MHB Geometric rv and exponential rv question

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The discussion focuses on calculating the probability for a geometric random variable (rv) and an exponential rv. The user initially attempts to find the probability using the probability density function (PDF) but realizes they may be on the wrong track. The correct approach involves using the cumulative distribution function (CDF) to find the probability \( P(k-1 \leq X < k) \). This is achieved by subtracting the CDF values at the specified points, leading to a simplified expression after canceling terms. The conversation highlights the importance of using the CDF for this type of calculation.
nacho-man
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Please refer to the attached image.for part
a) this is what i did:

$G = k$, $k-1< X < k$

so I substituted $k-1$ and $k$ into the given exponential rv,

this gave me

$\lambda e^{-\lambda(k-1)}$ and $\lambda e^{-\lambda k}$
$= \lambda e^{-\lambda(k-1)} + \lambda e^{-\lambda k}$
But I feel like I am on the wrong track.

This question is really hard for me to comprehend, could someone water it down for me a bit, or help me out?

Thanks in advance!
 

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To find $P(k-1\le X<k)$ where $X$ has exponential distribution, you need to subtract the values of CDF, not PDF:
\[
P(k-1\le X<k)=F(k)-F(k-1) =(1-e^{-\lambda k})-(1-e^{-\lambda (k-1)})
\]
After canceling 1's, factor out $e^{-\lambda(k-1)}$.
 
First trick I learned this one a long time ago and have used it to entertain and amuse young kids. Ask your friend to write down a three-digit number without showing it to you. Then ask him or her to rearrange the digits to form a new three-digit number. After that, write whichever is the larger number above the other number, and then subtract the smaller from the larger, making sure that you don't see any of the numbers. Then ask the young "victim" to tell you any two of the digits of the...

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