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Let us consider a town with $N$ inhabitants. We know that $30\%$ of them watch a certain TV show. Each night, the show makes 10 random phone calls to residents of the town. What is the probability that more than 8 of them are watching the show?

This is the statement I am having trouble with. Firstly, it is not specified whether a person can be called more than once — let us assume they cannot. Secondly, it is not stated whether the calls are made simultaneously, which might be relevant for the second approach to solving the problem.

My first instinct was to consider the hypergeometric distribution, namely:

$$ P(X>8)=P(X = 9) + P(X = 10) = \sum_{k=9}^{10} \frac{\binom{0.3N}{k} \binom{0.7N}{10-k}}{\binom{N}{10}} $$ Where $X$ is the random variable "number of people who were called and were watching the show".

However, I reviewed the solution provided by one of my colleagues, and he proposed the following: $X$ follows a binomial distribution with $n = 10$ and $p = 0.3$. Therefore,

$$ P(X > 8) = P(X = 9) + P(X = 10) = \binom{10}{9}(0.3)^9(0.7)^1 + \binom{10}{10}(0.3)^{10} $$

Based solely on the information I provided in the problem statement, which model do you think is more appropriate? I believe the key difference lies in the fact that the hypergeometric distribution accounts for the calls being made sequentially rather than simultaneously. To conclude, I believe that the computation using the binomial distribution is much more straightforward, as it does not involve factorials that depend on $0.3N$ or $0.7N$.

MiguelCG
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    We are dealing with a town. From this I would conclude that $N$ is large enough to go for binomial distribution. Moreover in the hypergeometric approach you can meet problems with $0.3N$ not being an integer so that approximations are not avoided anyway. – drhab May 06 '25 at 11:09

1 Answers1

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Without knowing $N$, the intended approach is likely to be to use the binomial distribution and your

$$P(X > 8) = P(X = 9) + P(X = 10) = \binom{10}{9}(0.3)^9(0.7)^1 + \binom{10}{10}(0.3)^{10}$$

is approximately $0.0001436859$, which to two significant figures is $0.00014$. It does not depend on $N$.

This is the answer if, rather than exactly $30\%$ are watching (requiring the population to be a multiple of $10$), the probability of a particular individual watching is $30\%$ independently of the others.

It is also the limiting value of the corresponding hypergeometric probability as $N \to \infty$. To see how fast the limit is approached, here are more precise values for various different values of $N$. This hypergeometric probability also rounds to $0.00014$ when $N\ge 1240$.

     N       P(X>8) hypergeometric

10       0
20       0
50       0.0000173455



100       0.0000595914
   200       0.0000955429
   500       0.0001228631
  1000       0.0001330054
  2000       0.0001382776
  5000       0.0001415062
 10000       0.0001425933
 20000       0.0001431389
 50000       0.0001434669
100000       0.0001435764
Henry
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