Probability help please. If throwing a coin 200 times, the probability of appearing tail is ? .(1 coin have 1 tail, 1 face) Ok so the answer books say it is 99.... How?and why not 100? Also it is a choice question so here are the choices: 1)103 2)99 3)95 4)91
+65 Well, we can eliminate a few answers right away. 103% is not possible, so it's not that. And 95 and 91 are clearly too low if you flip it 200 times (but who would have the patience to do that?). However, it's also pretty clear that it would round to 100% instead of 99%. The reason is because they might have truncated the percent instead of rounding it.
I actually proved 4 equations in 8th grade, and #4 helps out a lot with this.
#4: \(1-(1-c)^t\) the probability of an event with \(c\) chance occurring at least once after \(t\) times.
For your problem, \(t=200\) and \(c=\frac12\). \(1-(1-\frac12)^{200}=1-(\frac12)^{200}=1-\frac{1}{BIG}=\frac{BIG-1}{BIG}\). I substituted 'BIG' for the result of the power because the number itself is 61 decimal digits. If we transfer the fraction generated into a percent and truncate at 60 decimal digits, then this is the percent: \(99.999999999999999999999999999999999999999999999999999999999937\)%. If I round, it will without a doubt round to \(100\)%. But if I truncate the entire decimal, I get \(99\)% (if you are wondering, truncating a fraction is basically cutting off its decimal after a certain point. In this case, I cut off all of its decimal, and probably so did the textbook.).
