If you flip a fair coin four times, what is the probability that you get heads at least twice?

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Cenedesiidv · Accepted Answer

Explanation:Consider a general task of flipping N coins and the probability of exactly K times the heads are up. Let&#039;s use a symbol P(N, K) for this probability.Knowing this, we can use the result to evaluate P(4,2)+P(4,3)+P(4,4)which will answer the question of what is the probability of getting heads at lease 2 times out of flipping a coin 4 times.Since there are only 2 outcomes from a single flip, head or tail, for N flips we can get 2N different outcomes.The outcomes we are interested in are those that contain exactly K heads and N−K tails in any order. That is where combinatorics will come handy.Any outcome of the random experiment of flipping a coin N times can be represented as a string of N characters, each one being a letter H (to designate that the corresponding flip resulted in a head) or T (if it was a tail).The number of outcomes with exactly K heads out of N flips is the number of strings of the length N consisting of characters H and T, where H occurs K times and T occurs N−K times in any order.This number is, obviously, a number of combinations of K items out of N, which symbolically is represented as CNK (there are other notations as well) and is equal toCNK=N!K!⋅(N−K)!For all the theory behind this and other formulas of combinatorics we can refer you to a corresponding part of the advanced course of mathematics for high school at Unizor.The probability of having K heads out of N flips is equal to the ratio of the number of &quot;successful&quot; outcomes (those with exactly K heads) to a total number of outcomes mentioned above:P(N,K)=2NCNK=N!K!⋅(N−K)!⋅2NNow we can calculate the probability of at least two heads out of four flips (don&#039;t forget that 0≠1 by definition):P(4,2)+P(4,3)+P(4,4)==124⋅[4⋅3⋅2⋅1(1⋅2)⋅(1⋅2)+4⋅3⋅2⋅1(1⋅2⋅3)⋅(1)+4⋅3⋅2⋅1(1⋅2⋅3⋅4)⋅1]==6+4+116=1116

If you flip a fair coin four times, what is

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