answer 1

The Correct Option is B

To solve this problem, we will use the concept of conditional probability in combination with knowledge about geometric distribution since a fair die is involved.

Let \(X\) be a random variable representing the number of tosses needed to get a six. The probability of getting a six on a single toss, \(p\), is \(\frac{1}{6}\) while the probability of not getting a six, \(q\), is \(\frac{5}{6}\).

The probability mass function (PMF) of a geometric distribution for \(X = n\) is: 

\(P(X=n) = q^{n-1} \cdot p\)

We need to find \(P(X \ge 5 | X > 2)\). Using the definition of conditional probability, we have:

\(P(X \ge 5 | X > 2) = \frac{P(X \ge 5 \cap X > 2)}{P(X > 2)}\)

Since \(X \ge 5\) implies \(X > 2\), we have \(P(X \ge 5 \cap X > 2) = P(X \ge 5)\). Thus:

\(P(X \ge 5 | X > 2) = \frac{P(X \ge 5)}{P(X > 2)}\)

Now, calculate \(P(X \ge 5)\):

\(P(X \ge 5) = 1 - P(X < 5)\)

\(P(X < 5) = P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)\)

Substituting the PMF values:

• \(\(P(X = 1) = \frac{1}{6}\)\)
• \(\(P(X = 2) = \frac{5}{6} \cdot \frac{1}{6} = \frac{5}{36}\)
• \(\(P(X = 3) = \left(\frac{5}{6}\right)^2 \cdot \frac{1}{6} = \frac{25}{216}\)
• \(\(P(X = 4) = \left(\frac{5}{6}\right)^3 \cdot \frac{1}{6} = \frac{125}{1296}\)

Add these probabilities:

\(P(X < 5) = \frac{1}{6} + \frac{5}{36} + \frac{25}{216} + \frac{125}{1296} = \frac{671}{1296}\)

Thus, \(P(X \ge 5) = 1 - \frac{671}{1296} = \frac{625}{1296}\).

Now, calculate \(P(X > 2)\):

\(P(X > 2) = 1 - P(X \le 2)\)

\(= 1 - \left(P(X = 1) + P(X = 2)\right)\)

\(= 1 - \left(\frac{1}{6} + \frac{5}{36}\right) = \frac{25}{36}\)

Finally, substitute these values back into the conditional probability formula:

\(P(X \ge 5 | X > 2) = \frac{\frac{625}{1296}}{\frac{25}{36}} = \frac{625}{900} = \frac{25}{36}\)

Therefore, the conditional probability \(P(X \ge 5 | X > 2)\) is \(\frac{25}{36}\).