(i) A coin is tossed 7 times. What is the probability that head appears an odd number of times ?
(ii) A coin is tossed 7 times. What is the probability that tail appears an odd number of times ?
(iii) A coin is tossed 5 times. What is the probability that head appears an odd number of times ?

Answer

To solve the problem step by step, let's break down each part of the question systematically.

### (i) A coin is tossed 7 times. What is the probability that head appears an odd number of times?

**Step 1: Understand the total outcomes.**
When a coin is tossed 7 times, the total number of outcomes is \(2^7 = 128\).

**Step 2: Define the probability of heads and tails.**
The probability of getting heads (H) in one toss is \(p = \frac{1}{2}\), and the probability of getting tails (T) is \(q = \frac{1}{2}\).

**Step 3: Identify the odd outcomes.**
We want to find the probability of getting an odd number of heads. The odd outcomes can be 1, 3, 5, or 7 heads.

**Step 4: Use the binomial probability formula.**
The binomial probability formula is given by:
\[ P(X = r) = \binom{n}{r} p^r q^{n-r} \]
For our case, we can factor out \(p^n\) and \(q^n\) since they are the same for all outcomes:
\[ P(X \text{ is odd}) = \frac{1}{2^7} \left( \binom{7}{1} + \binom{7}{3} + \binom{7}{5} + \binom{7}{7} \right) \]

**Step 5: Calculate the combinations.**
Now we compute the combinations:
- \(\binom{7}{1} = 7\)
- \(\binom{7}{3} = 35\)
- \(\binom{7}{5} = 21\)
- \(\binom{7}{7} = 1\)

**Step 6: Sum the combinations.**
Now, add these values:
\[ 7 + 35 + 21 + 1 = 64 \]

**Step 7: Calculate the probability.**
Now, plug this back into the probability formula:
\[ P(X \text{ is odd}) = \frac{64}{128} = \frac{1}{2} \]

### (ii) A coin is tossed 7 times. What is the probability that tail appears an odd number of times?

**Step 1: Understand the relationship.**
The number of tails will also be odd when the number of heads is even, and vice versa. Since there are 7 tosses, the total outcomes remain the same.

**Step 2: Use the same logic as above.**
Since the probability of getting an odd number of heads is \(\frac{1}{2}\), the probability of getting an odd number of tails is also \(\frac{1}{2}\).

### (iii) A coin is tossed 5 times. What is the probability that head appears an odd number of times?

**Step 1: Total outcomes.**
When a coin is tossed 5 times, the total number of outcomes is \(2^5 = 32\).

**Step 2: Define the probability of heads and tails.**
The probability of heads (H) in one toss is \(p = \frac{1}{2}\), and the probability of tails (T) is \(q = \frac{1}{2}\).

**Step 3: Identify the odd outcomes.**
We want to find the probability of getting an odd number of heads. The odd outcomes can be 1, 3, or 5 heads.

**Step 4: Use the binomial probability formula.**
Using the same formula:
\[ P(X \text{ is odd}) = \frac{1}{2^5} \left( \binom{5}{1} + \binom{5}{3} + \binom{5}{5} \right) \]

**Step 5: Calculate the combinations.**
Now we compute the combinations:
- \(\binom{5}{1} = 5\)
- \(\binom{5}{3} = 10\)
- \(\binom{5}{5} = 1\)

**Step 6: Sum the combinations.**
Now, add these values:
\[ 5 + 10 + 1 = 16 \]

**Step 7: Calculate the probability.**
Now, plug this back into the probability formula:
\[ P(X \text{ is odd}) = \frac{16}{32} = \frac{1}{2} \]

### Summary of Answers:
1. Probability of heads appearing an odd number of times when tossed 7 times: \(\frac{1}{2}\)
2. Probability of tails appearing an odd number of times when tossed 7 times: \(\frac{1}{2}\)
3. Probability of heads appearing an odd number of times when tossed 5 times: \(\frac{1}{2}\)