Determine P (E|F) in : A die is thrown three times, E : 4 appears on the third toss, F: 6 and 5 appears respectively on first two tosses.Determine P(E∣F)

Explanation

To determine \( P(E|F) \), we will follow these steps:

### Step 1: Define the Events
- Let \( E \) be the event that a 4 appears on the third toss.
- Let \( F \) be the event that a 6 appears on the first toss and a 5 appears on the second toss.

### Step 2: Calculate the Total Sample Space
When a die is thrown three times, the total number of outcomes is:
\[ 6 \times 6 \times 6 = 216 \]
So, the total sample space \( S \) has 216 outcomes.

### Step 3: Determine \( P(F) \)
The event \( F \) consists of outcomes where the first toss is 6 and the second toss is 5. The third toss can be any of the 6 faces of the die. Therefore, the outcomes for event \( F \) are:
- \( (6, 5, 1) \)
- \( (6, 5, 2) \)
- \( (6, 5, 3) \)
- \( (6, 5, 4) \)
- \( (6, 5, 5) \)
- \( (6, 5, 6) \)

Thus, there are 6 outcomes in event \( F \). Therefore, the probability of \( F \) is:
\[ P(F) = \frac{6}{216} = \frac{1}{36} \]

### Step 4: Determine \( P(E \cap F) \)
The intersection \( E \cap F \) consists of outcomes where the third toss is 4, and the first two tosses are 6 and 5 respectively. The only outcome that satisfies both \( E \) and \( F \) is:
- \( (6, 5, 4) \)

Thus, there is 1 outcome in \( E \cap F \). Therefore, the probability of \( E \cap F \) is:
\[ P(E \cap F) = \frac{1}{216} \]

### Step 5: Calculate \( P(E|F) \)
Using the formula for conditional probability:
\[ P(E|F) = \frac{P(E \cap F)}{P(F)} \]
Substituting the values we found:
\[ P(E|F) = \frac{\frac{1}{216}}{\frac{1}{36}} = \frac{1}{216} \times \frac{36}{1} = \frac{36}{216} = \frac{1}{6} \]

### Final Answer
Thus, the probability \( P(E|F) \) is:
\[ \boxed{\frac{1}{6}} \]