A biased coin is tossed 10 times. The head is 2 times more likely to appear than the tail. The probability that `2^("nd")` tail and `4^("th")` tail occur at `4^("th")` and `10^("th")` tosses respectively is

Options

A. `(16)/(3^(9))`
B. `(320)/(3^(10))`
C. `(320)/(3^(9))`
D. `(160)/(3^(10))`

Answer

To solve the problem step by step, we need to find the probability that the 2nd tail occurs on the 4th toss and the 4th tail occurs on the 10th toss when tossing a biased coin 10 times. The probability of heads (H) is 2/3 and the probability of tails (T) is 1/3.

### Step 1: Define the probabilities
Given that the probability of heads (H) is 2/3 and the probability of tails (T) is 1/3, we can denote:
- P(H) = 2/3
- P(T) = 1/3

### Step 2: Determine the conditions
We need to ensure that:
- The 2nd tail occurs on the 4th toss.
- The 4th tail occurs on the 10th toss.

### Step 3: Analyze the first 4 tosses
For the first 4 tosses, we need to have exactly 2 tails, with the 2nd tail specifically on the 4th toss. This means that in the first 3 tosses, we must have exactly 1 tail.

The possible outcomes for the first 3 tosses can be represented as:
- T, H, H
- H, T, H
- H, H, T

### Step 4: Calculate the probability for the first 4 tosses
The probability of getting 1 tail (T) and 2 heads (H) in the first 3 tosses is calculated as follows:
- The probability of getting 1 tail and 2 heads in any order in the first 3 tosses is given by:

\[ \text{P(1 tail in first 3 tosses)} = \binom{3}{1} \cdot P(T)^1 \cdot P(H)^2 = 3 \cdot \left(\frac{1}{3}\right)^1 \cdot \left(\frac{2}{3}\right)^2 \]

Calculating this gives:
\[ = 3 \cdot \frac{1}{3} \cdot \frac{4}{9} = \frac{4}{9} \]

Now, since the 4th toss must be a tail (T), we multiply this by the probability of T:
\[ \text{Total probability for first 4 tosses} = \frac{4}{9} \cdot \frac{1}{3} = \frac{4}{27} \]

### Step 5: Analyze the last 6 tosses
Now, we need to ensure that the 4th tail occurs on the 10th toss. This means we need 2 more tails in the remaining 6 tosses (tosses 5 to 9).

### Step 6: Calculate the probability for the last 6 tosses
We need to find the probability of getting exactly 2 tails in the 6 tosses, with the last toss (10th) being a tail. The number of ways to choose 2 tails from the first 6 tosses is given by:
\[ \binom{6}{2} \cdot P(T)^2 \cdot P(H)^4 = \binom{6}{2} \cdot \left(\frac{1}{3}\right)^2 \cdot \left(\frac{2}{3}\right)^4 \]

Calculating this gives:
\[ = 15 \cdot \frac{1}{9} \cdot \frac{16}{81} = 15 \cdot \frac{16}{729} = \frac{240}{729} \]

### Step 7: Combine the probabilities
Now, we combine the probabilities from the first 4 tosses and the last 6 tosses:
\[ \text{Total probability} = \frac{4}{27} \cdot \frac{240}{729} \]

Calculating this gives:
\[ = \frac{4 \cdot 240}{27 \cdot 729} = \frac{960}{19683} \]

### Final Answer
Thus, the probability that the 2nd tail occurs on the 4th toss and the 4th tail occurs on the 10th toss is:
\[ \frac{960}{19683} \]