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Alja
2 months ago
6

Exclude leap years from the following calculations. ​(a) Compute the probability that a randomly selected person does not have a

birthday on March 14. ​(b) Compute the probability that a randomly selected person does not have a birthday on the 2 nd day of a month. ​(c) Compute the probability that a randomly selected person does not have a birthday on the 31 st day of a month. ​(d) Compute the probability that a randomly selected person was not born in February.
Mathematics
1 answer:
babunello [11.8K]2 months ago
6 0

Answer:

a) There is a 99.73% chance that a randomly picked individual does not celebrate their birthday on March 14.

b) There is a 96.71% chance that a randomly picked individual does not celebrate their birthday on the 2nd day of any month.

c) There is a 98.08% chance that a randomly picked individual does not celebrate their birthday on the 31st day of a month.

d) There is a 92.33% chance that a randomly picked individual was not born in February.

Step-by-step explanation:

The probability is calculated as the number of successful outcomes divided by the total outcomes.

A standard year comprises 365 days.

(a) Calculating the probability that a randomly selected individual does not have a birthday on March 14:

Excluding March 14 yields 365-1 = 364 days. Thus,

364/365 = 0.9973

So, there is a 99.73% chance that a randomly selected person does not celebrate their birthday on March 14.

(b) Calculating the probability that a randomly selected person does not celebrate their birthday on the 2nd day of a month:

With 12 months, there are 12 occurrences of the 2nd day.

Thus,

(365-12)/365 = 0.9671

Hence, a 96.71% chance that someone does not have a birthday on the 2nd day of any month.

(c) Calculating the probability that a randomly chosen individual does not have a birthday on the 31st day of any month:

Months with 31 days include January, March, May, July, August, October, and December.

This totals 7 instances of the 31st day.

Thus,

(365-7)/365 = 0.9808

In conclusion, there's a 98.08% chance that a randomly selected person does not celebrate their birthday on the 31st day of any month.

(d) Calculating the probability that a randomly selected person was not born in February:

February has 28 days in a non-leap year. Thus,

(365-28)/365 = 0.9233

So, a 92.33% chance that a randomly picked individual was not born in February.

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Step-by-step explanation:

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R displays antisymmetry, as whenever (a,b)∈R, it follows that a=b.

R lacks reflexivity since (1,1) ∉ R even though 1 ∈ A.

R is transitive; therefore, if (a,b)∈R and (b, c) ∈ R, then a=b=c and (a,c)=(a,a)∈R.

R fails to be a partial ordering due to its lack of reflexivity.

b): R = {(0,0),(1,1),(2,0),(2,2),(2,3),(3,3)}

R is antisymmetric because if (a,b)∈R and (b, a) ∈ R, then a must equal b (e.g., (2,0) ∈ R and (0,2) ∉ R; likewise, (2,3) ∈ R and (3,2) ∉ R).

R is reflexive since each (a,a) resides in R for all elements a ∈ A.

R is transitive; if (a,b)∈R and (b,c)∈R, it implies (a,c) exists in R or identical to (a,b) in R.

R qualifies as a partial ordering due to its reflexivity, antisymmetry, and transitivity.

c): R =  {(0,0),(1,1),(1,2),(2,2),(3,1),(3,3)}

R is reflexive as (a,a)∈R is true for every a ∈ A.

R is antisymmetric; if (a,b)∈R holds and if also (b,a)∈R, then a invariably equals b (e.g., (1,2)∈R while (2,1) ∉ R; similarly for (3,1) and (1,3)).  

R fails transitivity because (3,1) ∈ R and (1,2) ∈ R, but (3,2) ∉ R.

R is not a partial ordering due to transitivity not being satisfied.

d): R =  {(0,0),(1,1),(1,2),(1,3),(2,0),(2,2),(2,3), (3,0),(3,3)}

R exhibits reflexivity since (a,a)∈R for each element a ∈ A.

R displays antisymmetry, as if (a,b)∈R and (b,a)∈R then a must equal b (e.g., (1,2)∈R and (2,1)∉R; similarly validated for others).

R is not transitive because (1,2)∈R and (2,0)∈R, but (1,0)∉R.

R is not a partial ordering due to transitivity issues.

e):  R = { ( 0, 0 ), ( 0, 1 ), ( 0, 2 ), ( 0, 3 ), ( 1, 0 ), ( 1, 1 ), ( 1, 2 ), ( 1, 3 ), ( 2, 0 ), ( 2, 2 ), ( 3, 3 ) }

R proves to be reflexive, given that (a,a)∈R for all a∈A.

R is not antisymmetric since both (1,0)∈R and (0,1)∈R hold while 0 is distinct from 1.

R lacks transitivity, as (2,0)∈R and (0,3)∈R, while (2,3)∉R.

R cannot be classified as a partial ordering as it fails in both antisymmetry and transitivity.

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