# Probability question about geometric distribution Let X have a geometric distribution. Show that <math xmlns="http://www.w3.org/1998/Math/MathML"> <mi>P</mi> <mo stretchy="false">(</mo> <mi>X</mi> <mo>&gt;</mo> <mi>k</mi> <mo>+</mo> <mi>j</mi> <mrow class="MJX-TeXAtom-ORD"> <mo stretchy="false">|</mo> </mrow> <mi>X</mi> <mo>&gt;</mo> <mi>k</mi> <mo stretchy="false">)</mo> <mo>=</mo> <mi>P</mi> <mo stretchy="false">(</mo> <mi>X</mi> <mo>&gt;</mo> <mi>j</mi> <mo stretchy="false">)</mo> [/itex], where k and j are nonnegative integers.

Let X have a geometric distribution. Show that $P\left(X>k+j|X>k\right)=P\left(X>j\right)$, where k and j are nonnegative integers.
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Kyle Delacruz
Step 1
I will use the formula of conditional probability: $P\left(A|B\right)=\frac{P\left(A\cap B\right)}{P\left(B\right)}$.
Thus $P\left(X>k+j|X>k\right)=$

Step 2
As we are working with a geometric distribution, $P\left(X>\left(k+j\right)\right)={q}^{k+j}$ and $P\left(X>k\right)={q}^{k}.$
Then $P\left(X>\left(k+j\right)\right)/P\left(X>k\right)={q}^{k+j}/{q}^{k}={q}^{j}=P\left(X>j\right)$
$=P\left(X>j\right)$ when q is the probability of failure.