Table E1 shows the possible outcomes for three tosses.
Toss 1 |
Toss 2 |
Toss 3 |
|
Outcome 1 | H |
H |
H |
Outcome 2 | H |
H |
T |
Outcome 3 | H |
T |
H |
Outcome 4 | H |
T |
T |
Outcome 5 | T |
H |
H |
Outcome 6 | T |
H |
T |
Outcome 7 | T |
T |
H |
Outcome 8 | T |
T |
T |
The probability of at least one head is `7//8`.
In general, the probability of at least one head in `n` tosses is (the probability of a head on the first toss) plus (the probability of a tail on the first toss and a head on the second toss) plus `...` plus (the probability of tails on the first `n-1` tosses and a head on the `n`th toss). In symbols this is
`1/2+1/(2^2)+ cdots + 1/(2^k)+ cdots + 1/(2^n)`.
This is the same as `1` minus the probability of tails on all tosses. Since only one of the `2^n` outcomes has all tails, the probability of at least one head in `n` tosses is
`1-1/2^n = (2^n - 1)/2^n`.
In particular, for `n=5`, the probability of at least one head is `text[(]2^5 - 1text[)/]2^5 = 31text[/]32 = 0.96875`.