# Assume that the year has 366 days and all birthdays are equally likely. a) What is the probability that two people chosen at random were born on the same day of the week? b) What is the probability that in a group of n people chosen at random, there are at least two born on the same day of the week? c) How many people chosen at random are needed to make the probability greater than 1/2 that there are at least two people born on the same day of the week?

###### Question:

a) What is the probability that two people chosen at random were born on the same day of the week?

b) What is the probability that in a group of n people chosen at random, there are at least two born on the same day of the week?

c) How many people chosen at random are needed to make the probability greater than 1/2 that there are at least two people born on the same day of the week?

## Answers

Answer:The answers to the question area) [tex]\frac{1}{7}[/tex]b) Where n >7 the probability = 1 n = 7 it is 0.99388n = 6 it is 0.95716n = 5 it is 0.850062 n = 4 it is 0.65n = 3 it is 0.38776n = 2 it is 0.14286n = 1 it is 0c) n > 3 where n is an integerStep-by-step explanation:The probability for the first person to be born on a certain day is 1P(Week Birth Day) = 7/7 = 1The probability for the second person to be born on the same day as the first person is P(Week Birth Day Second l Week Birth Day) = 1/7By conditional probability, we have P(B|A) = [tex]\frac{P(AnB}{P(A)}[/tex] that is [tex]\frac{1*\frac{1}{7} }{1} = \frac{1}{7}[/tex] (b) Since there 7 days in a week, the probability that in a group of n people chosen at random, there are at least two born on the same day of the week where n ≥ 8 is 1 as there must be two people born on the same dayP(A) = 1 when n ≥ 8The probability that at least two were born on the same day when n < 8 then there are Number of ways to choose the birthday of first person = 7Number of ways to choose the birthday of second person = 6Number of ways to choose the birthday of third person =5and so on till we haveNumber of ways to choose the birthday of seventh person = 1P(At least 2 born on the same day) can then be found by the complement rule thus Let P(At least 2 born on the same day) = P(A)Then we have P(A) = 1 - P(A)' where P(A)' = P (No 2 born on the same day)Thus we have P(A) = 1 - [tex]\frac{7}{7} *\frac{6}{7}* \frac{5}{7} *\frac{4}{7} *\frac{3}{7}* \frac{2}{7} *\frac{1}{7}[/tex] = 1 - [tex]\frac{720}{117649}[/tex] = 0.99388n = 6 we have P(A) = 1 - [tex]\frac{7}{7} *\frac{6}{7}* \frac{5}{7} *\frac{4}{7} *\frac{3}{7}* \frac{2}{7}[/tex] = 1 - [tex]\frac{720}{117649}/\frac{1}{7}[/tex] =1 - [tex]\frac{720}{16807}[/tex] =0.95716n = 5 gives P(A) =1 - [tex]\frac{7}{7} *\frac{6}{7}* \frac{5}{7} *\frac{4}{7} *\frac{3}{7}[/tex] = 1 - [tex]\frac{720}{16807}/\frac{2}{7}[/tex] = 1- [tex]\frac{360}{2401}[/tex] = 0.850062 n = 4 P(A) =1 - [tex]\frac{7}{7} *\frac{6}{7}* \frac{5}{7} *\frac{4}{7}[/tex] = 1 - [tex]\frac{360}{2401} / \frac{3}{7}[/tex] = 1 - [tex]\frac{120}{343}[/tex] = [tex]\frac{223}{343}[/tex] = 0.65n = 3 P(A) =1 - [tex]\frac{7}{7} *\frac{6}{7}* \frac{5}{7}[/tex] = 1 - [tex]\frac{120}{343}*\frac{7}{4}[/tex] = 1 -[tex]\frac{30}{49}[/tex] = [tex]\frac{19}{49}[/tex] = 0.38776n = 2 P(A) =1 - [tex]\frac{7}{7} *\frac{6}{7}[/tex] = 1 -[tex]\frac{30}{49}*\frac{7}{5}[/tex] = [tex]\frac{1}{7}[/tex] = 0.14286(c) For the probability to be greater than half that there are at least two people born on the same day of the week is given by n = 4, P(A) = 0.65n = 3, P(A) = 0.38776Therefore for a probability greater than 4, n should be > 3n > 3