According to axioms, a probability is always less or equal to 1. So I must have miss something here.
Let X be the number of heads observe in three independant throws.
The binomial theorem : [MATH]P(X=k)={{N}\choose{k}}p^k(1-p)^{N-k}[/MATH]
In a game of head or tail, [MATH]p=\frac{1}{2}[/MATH] and [MATH]1-p=\frac{1}{2}[/MATH]
So [MATH]P(X=1)={{3}\choose{1}}\frac{1}{2}^1\frac{1}{2}^2 = 0.875[/MATH]and [MATH]P(X=2)={{3}\choose{2}}\frac{1}{2}^2\frac{1}{2}^1 = 0.875[/MATH]
So, the probability of having either 1 or 2 heads is
[MATH]P(X=1 \;OR\; X=2)=P(X=1)+P(X=2) - P(X=1 \; AND \; X=2)=0.875+0.875-0=1.75 > 1[/MATH]
But, that result does not make any sense.
Let X be the number of heads observe in three independant throws.
The binomial theorem : [MATH]P(X=k)={{N}\choose{k}}p^k(1-p)^{N-k}[/MATH]
In a game of head or tail, [MATH]p=\frac{1}{2}[/MATH] and [MATH]1-p=\frac{1}{2}[/MATH]
So [MATH]P(X=1)={{3}\choose{1}}\frac{1}{2}^1\frac{1}{2}^2 = 0.875[/MATH]and [MATH]P(X=2)={{3}\choose{2}}\frac{1}{2}^2\frac{1}{2}^1 = 0.875[/MATH]
So, the probability of having either 1 or 2 heads is
[MATH]P(X=1 \;OR\; X=2)=P(X=1)+P(X=2) - P(X=1 \; AND \; X=2)=0.875+0.875-0=1.75 > 1[/MATH]
But, that result does not make any sense.
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