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# Geometric Distributions S1 watch

1. 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
2. Sorry you've not had any responses about this. Are you sure you've posted in the right place? Here's a link to our subject forum which should help get you more responses if you post there.

Just quoting in Fox Corner so she can move the thread if needed
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3. (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 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!

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