A coin is tossed until a head appears or tail appears 4 times in succession. Find the probability getting a head it is known that tail occured at least twice .

Answer

To solve the problem of finding the probability of getting a head given that tails have occurred at least twice, we can follow these steps:

### Step 1: Define the Events
We need to consider the possible sequences of coin tosses that lead to either a head appearing or tails appearing four times in succession. The events we are interested in are:
- Getting a head after at least two tails.
- Getting tails four times in a row, which ends the game.

### Step 2: Identify the Sample Space
The sample space consists of the sequences that can occur before the game ends. The relevant sequences are:
1. TTH (Two tails followed by a head)
2. TTTH (Three tails followed by a head)
3. TTTT (Four tails in succession, which ends the game)

### Step 3: Calculate the Probabilities
1. **Probability of TTH**:
- The probability of getting tails twice followed by heads is:
\[ P(TTH) = P(T) \times P(T) \times P(H) = \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{8} \]

2. **Probability of TTTH**:
- The probability of getting tails three times followed by heads is:
\[ P(TTTH) = P(T) \times P(T) \times P(T) \times P(H) = \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{16} \]

3. **Probability of TTTT**:
- The probability of getting tails four times is:
\[ P(TTTT) = P(T) \times P(T) \times P(T) \times P(T) = \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{16} \]
However, this event does not lead to a head, so we will not consider this in our final probability calculation.

### Step 4: Total Probability of Getting a Head
Now, we can sum the probabilities of the sequences that result in a head:
\[ P(\text{Head | At least 2 Tails}) = P(TTH) + P(TTTH) = \frac{1}{8} + \frac{1}{16} \]
To add these fractions, we need a common denominator:
\[ \frac{1}{8} = \frac{2}{16} \]
Thus,
\[ P(\text{Head | At least 2 Tails}) = \frac{2}{16} + \frac{1}{16} = \frac{3}{16} \]

### Final Answer
The probability of getting a head given that tails occurred at least twice is:
\[ \frac{3}{16} \]