Probability?

Question

From an alphabet of 8 symbols, say {A,B,C,D,E,F,G,H}, I randomly pick 86 symbols. What is the probability of any symbol being picked 25 times or more?

Answer 1

Let X be the number of times that one of the symbols is picked out the the 8. X has the binomial distribution with n = 86 trials and success probability 1/8

P(X ≥ 25) = 0.00003511022


there are 8 different ways this could happen, so:

8 * 0.00003511022 = 0.0002808818

that is the probability that you would pick one of the symbols at least 25 times. This is for only seeing one of the symbols 25 times. it is possible to see up to three symbols at least 25 times in one experiment. however, the probabilities of that are so low that i will ignore them and estimate the probability to be 0.00028 as good enough.

Answer 2

This is a binomial distribution so the expected value is np and the standard deviation is sqrt(npq).

The expected number of each symbol is: 86/8 = 10.75
Standard deviation = sqrt(86 * 1/8 * 7/8) = 9.41

Standardize the random variable:
P(X > 25)
= P([(X - exp) / stdev] > [(25 - exp) / stdev])
= P(z > [(25 - 10.75) / 9.41])
= P(z > 1.514)
Use a z-score table:
= 0.065

So the probability of picking any symbol 25 times or more is 6.5%.

EDIT: I caught my mistake (thanks to Dr. D!). I forgot to take the square root of the variance to find standard deviation. If you do that, you may get more desirable results - the probability becomes almost 0.

Answer 3

I don't know if you have a more formal way of doing this, but I agree with Kyle that using the normal distribution is the best way to estimate the answer.

Just a few corrections though.
A = no. of times A appears
A ~ B(86, 1/8)
which approximates to
A ~ N(10.75, 3.067^2)

P(A >= 25) = P(A > 24.5) using hte continuity correction, not that it really matters.
= P(Z > [24.5 - 10.75] / 3.067)
= P(Z > 4.483)
= 1 - 0.999996 = 4e-6

There are 8 ways this can happen so the final estimate is:
3.2e-5 or 1 in 31250.

NOTE we have neglected the probability that two or more letters appear 25 or more times. But it is clear that this does not significantly affect the accuracy.

*EDIT*
I think merlyn's answer is more accurate since it works directly with the binomial, whereas the normal approximation introduces errors especially for probabilities of this magnitude.