A biased coin is tossed repeatedly until a tail appears for the first time. Heads is 3 times as likely to appear as tails. Let x be the number of tosses required. Assume that all the trials of tossing a biased coin are independent. Then, the conditional probability that ` xge6`, gives that `x gt3`, is equal to

Options

A. `(9)/(16)`
B. `(1)/(16)`
C. `(1)/(2)`
D. `(3)/(5)`

Answer

To solve the problem, we need to find the conditional probability \( P(X \geq 6 | X > 3) \), where \( X \) is the number of tosses required until the first tail appears.

### Step-by-Step Solution:

1. **Define the probabilities of heads and tails:**
Since heads is 3 times as likely to appear as tails, we can denote:
- Let the probability of tails \( P(T) = p \).
- Then, the probability of heads \( P(H) = 3p \).
- Since the total probability must equal 1, we have:
\[ P(H) + P(T) = 1 \implies 3p + p = 1 \implies 4p = 1 \implies p = \frac{1}{4} \]
- Therefore, \( P(T) = \frac{1}{4} \) and \( P(H) = \frac{3}{4} \).

2. **Calculate \( P(X \geq 6) \):**
The event \( X \geq 6 \) means that we have at least 5 heads before the first tail appears. The probability of getting 5 heads followed by a tail is:
\[ P(X \geq 6) = P(H)^5 \cdot P(T) = \left(\frac{3}{4}\right)^5 \cdot \left(\frac{1}{4}\right) \]
- This can be simplified:
\[ P(X \geq 6) = \left(\frac{3}{4}\right)^5 \cdot \frac{1}{4} = \frac{243}{1024} \cdot \frac{1}{4} = \frac{243}{4096} \]

3. **Calculate \( P(X > 3) \):**
The event \( X > 3 \) means that we have at least 3 heads before the first tail appears. The probability of getting 3 heads followed by a tail is:
\[ P(X > 3) = P(H)^3 \cdot P(T) + P(H)^4 \cdot P(T) + P(H)^5 \cdot P(T) + \ldots \]
- This can be expressed as a geometric series:
\[ P(X > 3) = P(T) \sum_{n=3}^{\infty} P(H)^n = \frac{1}{4} \sum_{n=3}^{\infty} \left(\frac{3}{4}\right)^n \]
- The sum of the infinite series \( \sum_{n=3}^{\infty} r^n = \frac{r^3}{1 - r} \) where \( r = \frac{3}{4} \):
\[ \sum_{n=3}^{\infty} \left(\frac{3}{4}\right)^n = \frac{\left(\frac{3}{4}\right)^3}{1 - \frac{3}{4}} = \frac{\frac{27}{64}}{\frac{1}{4}} = \frac{27}{16} \]
- Therefore:
\[ P(X > 3) = \frac{1}{4} \cdot \frac{27}{16} = \frac{27}{64} \]

4. **Calculate the conditional probability \( P(X \geq 6 | X > 3) \):**
Using the definition of conditional probability:
\[ P(X \geq 6 | X > 3) = \frac{P(X \geq 6 \cap X > 3)}{P(X > 3)} = \frac{P(X \geq 6)}{P(X > 3)} \]
- We already calculated \( P(X \geq 6) = \frac{243}{4096} \) and \( P(X > 3) = \frac{27}{64} \):
\[ P(X \geq 6 | X > 3) = \frac{\frac{243}{4096}}{\frac{27}{64}} = \frac{243}{4096} \cdot \frac{64}{27} = \frac{243 \cdot 64}{4096 \cdot 27} \]
- Simplifying gives:
\[ = \frac{243 \cdot 64}{4096 \cdot 27} = \frac{243 \cdot 64}{4096 \cdot 27} = \frac{9}{16} \]

### Final Answer:
Thus, the conditional probability \( P(X \geq 6 | X > 3) = \frac{9}{16} \).