Question

If a fair coin is flipped 15 times, what is the probability that there are more heads than tails?

Answer 1

Answer:

The probability that there are more heads than tails is equal to $\dfrac{1}{2}$.

Step-by-step explanation:

Since the number of flips is an odd number, there can't be an equal number of heads and tails. In other words, there are either

• more tails than heads, or,
• more heads than tails.

Let the event that there are more heads than tails be $A$. $\lnot A$ (i.e., not A) denotes that there are more tails than heads. Either one of these two cases must happen. As a result, $P(A) + P(\lnot A) = 1$.

Additionally, since this coin is fair, the probability of getting a head is equal to the probability of getting a tail on each toss. That implies that (for example)

• the probability of getting 7 heads out of 15 tosses will be the same as
• the probability of getting 7 tails out of 15 tosses.

Due to this symmetry,

• the probability of getting more heads than tails (A is true) is equal to
• the probability of getting more tails than heads (A is not true.)

In other words $P(A) = P(\lnot A)$.

Combining the two equations:

$\left\{\begin{aligned}&P(A) + P(\lnot A) = 1 \cr &P(A) = P(\lnot A)\end{aligned}\right.$,

$P(A) = P(\lnot A) = \dfrac{1}{2}$.

In other words, the probability that there are more heads than tails is equal to $\dfrac{1}{2}$.

This conclusion can be verified using the cumulative probability function for binomial distributions with $\dfrac{1}{2}$ as the probability of success.

$\begin{aligned}P(A) =& P(n \ge 8) \cr =& \sum \limits_{i = 8}^{15} {15 \choose i} (0.5)^{i} (0.5)^{15 - i}\cr =& \sum \limits_{i = 8}^{15} {15 \choose i} (0.5)^{15}\cr =& (0.5)^{15} \left({15 \choose 8} + {15 \choose 9} + \cdots + {15 \choose 15}\right) \cr =& (0.5)^{15} \left({15 \choose (15 - 8)} + {15 \choose (15 - 9)} + \cdots + {15 \choose (15 - 15)} \right) \cr =& (0.5)^{15} \left({15 \choose 7} + {15 \choose 6} + \cdots + {15 \choose 0}\right)\end{aligned}$

$\begin{aligned}\phantom{P(A)} =& \sum \limits_{i = 0}^{7} {15 \choose i} (0.5)^{15}\cr =& P(n \le 7) \cr =& P(\lnot A)\end{aligned}$.