Trial # Number of people (day) 1 2(214) 2 2(190) 3 2(54,117,256) 4 2(5,295) 5 2(

Question

Trial #

Number of people (day)

1

2(214)

2

2(190)

3

2(54,117,256)

4

2(5,295)

5

2(182,275)

6

2(148,205)

7

0

8

0

9

2(80,220)

10

2(300,353)

For the above the table shows there are 10 trials and in each trial there are 35 people. The right side shows how many people in that trial have the same birthday - for exampel trial 1 shows there are 2 people sharing the 214th day of the year as their birthdays.

a)Based on this table then, what is the estimate for the probabilty of at least 2 people in a group of 35 sharing the same birthday (Hint the relative frequency of trialks with one or more birthday match is an estimate of the probability).

b) This part requires the software R so if anyone has it can you please help me with this one as well....I am to compute the exact prob of at lest 2 people in a group of 35 sharing the same birthday. Note that R functon prod(x) computes the product of all elements in vector x. Then compare the estimate that you found in part a) with this exact probability. You should refer your explanation to "number of trials" instead of "sample size"

Thanks!

Trial #

Number of people (day)

1

2(214)

2

2(190)

3

2(54,117,256)

4

2(5,295)

5

2(182,275)

6

2(148,205)

7

0

8

0

9

2(80,220)

10

2(300,353)

Explanation / Answer

PART A

As given here out of 10 trials 8 trials have at least 2 birthdays on same day out of 10 randomly chosen trials.

so required probabilty of at least 2 people in a group of 35 sharing the same birthday is 8/10 i.e 0.8

PART B

To find the exact probability of the above event it is easier to compute 1 - probability that all have their birthday on diffrent days.

here favorable ways are firstly choosing 35 out of 365 days then arranging them in 35 factorial ways

and number of total trial possible is 365^35.

i.e 1 - (365 Choose 35 )*(35 !) / 365^35.

computing it with r the required answer is... 1-0.1856 equals to 0.8144

it is clearly evident from the computation that our expected probability is is significantly close to the exact one .

CODE used in R is:

> x<-c(0:34)
> x<-365-x
> y<-c(1:35)
> y<-y*0
> y<-y+365
> prod(x/y)
[1] 0.1856168

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