An unbiased coin is tossed eight times what is the probability of getting less than 4 heads.

**Correct Answer**

\(\frac{1}{64}\)

The probability of getting less than 4 heads can also be expressed as the sum of probabilities of getting 0, 1, 2, or 3 heads.

So, the required probability is = \((\frac{1}{2})^8 + (\frac{1}{2})^8 + (\frac{1}{2})^8 + (\frac{1}{2})^8 = 4(\frac{1}{2})^8 = \frac{1}{2^6} = \frac{1}{64}\)