Probability That No One Has A Specific Birthday and At Least One Person Does

algebra-precalculusprobability

The question is as follows:

What is the probability that $1017$ persons in a certain building will say that they were not born on July $15$? What is the probability that someone (meaning at least one person) will say July $15$?

My approach…

The probability that one person does not have their birthday fall on July $15$ is $\frac{364}{365}$, assuming that the year is not a leap year. For all 1017 persons, I thought of raising $\frac{364}{365}$ to the $1017$th power: $(\frac{364}{365})^{1017}$.

The second portion of the question made me think that if there is someone who has a birthday on July $15$, then the remaining 1016 will have the probability of $\frac{364}{365}$ and the one person will have the probability of $\frac{1}{365}$. However, I am unsure because the equation $(\frac{364}{365})^{1016}(\frac{1}{365})$ will represent the probability for exactly one person to have their birthday on that date, not at least one person.

Any help will be greatly appreciated.

Best Answer

Your solution to the first part is correct. As for the second part just note that "at least one person was born that day" is the complement of "none of them were born on that day". So the answer is just $1-(\frac{364}{365})^{1017}$.

Related Solutions

[Math] probability that at least $2$ people have same birthday from group of $N$ people

Let's rephrase the question.

Find the probability that any $2$ people have their birthday in the same day of the year in a group $4n$ people, of which $n$ born were in the leap year and $3n$ people were born in the regular year.

The complement of the event "any 2 people have their birthday" is "all people have birthday on different days of the year".

Consider two mutually exclusive possibilites: $\mathcal{A}_1$ - one of these birthdays falls on "Feb. 29", and $\mathcal{A}_2$ -- none of these birthdays fall on the leap day.

The number of configurations of the second kind is $$ N\left(\mathcal{A}_2\right) = \prod_{k=1}^{4 n} \left( 365 - (k-1) \right) $$ and the number of configurations of the first kind is $$ N\left(\mathcal{A}_1\right) = n \prod_{k=1}^{4 n-1} \left( 365 - (k-1) \right) $$ The total configuration size is $N_T=366^n 365^{3n}$, hence the probability of having no birthday duplicates is

$$ p_c = \frac{ N\left(\mathcal{A}_1\right) + N\left(\mathcal{A}_2\right)}{N_T} $$

Here is a confirmation in Mathematica:

(* computed probability expression *)
p[n_Integer] = (Product[365 - (k - 1), {k, 1, 4 n}] + 
     n Product[365 - (k - 1), {k, 1, 4 n - 1}])/(366^n 365^(3 n));

(* simulation result *) 
In[45]:= With[{n=7,reps=10^8},
   Sum[Boole[DuplicateFreeQ[Join[
       RandomChoice[Range[1,366],n],
       RandomChoice[Range[1,365],3n]]]],{k,1,reps}]/reps]//N

Out[45]= 0.34599694

In[53]:= N[p[7]]

Out[53]= 0.3460041650022472

(* verify that computed value belongs to the sampling 
    confidence interval *)
In[55]:= With[{pSample = Out[45], pExpected = p[7]},
 IntervalMemberQ[
  Interval[pSample + {-1, 1} Sqrt[(pSample (1 - pSample))/10^8]], 
  pExpected]]

Out[55]= True

[Math] Smallest number of people that has birthday today exceeds 1/2

A few steps: $$1−(364/365)^n\geq 1/2$$ $$1/2=1−1/2\geq (364/365)^n$$ $$\log(1/2)\geq n\log(364/365)$$ $$\log(2)\leq n(\log(365/364))$$ Can you take it form here?

Best Answer

Related Solutions

[Math] probability that at least $2$ people have same birthday from group of $N$ people

[Math] Smallest number of people that has birthday today exceeds 1/2

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