Algebra

Probability

Medium

applied

A fair coin is tossed 100 times. The probability of getting head an odd number of times is

$\frac{1}{2}$

$\frac{1}{8}$

$\frac{3}{8}$

$\frac{80}{100}$

✅ Correct Option: 1

A fair coin is tossed 100 times. After 100 tosses, heads can appear either an even number of times (0, 2, 4, 6, ..., 98, 100) or an odd number of times (1, 3, 5, 7, ..., 97, 99).

These are the only two possibilities, so:

$P(\text{Even heads}) + P(\text{Odd heads}) = 1$

Since the coin is fair, there is perfect symmetry between getting heads and tails. Each has probability $\frac{1}{2}$ .

Every outcome with an even number of heads has a corresponding outcome with an odd number of heads. There is no reason for even or odd to be more likely than the other.

Therefore:

$P(\text{Even heads}) = P(\text{Odd heads})$

Since both probabilities are equal and must add up to 1:

$P(\text{Odd heads}) + P(\text{Odd heads}) = 1$

$2 \times P(\text{Odd heads}) = 1$

$P(\text{Odd heads}) = \dfrac{1}{2}$

For any fair coin tossed any number of times, the probability of getting heads an odd number of times is always $\dfrac{1}{2}$ due to symmetry.

Therefore, the probability is $\dfrac{1}{2}$ .

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Solution

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