91Ó°ÊÓ

Given in the question that consider a random collection of $n$individuals. In approximating the probability that no $3$of these individuals share the same birthday, a better Poisson approximation than that obtained in the text (at least for values of $n$between 80 and 90 ) is obtained by letting $E_i$be the event that there are at least 3 birthdays on day $i,i=1,…,365$.

Define random variable $X$that marks the number of people that have birthday on $ith$day. Hence, the event $E_i$is equivalent to the event $Xâ‰\yen 3$. We have that

$P\left(E_i\right)=P(Xâ‰\yen 3)=1-P(X=0)-P(X=1)-P(X=2)$

Observe that $X$can be approximated with Poisson distribution with parameter $\mathrm{λ}=\frac{n}{365}$. Hence

$P(X=0)=e^{-n/365},P(X=1)=\frac{n}{365}{e}^{-n/365},P(X=2)=\frac{(n/365{)}^2}{2}{e}^{-n/365}$

thus,

$P\left(E_i\right)=1-e^{-n/365}-\frac{n}{365}{e}^{-n/365}-\frac{(n/365{)}^2}{2}{e}^{-n/365}$

$P\left(E_i\right)=1-e^{-n/365}-\frac{n}{365}{e}^{-n/365}-\frac{(n/365{)}^2}{2}{e}^{-n/365}$

Given in the question that,consider a random collection of n individuals. In approximating the probability that no $3$of these individuals share the same birthday, a better Poisson approximation than that obtained in the text (at least for values of n between $80$ and $90$) is obtained by letting $Ei$be the event that there are at least $3$ birthdays on day$i,i=1,...,365.$

Define $p$as the number calculated in (a). Define$Y$as the random variable that counts the number of days that have the property that at least three people have birthday on that day. Hence, $Y~Binom(365,p)$, which can be approximated with Pois $(365p)$. The event that no three people share birthday is equivalent to the event $Y=0$. Hence

$P(Y=0)=e^{-365p}$

$P(Y=0)=e^{-365p}$

Given in the question that,consider a random collection of $n$individuals. In approximating the probability that no$3$ of these individuals share the same birthday, a better Poisson approximation than that obtained in the text (at least for values of $n$ between$80$ and$90$) is obtained by letting $Ei$ be the event that there are at least $3$ birthdays on day $i,i=1,...,365$.

We are required to calculate $P(Y=0)$for $n=88$. Let's find $p$for that $n$. We have that

$p=1-e^{-88/365}-\frac{88}{365}{e}^{-88/365}-\frac{(88/365{)}^2}{2}{e}^{-88/365}=1.9515·{10}^{-}3$

So we have

$P(Y=0)=e^{-365p}=e^{-0.7122965}≈0.5$

$P(Y=0)≈0.5$