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 :
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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.
Step 1: Understanding the Question
This is a problem based on the Geometric distribution. The random variable \(X\) is the number of trials required to get the first success. A success is getting a '6' on a fair die.
The probability of success is \(p = 1/6\).
The probability of failure is \(q = 1 - p = 5/6\).
The probability mass function is \(P(X=k) = q^{k-1}p\). Step 2: Key Formula or Approach
We need to find the conditional probability \(P(X \ge 5 | X>2)\).
The formula for conditional probability is \(P(A|B) = \frac{P(A \cap B)}{P(B)}\).
Let \(A\) be the event \(X \ge 5\) and \(B\) be the event \(X>2\).
The intersection \(A \cap B\) is the event that \(X\) is both greater than or equal to 5 AND greater than 2. This simplifies to just \(X \ge 5\).
So, we need to calculate \(\frac{P(X \ge 5)}{P(X>2)}\). Step 3: Detailed Explanation
The event \(X>k\) means that the first \(k\) tosses were failures. The probability of this is \(q^k\).
So, \(P(X>2)\) is the probability of not getting a 6 in the first two tosses, which is \(q^2 = (5/6)^2\).
The event \(X \ge 5\) is equivalent to the event \(X>4\). This means that the first 4 tosses were failures. The probability of this is \(q^4 = (5/6)^4\).
Now, we calculate the conditional probability:
\[ P(X \ge 5 | X>2) = \frac{P(X \ge 5)}{P(X>2)} = \frac{P(X>4)}{P(X>2)} = \frac{q^4}{q^2} = q^2 \]
\[ = \left(\frac{5}{6}\right)^2 = \frac{25}{36} \]
Alternative approach (Memoryless Property):
The geometric distribution has a memoryless property, which states that \(P(X>m+n | X>m) = P(X>n)\).
Let \(m=2\). We need \(P(X \ge 5 | X>2)\), which is \(P(X>2+2 | X>2)\).
Here \(n=2\). According to the property, this is equal to \(P(X>2)\).
\(P(X>2) = q^2 = (5/6)^2 = 25/36\). Step 4: Final Answer
The required probability is \(\frac{25}{36}\).
