# probability of an event occuring with numerous attempts

by Matthew Oakes   Last Updated October 10, 2019 03:20 AM - source

If I have to pull ball out of a bag and I have a $$4\%$$ chance of picking the winning ball. I have $$3$$ attempts to pick out the winning ball (each time the odds are $$4\%$$ of success as the loosing ball is returned) What is my overall chance of picking out a winning ball? I would have believed it still to be $$4\%$$ overall but I am being told differently?

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You have a $$96\%$$ chance of losing each attempt, correct?

The chances of losing three attempts in a row is therefore $$96\%\cdot96\%\cdot96\% = 88.4736\%$$

So your chances of winning at least once is $$11.5264\%$$

And it kind of makes sense, doesn't it? The probability is just under $$12\%$$, and you have three attempts each at $$4\%$$.

The real concept question here is, why isn't it $$12\%$$?

SquishyRhode
October 10, 2019 02:37 AM

Your chance of failing is $$0.96$$ each time. Since the tries are independent, the chance of failing three times is $$.96^3=0.884736$$, so your probability of success $$=0.115264$$.

herb steinberg
October 10, 2019 02:41 AM

Often in calculating probabilities, it is sometimes easier to calculate the probability of the 'opposite', the technical term being the complement. Because if something happens with probability $$p$$, then it does not happen with probability $$1-p$$, e.g. if something happens with probability $$0.40$$ ($$40\%$$) then it does not happen with probability $$1-0.40=0.60$$ ($$60\%$$).

The 'opposite' (complement) of winning at least once is never winning at all. The probability of not picking the winning ball the first time is $$1-0.04=0.96$$, i.e. $$96\%$$. But you also want this to happen the second time and the third time. So you do not win with probability $$0.96 \cdot 0.96 \cdot 0.96= 0.884736,$$ i.e. $$88.4736\%$$. But then this is the probability that you do not win so that the probability that you win is $$1-0.884736= 0.115264$$, or $$11.5264\%$$.

mathematics2x2life
October 10, 2019 02:42 AM

You can pick a winning ball in one of 3 disjoint events:

1. pick a winning ball in the first attempt, subsequent attempts are irrelevant.
2. fail in the first attempt but pick a winning ball in the second attempt, third attempt is irrelevant.
3. fail in the first two attempts but pick a winning ball in the 3rd attempt.

Therefore, sum of the probabilities of these 3 disjoint events is $$0.04 + 0.96*0.04 + 0.96^2 * 0.04$$

Black
October 10, 2019 02:43 AM