Answer:
The probability that there are more heads than tails is equal to [tex]\dfrac{1}{2}[/tex].
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
Let the event that there are more heads than tails be [tex]A[/tex]. [tex]\lnot A[/tex] (i.e., not A) denotes that there are more tails than heads. Either one of these two cases must happen. As a result, [tex]P(A) + P(\lnot A) = 1[/tex].
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)
Due to this symmetry,
In other words [tex]P(A) = P(\lnot A)[/tex].
Combining the two equations:
[tex]\left\{\begin{aligned}&P(A) + P(\lnot A) = 1 \cr &P(A) = P(\lnot A)\end{aligned}\right.[/tex],
[tex]P(A) = P(\lnot A) = \dfrac{1}{2}[/tex].
In other words, the probability that there are more heads than tails is equal to [tex]\dfrac{1}{2}[/tex].
This conclusion can be verified using the cumulative probability function for binomial distributions with [tex]\dfrac{1}{2}[/tex] as the probability of success.
[tex]\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}[/tex]
[tex]\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}[/tex].
