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# TI-84 geometpdf and geometcdf functions

Using a TI-84 (very similar for TI-85 or TI-89) calculator for making calculations regarding geometric random variables.

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• using scipy.stats:
from scipy.stats import geom
geom.pmf(5,1/13)
geom.cdf(9,1/13)
geom.cdf(12,1/13)
• How would I solve a problem that was asking "how many shots until he misses" or "how many card draws until he picks a king"
• You are talking about a geometric distribution (of a geometric variable).
If we are given that someone has a free throw probability of 0.75 (of making it), then we can't know for sure when he will miss, but we can calculate the expected value of a geometric value.

Sal derives the expected value of a geometric variable X, as E(x) = 1/p in another video, where p is the probability of success.

So in our case that would be 1/0.75 = 1/(3/4) = 4/3
Of course expected valued as not always integers so this is just in 'theory'. All this means is that if we show until we made it n times, then we expect to throw a total of n*(4/3) shots.

Hope this helps!
- Convenient Colleague
• How would one enter this on Desmos? Or should I just calculate it manually?
• At , Sal says that the approximate answer is 38.3%, but the calculator says 0.382 percent.
(1 vote)
• The calculator actually says the probability is 0.3826967067
A probability of 0.383 is 38.3%, just like a probability of 0.5 is 50%
• Can someone clear this up for me.

On the previous lessons of cumulative geometric probabilities for lesser than and greater than. I devised the following definitions to make calculating problems easier.

F = a given event happening assuming a prob. of n

A prob. of a thing occurring in # trials greater than x;
P(F > x) = P(F not happening @ ≤ x) = (1 - n)^x

A prob. of a thing occurring in # trials less than x;
P(F < x) = 1 - P(F not happening @ ≤ x-1) = 1 - (1-n)^x-1

When this lesson came up, I went ahead and tried the questions by hand before the lesson started. I got the correct answers in the same decimal range, but the calculators cumulative function seems to always use the success probability to derive its answer.
I seem to be doing the reverse and cannot find an inverse formula to use the probability of success to find the correct answers.
• Your not communication yourself well. Generally speaking n is used for number of trials and p is used for probability and X is the random variable.

So you could say something like X ~ Geo(p).

But what you seem to be referring to is complementary probability.
• I'm very confused. In Excel I'm trying to calculate cumulative geometric probability using the NEGBINOM.DIST function for a question about 15% of emergency room visitors are uninsured. "Find the probability that it takes at least 6 patients until it receives the first uninsured patient." 1 - NEGBINOM.DIST(5,1,0.15,TRUE) does not work, but 1 - =NEGBINOM.DIST(4,1,0.15,TRUE) seems to give the correct answer. Why would 4 work and not 5??
(1 vote)
• May i clarify, first part (to get a king within 5 cards) would be 33% instead of 5.6% (this would instead be getting a king in the 5th card pick) correct?
(1 vote)
• I was using Matlab to calculate pdf and cdf and it needs different inputs to get the same answers as TI-84 in the video above. And my question is why? For example :
1. To calculate probability of picking less than 10 cards :
y = geopdf(8, 1/13)
y = 0.513
so i need to say that x=8 to get same probability of 0.513 (as in video) be:cause it gives higher probability with x=9
2. What is the probability of picking more than 12 cards :
y=1-geocdf(11,1/13)
y =0.3827
So here i also need to give 1 integer less than the one in question.
Where could these differences come from?
THank you :)