Re: Can someone answer this simple question for me?


chapmjw wrote:
> Probability question.
>
> Lets say I'm a 50% free throw shooter.
> What are the odds of me making at least one free throw in 4 attempts.
>
> Please let me know how this is calculated.

The easy way to attack "at least one" calculations
is to calculate the complement, i.e., the probability
of missing all four times.

P(4 misses) = (1/2)^4 = 1/16.

So the probability of not missing all, i.e. of getting
at least one success, is 15/16.

- Randy

.

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