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Geometric Distribution

r=1r = 1

X=#ย ofย trialsย untilย theย 1stย successX = \text{\# of trials until the 1st success}

xโˆˆZ+x \in \mathbb{Z}^{+}

P=Probabilityย ofย SuccessP = \text{Probability of Success}


P(X=x)=(1โˆ’p)xโˆ’1pP(X=x) = (1-p)^{x-1} p

Where (1โˆ’p)xโˆ’1(1-p)^{x-1} is the xโˆ’1x-1 trials that failed and pp is the xthx^{th} trial that succeeded. Then,


P(X>x)=โˆ‘j=x+1โˆžP(x=j)=โˆ‘j=x+1โˆž(1โˆ’p)xโˆ’1pP(X>x) = \sum\limits_{j=x+1}^{\infty} P(x=j) = \sum\limits_{j=x+1}^{\infty} (1-p)^{x-1} p

Define q=1โˆ’pq=1-p


=โˆ‘j=x+1โˆžqjโˆ’1p=pqโˆ‘j=x+1โˆžqj=pqqx+11โˆ’q=qx=\sum\limits_{j=x+1}^{\infty} q^{j-1} p = {p \over q}\sum\limits_{j=x+1}^{\infty} q^{j} = {p \over q} {q^{x+1} \over {1-q}} = q^x

P(Xโ‰คx)P(X \leq x)โ€‹

P(Xโ‰คx)=1โˆ’P(X>x)=1โˆ’qxP(X \leq x) = 1 - P(X>x) = 1-q^x

โˆ‘x=1โˆž(1โˆ’p)xโˆ’1p=p1โˆ’pโˆ‘x=1โˆž(1โˆ’p)x=1\sum\limits_{x=1}^{\infty} (1-p)^{x-1} p = {p \over {1-p}} \sum\limits_{x=1}^{\infty} (1-p)^x = 1