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For a group of $100$ random people, find:

  1. The expected number of days that are the birthday of at least 3 people, birthyear not being significant.

  2. The expected number of distinct birthday days.

(Consider the year as having $365$ days so assume nobody has a birthday on Feb 29th)

My attempt:

For the first question, my try consists of writing a random variable that counts the number of days that are the birthday of at least $3$ people as a sum of r.v

$$X=X_1+...+X_{365}$$

Such that

$$X_i=1 \text{ if at least $3$ people have birthday on the ith day}$$ $$X_i=0,\text{ otherwise}$$

My problem is to find the probability of $X_i=1$. I've tried to find the probability of nobody having a birthday on day i and exactly $1$ and $2$ having a birthday on day i and grab the complement, but it didn't work. I thought, for example, that the probability of exactly $2$ people having a birthday on a pre-established day i is$$\frac{\binom{100}{2} (364)^{98}}{(365)^{100}}$$is that correct?

The answer is $0.9301$

For the second, I've tried something similary, but in this case

$$Y_i=1 \text{ if exactly 1 people have birthday on day i}$$

And

$$P(Y_i=1) = 100 \frac{1}{365} (\frac{364}{365})^{99}$$

Didnt work.

Thanks in advance.

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  • $\begingroup$ If you define $X$ like that, you're going to run into problems, as the $X_i$ are not independent. $\endgroup$
    – Mathmo123
    Commented Oct 12, 2014 at 13:07
  • $\begingroup$ But the expectation of the sum isnthe sum of the expectations even if they are dependent $\endgroup$
    – Giiovanna
    Commented Oct 12, 2014 at 13:10
  • $\begingroup$ Ah right - I misread. $\endgroup$
    – Mathmo123
    Commented Oct 12, 2014 at 13:14
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    $\begingroup$ I get 0.002727474681507630629688570272749126876363690060841707485327... (using Wolfram alpha) for the probability for $X_i$. $\endgroup$ Commented Oct 12, 2014 at 13:25
  • $\begingroup$ For the second one, I'd define $Y_i$ as the event that day $i$ is a birthday for someone. $\endgroup$ Commented Oct 12, 2014 at 13:29

3 Answers 3

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For (1) you could try a Poisson approximation: Number of birthdays that fall on a given day is a random variable that is approximately Poisson$\left(\lambda=\frac{100}{365}\right)$.

For (2) you could use the same approach and take $365-$ (expected number of days with no birthdays).

I think these would give you good approximations to the exact values.

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  • $\begingroup$ Probably the author considered the poisson approximation, since I got a similar number as the answer of the first one $\endgroup$
    – Giiovanna
    Commented Oct 12, 2014 at 13:37
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    $\begingroup$ So if $N_i \sim Poisson(100/365)$, then $P(X_1=1)=1-P(N_i=0)-P(N_i=1)-P(N_i=2)$? $\endgroup$
    – Giiovanna
    Commented Oct 12, 2014 at 13:47
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Two hints:

For 1, the number of people who have a birthday on January 1 has a Binomial distribution.

For 2, try defining $$ X_i = \begin{cases} 1 &\text{if at least one person has a birthday on day i}\\ 0 &\text{otherwise} \end{cases}$$ then consider $\sum_{i=1}^{365} X_i$.

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  • $\begingroup$ For 2, the case when nobody have birhday on that day is computed as a "different birthday", which I think is not correct. For 1, note that I used binomial for compute the case when exactly 2 people have the birthday on the same day i preestabilished. As you can see:m.wolframalpha.com/input/?i=%281-%28364*364%2B100*364%2B%28100+choose+2%29%29364%5E98%2F%28365%29%5E100%29365&x=0&y=0 $\endgroup$
    – Giiovanna
    Commented Oct 12, 2014 at 13:34
  • $\begingroup$ Oh, I misread your suggestion for 2. I will consider it. $\endgroup$
    – Giiovanna
    Commented Oct 12, 2014 at 13:35
  • $\begingroup$ Right, the second one givesnthe book answer. But what is wrong with my approach to the first one? Did I make some wrong caculation? $\endgroup$
    – Giiovanna
    Commented Oct 12, 2014 at 13:42
  • $\begingroup$ @Giiovanna, What was your answer for #2? Did you just take the probability that exactly one person has a birthday on day i, and then multiply by 365? If so, what about the days where more than one person has a birthday on that day? $\endgroup$
    – awkward
    Commented Oct 12, 2014 at 20:56
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  • You have $\frac{\binom{100}{2} 364^{98}}{365^{100}} \approx 0.02839582 $ as the probability exactly $2$ out of $100$ share day $i$

  • Similarly $\frac{\binom{100}{1} 364^{99}}{365^{100}} \approx 0.20880964$ is the probability exactly $1$ out of $100$ has birthday $i$

  • and $\frac{\binom{100}{0} 364^{100}}{365^{100}} \approx 0.76006707$ is the probability that $0$ out of $100$ has birthday $i$

  • Adding these up and subtracting from $1$ gives about $0.00272747$ for the probability at least $3$ out of $100$ share day $i$, as Henno Brandsma commented

  • Multiplying this by $365$ gives about $0.9955$ expected days that at least $3$ out of $100$ share as birthdays

  • Similarly the expected number of distinct birthdays (i.e the days that at least $1$ out of $100$ has as a birthday is about $365(1-0.76006707) \approx 87.5755$

A Poisson approximation on the probabilities might suggest these numbers could be close to about $0.02853642$, $0.20831586$, $0.76035291$, $0.00279481$, $1.0201$ and $87.4712$ respectively, which are not that far away

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