A fair coin is tossed 5 times in succession. What is the probability that the number of heads obtained is an odd number (that is, 1, 3 or 5)?

Introduction / Context:
This question examines the binomial distribution for repeated coin tosses. We toss a fair coin five times and want the probability that the number of heads is odd. The possible odd counts in 5 tosses are 1, 3 or 5. There is a useful symmetry result for fair coins: the probability of an odd number of heads equals the probability of an even number of heads, so each is 1/2. We can use symmetry or compute directly using binomial coefficients.

Given Data / Assumptions:

A fair coin is tossed 5 times.

Each toss has two possible outcomes: head or tail.

The probability of head in each toss is 1/2, and tail is 1/2.

We seek the probability of getting an odd number of heads, namely 1, 3 or 5 heads.

Concept / Approach:
The total number of possible outcomes for 5 tosses is 2^5. The number of ways to obtain exactly k heads is given by 5Ck. Using the binomial distribution, P(k heads) = 5Ck * (1/2)^5. Since the coin is fair, we can either calculate P(1 head) + P(3 heads) + P(5 heads) directly or use the symmetry that the probability of an odd number of heads equals the probability of an even number of heads, which must be 1/2 each.

Step-by-Step Solution (Direct Method):
Step 1: Total number of outcomes = 2^5 = 32.Step 2: Probability of exactly 1 head = 5C1 * (1/2)^5 = 5 * (1/32) = 5/32.Step 3: Probability of exactly 3 heads = 5C3 * (1/2)^5 = 10 * (1/32) = 10/32.Step 4: Probability of exactly 5 heads = 5C5 * (1/2)^5 = 1 * (1/32) = 1/32.Step 5: Probability of an odd number of heads = 5/32 + 10/32 + 1/32 = 16/32 = 1/2.

Verification / Symmetry Argument:
For a fair coin, the distribution of heads and tails is symmetric. If we flip the coin 5 times, every outcome with k heads has a corresponding outcome with 5 - k heads obtained by swapping heads and tails in each toss. Thus, the probability of even heads equals the probability of odd heads. Since these two cases cover all possibilities, each must have probability 1/2. This symmetry matches our detailed calculation of 16 favourable outcomes out of 32.

Why Other Options Are Wrong:
The value 1/3 would correspond to about 10.67 favourable outcomes out of 32, which is not an integer count. The value 2/3 is larger than 1/2 and would require more odd head combinations than are actually possible. The value 1 would mean that an odd number of heads occurs every time, which is clearly not true. The fraction 3/8 is 12/32, less than the correct 16 favourable outcomes and misses part of the distribution.

Common Pitfalls:
Some learners forget to include all odd cases (for example, they might account for 1 and 3 heads but forget 5 heads). Others miscalculate the binomial coefficients 5C3 or 5C1. A third common issue is treating the events of 1 head, 3 heads and 5 heads as overlapping, which they are not. Using the binomial formula carefully and checking the symmetry between heads and tails helps avoid these mistakes.

Final Answer:
The probability that heads appear an odd number of times in five tosses is 1/2.