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

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2 votes

Explanation:

User DAIvd
by
7.4k points
5 votes

Answer:

The probability that there are more heads than tails is equal to
(1)/(2).

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) = (1)/(2).

In other words, the probability that there are more heads than tails is equal to
(1)/(2).

This conclusion can be verified using the cumulative probability function for binomial distributions with
(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}.

User Serge Harnyk
by
8.9k points

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