In agracy2246/MATH4753grac0009: Adam Gracy MATH4753

knitr::opts_chunk$set(echo = TRUE, message=FALSE)

Introduction

In class we gave the example of calculating the probability of event $A$, at least 2 people having the same birthday when the class size is $n$.

In this minilab we will make a function that will calculate the probability using the formula given in class.

$$P(A^C)=\frac{365\times 364\times \cdots \times (365-n+1)}{365^n}$$

This means that the $P(A)$ can be calculated indirectly:

$$P(A)=1-P(A^C)$$

Questions

Complete mybirthday() function that will create the probability that 2 or more people in a group of size $n$ will have the same birthday.

More specifically:

1. The function will create a table of $n$ and $P(A)$ values
2. It will also plot a pie chart of $P(A)$ Vs $n$ where $n\in \{1,\cdots , n_{max}\}$
3. The value of n is an upper value of the number of people in the room.

Please fill in the gaps ### and make it produce the output shown in the html document.

mybirthday = function(n)
{
  # One problem is the size of the numbers
  # Use log to make calculation manageable lfactorial is the log of the factorial
  # Below is log(P(AC))
  logpAc = lfactorial(365)-lfactorial(365-(1:n))-(1:n)*log(365)
  pAc=exp(logpAc)
  pA = 1- pAc
  pA
  names(pA)=1:n #C

  pie(pA, col=rainbow(n))

  mat = matrix(c(round(pA,4)), nr=n, nc=1, byrow=FALSE,
               dimnames=list("Number in room"=1:n, c("pA")))
  as.table(mat)
}

mybirthday(50)