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There are $5$ balls in a bag marked A, B, C, D, E. Each ball has equal probability of getting picked. As soon as any one ball is picked, it is replaced back in the bag. What is the probability that in $6$ consecutive picks, ball A gets picked at least once? (This is not a homework question... I am not in school anyway)

One method I can think of is $$1 - (\text{probability of A not getting picked at all}) = 1 - \left(\frac{1}{5} \cdot 4\right)^6 = 0.74$$

Another method I thought would work is: sum of probability of picking A exactly once, exactly 2 times, ..., exactly 6 times. This seems to give: $$(0.2^6) + (0.2^5)0.8 + (0.2^4)(0.8^2) + (0.2^3)(0.8^3) + (0.2^2)(0.8^4) + (0.2)(0.8^5) = 0.087360$$

If the first method is correct (I think so, not sure), then there is definitely something missing from the second method. Can anyone please point out what am I missing here?

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First method is correct, and far easier.

What is missing from your second method are some coefficients. For example, your term for picking A 5 times out of 6 is $0.2^{5} *0.8$. That's not correct. The actual term you need is $\left(\array{6\\5}\right)*0.2^{5} *0.8$. The factor $\left(\array{6\\5}\right)$ accounts for the fact the balls can come in different orders - the non-A ball could be the 1st, 2nd, through to 6th ball you pick. So you need to actually perform the sum:

$$\sum_{n=1}^{6}\left[\left(\array{6\\n}\right) * 0.2^{n} * 0.8^{6-n}\right]$$ Since you only have 6 picks, it's not too bad. But imagine if you want to find the probability you don't pick A in 100 consecutive picks, or 1000.

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  • $\begingroup$ Ah! You're right. I missed that part. Thanks. $\endgroup$ – Imagi Neshun Aug 10 '17 at 10:28

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