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# If you flip a fair coin four times, what is

explanation :
Consider a general tax of flipping N coins and the probability of precisely K times the heads are up. Let ‘s use a symbol P ( N, K ) for this probability.
Knowing this, we can use the leave to evaluate

P(4,2)+P(4,3)+P(4,4)

which will answer the interview of what is the probability of getting heads at lease 2 times out of flipping a mint 4 times.
Since there are only 2 outcomes from a single flip, head or tail, for N flips we can get

2N

unlike outcomes.
The outcomes we are concerned in are those that contain precisely K heads and

N−K

tails in any order. That is where combinatorics will come handy.
Any result 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 comparable flip resulted in a head ) or T ( if it was a buttocks ).
The number of outcomes with precisely K heads out of N flips is the number of strings of the length N consist 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 to

CNK=N!K!⋅(N−K)!

For all the hypothesis behind this and early formulas of combinatorics we can refer you to a correspond part of the advanced course of mathematics for high educate at Unizor.
The probability of having K heads out of N flips is equal to the ratio of the number of “ successful ” outcomes ( those with precisely K heads ) to a sum number of outcomes mentioned above :

P(N,K)=2NCNK=N!K!⋅(N−K)!⋅2N

now we can calculate the probability of at least two heads out of four flips ( do n’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

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