How can the number of times an event occurs a given number of times in a decision making tree be calculated when the odds of the occurrences is known?

Question

stackexchangers

I am attempting to calculate the probability that a subject will experience an event a minimum of 7 times over ten periods when the probability of the (independent) event occurring is 0.5.

I think I need to sum the probabilities that the event will happen 7 times, 8 times, 9 times and 10 times. Without having drawn a tree, I can envision that the probability that it will occur 10 times is $(\frac{1}{2})^{10}$ since there is only one branch-path (not sure what the correct terminology for this is, so I’m calling it that for now) in which that happens.

I need to calculate how many times the other branch-paths that have the event occurring 7, 8 and 9 times occur and then sum all of the probabilities together for all paths meeting the condition of events $\geq$ 7

I’m not even sure how to approach this. Maybe i need to draw a table and count the occurrences? Although this could be infeasible later if the cardinality of the periods increases.

Thanks @trueblueanil this appears to be what I’m looking for to determine the cardinality of branch-paths where 7, 8, 9 or 10 events occur. Hopefully the rest of my logic is sound s well. — duckegg, Feb 12 '22 at 08:29

score 1 · Accepted Answer · answered Feb 12 '22 at 05:34

The probability to have $k$ successes in $n$ independent trials when each success occurs with probability $1/2$ is given by $$ {1\over 2^n}\pmatrix{n \cr k} $$ where $\pmatrix{n \cr k} = {n! \over k!(n-k)!}$ is the binomial coefficient that counts the number of subsets with k elements in a set with n elements. So you want to compute $${1\over 2^{10}}\left(\pmatrix{10 \cr 7}+\pmatrix{10 \cr 8}+\pmatrix{10 \cr 9}+\pmatrix{10 \cr 10}\right) $$ My computer tells me that this is equal to $11/64 \simeq {\bf 0.17}$.

This can be also computed by hand by noting that $\pmatrix{n \cr k} = \pmatrix{n \cr n-k}$. This gives $$ \pmatrix{10 \cr 7} = \pmatrix{10 \cr 3} = {10\times 9 \times 8 \over 3\times 2 \times 1}, \quad \pmatrix{10 \cr 8} = \pmatrix{10 \cr 2} = {10\times 9 \over 2 \times 1}, \quad \pmatrix{10 \cr 9} = \pmatrix{10 \cr 1} = {10\over 1}, \quad \pmatrix{10 \cr 10} = 1. $$ The binomial coefficients can also be computed by hand using Pascal's triangle.

Hello Mr/Ms Coudy. At first I didn’t understand your answer, but later after reading @true blue anils hint about the binomial distribution, I watched the khan academy video about how to calculate it and found my self arriving at the same answer that you did, 11/64. — duckegg, Feb 12 '22 at 08:32

How can the number of times an event occurs a given number of times in a decision making tree be calculated when the odds of the occurrences is known?

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