Mathematics, 04.07.2019 12:00 saraderrico

Suppose there are 30 people at a party. do you think any two share the same birthday? let's use the random-number table to simulate the birthdays of the 30 people at the party. ignoring leap year, let's assume that the year has 365 days. number the days, with 1 representing january 1, 2 representing january 2, and so forth, with 365 representing december 31. draw a random sample of 30 days (with replacement). these days represent the birthdays of the people at the party. would you expect any two of the birthdays to be the same?

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In this question, you are asked the probability for any of the 30 person to have the same birthday. To answer this it will be easier to calculate how much the probability for no one has same birthday. Let say the first person birthday is 1. Then the next person birthday should be other than 1, which mean 364 possible days out of 365 days. The next person should be 363 possible days out of 365 days
Then the calculation for 30 people would be:

(365!/365-30!)/(365^(30)= (365!/335!)/ 365^30= 29.4%
Then the probability of at least two person have same birthday would be: 100%-29.4%= 70.6%

Answer from: 21stcenturycyber

The probability that at least 2 people have the same birthday is 29.37%

Further explanation

Probability is the likelihood of an event occurring. Probability is the number of ways of achieving success. Probability is also the total number of possible outcomes.

Suppose there are 30 people at a party. Do you think any two share the same birthday?

Let's use the random-number table to simulate the birthdays of the 30 people at the party, ignoring leap year.

Let's assume that the year has 365 days. number the days, with 1 representing January 1, 2 representing January 2, and so forth, with 365 representing December 31.

Draw a random sample of 30 days (with replacement). These days represent the birthdays of the people at the party. Would you expect any two of the birthdays to be the same?

$1^{st} people = \frac{365}{365} \\ 2^{nd} people = \frac{364}{365} \\ 3^{rd} people = \frac{363}{365}$

For 30 people

365! = 365*364*363*...336

$= \frac{365*364*363*...336}{(365^{30})} = \frac{365!}{(365^{30})}$

$\frac{365!}{(365^{30})}$ $= \frac{365!/(365-30)!}{365^{30}}$

$= \frac{365!/335!}{365^{30}} \\ = 0.2937 = 29.37%$

The probability that at least 2 people have the same birthday is 29.37%

Learn moreLearn more about the same birthday linkLearn more about probability linkLearn more about random sample linkAnswer details

Grade: 9

Subject: mathematics

Chapter: probability

Keywords: the same birthday, probability, random sample, party, simulate

Answer from: Quest

horizontal translation of 2 units in the positive x direction

followed by

vertical translation of 3 units in the positive y direction

Answer from: Quest

step-by-step explanation:

a rhombus is a parallelogram enjoying his symmetry.

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