Best Answer:
MVB has given the exact answer:

The chance of not happening is P=(0.99)^100=36.60%

The chance of happening at least once=1-P=63.40%

You might be interested to know that there is a more general pattern here. If something has 1 chance in N of happening, what's the chance of it not happening after N times? For large N, the answer is e^(-1)=36.79%.

Note that this value is very close to the actual value (36.60%) for N=100. If something has 1 chance in a 1000 of happening, the chance that it doesn't happen after 1000 times is 36.77% -- almost exactly the same as e^(-1).

("e" is the base of natural logarithms -- namely, 2.71828.... Also, e^(-1) is 1/e, or 0.36787...)

There's more: What's the chance of it not happening after k*N times? Again for large N, the answer is e^(-k). (This is true for any non-negative value of k, fractional as well as integral.)

Thus, if something has 1 chance in N of happening (for large N), the probability of it not happening after T tries is as follows:

T=N prob=36.79%

T=2N prob=13.53%

T=3N prob=4.98%

Example: Suppose you spin a big wheel marked with 1000 numbers, 1 to 1000. What's the chance that number 500 comes up? On the average, it appears once per 1000 spins. If you make 3000 spins, you expect this number to show up 3 times on the average, but there's a 5% chance that it doesn't show up at all.

Example: Suppose a no-hitter occurs in one baseball game out of 1500, and that you attend all the home games (81) of your favorite team one season. What's the chance of seeing a no-hitter?

In this example, N is 1500 and k=81/1500=0.054. The chance of a no-hitter is as follows:

probability of *no* no-hitter = e^(-0.054)=94.74%

probability of seeing at least one no-hitter =

1-94.74% = 5.26%

Finally, instead of asking what is the chance of seeing one or more no-hitters, you might be curious about the chance of seeing exactly one, or exactly two. I won't go into the details here; but if you look up "Poisson distribution" on the web, you'll see how to calculate this. Here are the results (rounded to 0.01%):

probability of exactly 0: 94.74%

probability of exactly 1: 5.12%

probability of exactly 2: 0.14%

probability of exactly 3 or more: very small

(I want to be clear that the Poisson distribution is strictly true only in the limit of large N. But for an example like a baseball no-hitter, where N=1500, the Poisson distribution is extremely close. If you want the exact answer, you need to use the binomial distribution, which you can also look up on the web. As N increases, the binomial distribution looks more and more like a Poisson distribution.)

-- edit

Finally, I'll get back to your original question (1% chance, 100 tries), and calculate the probability that the event will happen exactly M times.

The exact probability according to the binomial distribution is

prob = C(100,M) * (0.01^M) * (0.99^(100-M))

where C(N,M) is the combination function:

C(N,M) = N! / (M! (N-M)!)

Here are the probabilities (exact, but rounded to 0.01%):

M=0: prob=36.60% (This is the result mentioned in the first paragraph of the answer.)

M=1: prob=36.97%

M=2: prob=18.49%

M=3: prob=6.10%

M=4: prob=1.49%

M=5: prob=0.29%

M=6: prob=0.05%

M=7: prob=0.01%

M=8 or more: prob < 0.001%

The sum of the probabilities is 100%. The mean of this distribution is 1, and the mode is 1 (just barely, since the chance of 0 events is almost as much as the chance of 1).

If I had used the Poisson distribution instead, the answers would have been close to these values, but slightly in error.

The assumption throughout is that each trial is independent (e.g., you spin a wheel 100 times, and the outcome of one spin doesn't depend on the outcome of the previous ones). The example given by Kate Z involves non-independent trials. Once you get eaten by the shark, there are no more trials! Or maybe you have a large number of volunteers to replace the one who was eaten, but at that point the shark is no longer hungry.

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