How do you calculate the probability of getting tails on a coin "n" number of times in a row?
2006-09-23 08:15:34 · 6 answers · asked by Bob 3 in Science & Mathematics ➔ Mathematics
probability of tails = 1/2 so the formula would be (1/2)^n
2006-09-23 08:17:54 · answer #1 · answered by anonymous_20003 3 · 0⤊ 0⤋
Probability equals the number of successes divided by the total number of possible outcomes (P(s) = n(s)/N).
If your coin is fair (not loaded or two-headed), one can expect tails to come up, on average, half of the tosses. In math talk that's P(s = tail) = .5N/N = .5. So each time we toss a fair coin, there is a 50-50 chance we will get a tail.
So the chance of getting a tail on our first toss is P(s = tail) = .5 Similarly, the chance of getting a tail on the second toss is also P = .5 and so on. Each toss outcome is independent of the previous outcome.
Now, to your question, what are chances of throwing a tail n times in a row? For independent events, like throwing the coin, that is simply the multiple of each individual probability. So for n = 3 tosses, for example, we have P(s = tail AND s = tail AND s = tail) = P(s = tail) X P(s = tail) X P(s = tail) = .5 X .5 X .5 = .125
2006-09-23 08:34:44 · answer #2 · answered by oldprof 7 · 0⤊ 0⤋
first of all the probability of getting tails is 1/2, so all u have to do is multiply n by 1/2
2006-09-23 08:18:46 · answer #3 · answered by Gina C 2 · 0⤊ 0⤋
1/(2x2x2x...2), where there are n 2s, i.e. 1/(2^n) or 2 to the power of n
So
n=1, p=1/2 =0.5
n=2, p=1/4 = 0.25
n=3, p=1/8 =0.125
n=4, p=1/16 = 0.0625
.
.
.
and so on.
2006-09-23 08:20:36 · answer #4 · answered by Friseal 3 · 2⤊ 0⤋
P(heads on a toss) = 0.5
P(tails on a toss) = 0.5
P(tails) and P(tails) and P(tail) .... n times =
P(t)*P(t)*P(t)... n times = P(tails)^n = 0.5 ^ n
2006-09-23 08:19:40 · answer #5 · answered by holden 4 · 0⤊ 0⤋
0,5 to the power n
2006-09-23 08:22:18 · answer #6 · answered by iyiogrenci 6 · 0⤊ 0⤋