# Approximately 80,000 marriages took place in the state of New

Approximately 80,000 marriages took place in the state of New York last year. Estimate the probability that, for at least one of these couples, (a) both partners were born on April 30; (b) both partners celebrated their birthday on the same day of the year. State your assumptions.
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William Appel
In both of these cases we will use Poisson approximation of Binomial, since $n=80000$ is extremely large and p is relatively small such that np is relatively normal number.
a) The probability that both in a couple are born on April 30 is simply ${\left(\frac{1}{365}\right)}^{2}$. So, use parameter $\lambda =80000\cdot {\left(\frac{1}{365}\right)}^{2}=0.6$ to conclude that the probability that at least one couple has that property is $P\left(X\ge 1\right)=1-P\left(X=0\right)=1-{e}^{-\lambda }=0.45$
b) The probability that both in a couple are born on the same day is simply $\frac{1}{365}$. So, use parameter $\lambda =80000\cdot \frac{1}{365}=219.17$ to conclude that the probability that at least one couple has that property is $P\left(X\ge 1\right)=1-P\left(X=0\right)=1-{e}^{-\lambda }\approx 1$

a)0,45119
b)1
Step-by-step explanation: For part A of the problem we must first find the probability that both people in the couple have the same birthday (April 30)
$P=\frac{1}{365}\cdot \frac{1}{365}=\frac{1}{133225}$
Now the poisson approximation is used
$\lambda =nP=80000\cdot \frac{1}{133225}=0,6$
Now, let X be the number of couples that birth April 30
$P\left(X\ge 1\right)=$
$1-P\left(X=0\right)=$
$1-\frac{{e}^{-}0.6}{\cdot }{\left(-0,6\right)}^{0}\right\}\left\{0!\right\}$
$P\left(X\ge 1\right)=0,45119$
B) Now want to find the probability that both partners celebrated their birthday on th, assuming that the year is 52 weeks and therefore 52 thursday
$P=52\cdot \frac{1}{365}\cdot \frac{1}{365}=\frac{52}{133225}$
Now the poisson approximation is used
$\lambda =nP=80000\cdot \frac{52}{133225}=31.225$
Now, let X be the number of couples that birth same day
$P\left(X\ge 1\right)=$
$1-P\left(X=0\right)=$
$1-\frac{\left({e}^{-}31.225\right)\cdot {\left(-31.225\right)}^{0}}{0!}$
$P\left(X\ge 1\right)=1$

nick1337

a) There is a 45.12% probability that, for at least one of these couples, both partners were born on April 30.
b) There is a 100% probability that, for at least one of these couples, both partners were born on the same day.
Step-by-step explanation:
For each couple, either they have the same birth date, or they do not. This means that we solve this problem using the binomial probability distribution.
Binomial probability distribution
The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.
$P\left(X=x\right)={C}_{n,x}\cdot {p}^{x}\cdot \left(1-p{\right)}^{n-x}$
In which ${C}_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.
${C}_{n,x}=\frac{n!}{x!\left(n-x\right)!}$
And p is the probability of X happening.
In this problem we have that:
There are 80000 couples, so $n=80000.$
The probability that a couple was born on April 30 is $p=\frac{1}{365}\cdot \frac{1}{365}=0.0000075$
Estimate the probability that, for at least one of these couples, a) both partners were born on April 30;
This is $P\left(X>0\right)=1-P\left(X=0\right)$
So, $P\left(X=x\right)={C}_{n,x}\cdot {p}^{x}\cdot \left(1-p{\right)}^{n-x}$
$P\left(X=0\right)={C}_{80000,0}\cdot \left(0.0000075{\right)}^{0}\cdot \left(1-0.0000075{\right)}^{80000}=0.5488$
$P\left(X>0\right)=1-P\left(X=0\right)=1-0.5488=0.4512$
There is a 45.12% probability that, for at least one of these couples, both partners were born on April 30.
b) both partners celebrated their birthday on the same day of the year.
There are 365 days on the year. So the probability that a couple was born on the same day is $p=365\cdot \frac{1}{365}\cdot \frac{1}{365}=365\cdot 0.0000075=0.0027$
So: $P\left(X>0\right)=1-P\left(X=0\right)$
$P\left(X=x\right)={C}_{n,x}\cdot {p}^{x}\cdot \left(1-p{\right)}^{n-x}$
$P\left(X=0\right)={C}_{80000,0}\cdot \left(0.0027{\right)}^{0}\cdot \left(1-0.0027{\right)}^{80000}=1×{10}^{-94}=0$
$P\left(X>0\right)=1-P\left(X=0\right)=1-0=1$
There is a 100% probability that, for at least one of these couples, both partners were born on the same day.