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Geometric Distributions S1

I have a question:
The random variable T can take values 1, 2, 3,...) and has a geometric distribution. It is given that P(T= 1 or 2) = 0.4375. Find the value of P(T=1). the answer is 0.25. I don't know how to do the geometric distribution and I don't know how to work out the probability of success. Please can you help me. Thanks
Sorry you've not had any responses about this. :frown: Are you sure you've posted in the right place? :smile: Here's a link to our subject forum which should help get you more responses if you post there. :redface:


Just quoting in Fox Corner so she can move the thread if needed :wizard:

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Original post by Sibghy
I have a question:
The random variable T can take values 1, 2, 3,...) and has a geometric distribution. It is given that P(T= 1 or 2) = 0.4375. Find the value of P(T=1). the answer is 0.25. I don't know how to do the geometric distribution and I don't know how to work out the probability of success. Please can you help me. Thanks


Well then, first step would be to look up what the geometric distribution is. T having a geometric distribution means P(T=t)=(1−p)t−1p P(T=t) = (1-p)^{t-1}p where p is the probability parameter (usually it is the probability of a Bernoulli trial, and the geometric distribution usually measures the probability of how many failures you have before your first success - read the Wiki page).

The events are mutually exclusive. Therefore P(T=1 or 2) = P(T=1) + P(T=2).

Using the distribution above, you get P(T=1 or 2) = p + (1-p)p = 0.4375. This will give you a quadratic you can solve for p (almost certainly one value is negative and the other is positive - so you know which value is the solution you care about).

Since P(T=1) = p, once you've found p, you're done!
(edited 7 years ago)

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