Mathematics, 07.03.2020 05:44 Dhejxbdbbxbdbbx

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?

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Answer from: Aviannakelly

The answers to the question are

a) $\frac{1}{7}$

b) Where

n >7 the probability = 1

n = 7 it is 0.99388

n = 6 it is 0.95716

n = 5 it is 0.850062

n = 4 it is 0.65

n = 3 it is 0.38776

n = 2 it is 0.14286

n = 1 it is 0

c) n > 3 where n is an integer

Step-by-step explanation:

The probability for the first person to be born on a certain day is 1

P(Week Birth Day) = 7/7 = 1

The 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) = 1/7

By conditional probability, we have

P(B|A) = $\frac{P(AnB}{P(A)}$ that is $\frac{1*\frac{1}{7} }{1} = \frac{1}{7}$

(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 day

P(A) = 1 when n ≥ 8

The 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 = 7

Number of ways to choose the birthday of second person = 6

Number of ways to choose the birthday of third person =5

and so on till we have

Number of ways to choose the birthday of seventh person = 1

P(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 - $\frac{7}{7} *\frac{6}{7}* \frac{5}{7} *\frac{4}{7} *\frac{3}{7}* \frac{2}{7} *\frac{1}{7}$ = 1 - $\frac{720}{117649}$ = 0.99388

n = 6 we have P(A) = 1 - $\frac{7}{7} *\frac{6}{7}* \frac{5}{7} *\frac{4}{7} *\frac{3}{7}* \frac{2}{7}$ = 1 - $\frac{720}{117649}/\frac{1}{7}$ =1 - $\frac{720}{16807}$ =0.95716

n = 5 gives P(A) =1 - $\frac{7}{7} *\frac{6}{7}* \frac{5}{7} *\frac{4}{7} *\frac{3}{7}$ = 1 - $\frac{720}{16807}/\frac{2}{7}$ = 1- $\frac{360}{2401}$ = 0.850062

n = 4 P(A) =1 - $\frac{7}{7} *\frac{6}{7}* \frac{5}{7} *\frac{4}{7}$ = 1 - $\frac{360}{2401} / \frac{3}{7}$ = 1 - $\frac{120}{343}$ = $\frac{223}{343}$ = 0.65

n = 3 P(A) =1 - $\frac{7}{7} *\frac{6}{7}* \frac{5}{7}$ = 1 - $\frac{120}{343}*\frac{7}{4}$ = 1 - $\frac{30}{49}$ = $\frac{19}{49}$ = 0.38776

n = 2 P(A) =1 - $\frac{7}{7} *\frac{6}{7}$ = 1 - $\frac{30}{49}*\frac{7}{5}$ = $\frac{1}{7}$ = 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) = 0.65

n = 3, P(A) = 0.38776

Therefore for a probability greater than 4, n should be > 3

n > 3

Answer from: Quest

hi there! my name is zalgo and i am here to you out today. the answer is pretty simple (in a way). this equation equals to 25.2!

i hope this answer !

"stay and stay proud! " - zalgo

Answer from: Quest

what graph

step-by-step explanation:

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Assume that the year has 366 days and all birthdays are equally likely.
a) What is the probab...