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What is the probability that I would get either all heads or tails in n coin tosses.

I know, for example, in $4$ tosses, the probability is $\left(\frac12\right)^4=\frac1{16}$ to get either all heads or tails. But how do I generalize it?

Ty for the assistance!

Commander Shepard
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  • $$P(A\cup B)=P(A)+P(B)-P(AB)$$ Here $AB=\phi$ – lab bhattacharjee Aug 03 '13 at 04:39
  • These sites are relevant to the question https://math.stackexchange.com/questions/234062/probability-of-tossing-a-fair-coin-with-at-least-k-consecutive-heads?rq=1 https://math.stackexchange.com/questions/1946899/probability-of-4-consecutive-heads-in-10-coin-tosses – Blair Apr 04 '18 at 00:57

1 Answers1

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First, the probability of tossing 4 coins and get either all heads or tails would be:

$$\frac12\times\frac12\times\frac12\times\frac12\times2=\frac1{8}$$

since you can either get all heads or tails.

The "general probability" depends on how many tosses you're going to do:

Let $n$ be the total coin toss, then the probability of getting either all heads or tails would be:

$$\left(\frac12\right)^n\times2=\frac1{2^{n-1}}$$