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

Arthur Pratt

Arthur Pratt

Answered question

2021-12-16

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.

Answer & Explanation

William Appel

William Appel

Beginner2021-12-17Added 44 answers

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 (1365)2. So, use parameter λ=80000(1365)2=0.6 to conclude that the probability that at least one couple has that property is P(X1)=1P(X=0)=1eλ=0.45
b) The probability that both in a couple are born on the same day is simply 1365. So, use parameter λ=800001365=219.17 to conclude that the probability that at least one couple has that property is P(X1)=1P(X=0)=1eλ1
Nadine Salcido

Nadine Salcido

Beginner2021-12-18Added 34 answers

Answer:
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=13651365=1133225
Now the poisson approximation is used
λ=nP=800001133225=0,6
Now, let X be the number of couples that birth April 30
P(X1)=
1P(X=0)=
1e0.6(0,6)0}{0!}
P(X1)=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=5213651365=52133225
Now the poisson approximation is used
λ=nP=8000052133225=31.225
Now, let X be the number of couples that birth same day
P(X1)=
1P(X=0)=
1(e31.225)(31.225)00!
P(X1)=1

nick1337

nick1337

Expert2021-12-27Added 777 answers

Answer:
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(X=x)=Cn,xpx(1p)nx
In which Cn,x is the number of different combinations of x objects from a set of n elements, given by the following formula.
Cn,x=n!x!(nx)!
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=13651365=0.0000075
Estimate the probability that, for at least one of these couples, a) both partners were born on April 30;
This is P(X>0)=1P(X=0)
So, P(X=x)=Cn,xpx(1p)nx
P(X=0)=C80000,0(0.0000075)0(10.0000075)80000=0.5488
P(X>0)=1P(X=0)=10.5488=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=36513651365=3650.0000075=0.0027
So: P(X>0)=1P(X=0)
P(X=x)=Cn,xpx(1p)nx
P(X=0)=C80000,0(0.0027)0(10.0027)80000=1×1094=0
P(X>0)=1P(X=0)=10=1
There is a 100% probability that, for at least one of these couples, both partners were born on the same day.

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