If an unbiased coin is tossed 10 times, probability of obtaining more head than tail is:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(\displaystyle\frac{193}{512}\)
Let X be the number of heads. Then X is binomial with n = 10, \(p = \frac{1}{2}\). More heads than tails means X = 6, 7, 8, 9 or 10.
The probability of equal heads and tails (X = 5):
\( \displaystyle P(X = 5) = {}^{10}C_{5}\left(\frac{1}{2}\right)^{10} = \frac{252}{1024} \)
\( \displaystyle = \frac{63}{256} \)
Since \(p = \frac{1}{2}\), by symmetry “more heads” and “more tails” have the same probability, and together with X = 5 they cover every outcome:
\( 2\,P(X \gt 5) + P(X = 5) = 1 \)
\( \displaystyle P(X \gt 5) = \frac{1}{2}\left(1 - \frac{63}{256}\right) \)
\( \displaystyle = \frac{1}{2} \times \frac{193}{256} = \frac{193}{512} \)
Check by direct addition: \({}^{10}C_{6} + {}^{10}C_{7} + {}^{10}C_{8} + {}^{10}C_{9} + {}^{10}C_{10} = 210 + 120 + 45 + 10 + 1 = 386\), and \(\frac{386}{1024} = \frac{193}{512}\).
Why the other options are incorrect
Option (1) \(\displaystyle\frac{63}{256}\) — Incorrect. This is P(X = 5), the probability of equal numbers of heads and tails.
Option (2) \(\displaystyle\frac{63}{512}\) — Incorrect. This is half of P(X = 5); it is not the probability of any of the events here.
Option (4) \(\displaystyle\frac{193}{256}\) — Incorrect. This is \(1 - P(X = 5)\), the probability that heads and tails are unequal; it includes the cases with more tails, so it must be halved.