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STALIN [3.7K]
3 years ago
5

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 simul

ate 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?
Mathematics
2 answers:
anzhelika [568]3 years ago
5 0

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

<h3>Further explanation </h3>

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

So

= \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%

<h3>Learn more</h3>
  1. Learn more about the same birthday brainly.com/question/4538530
  2. Learn more about probability brainly.com/question/12448653
  3. Learn more about random sample brainly.com/question/12384344

<h3>Answer details</h3>

Grade:  9

Subject:  mathematics

Chapter:  probability

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

Arlecino [84]3 years ago
4 0
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%
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PLEASE HELP!!!!<br> I need to solve for a, b, and c
nirvana33 [79]

Answer:

  • 25000 seats in section A
  • 14100 seats in sectin B
  • 10900 seats in section C

Step-by-step explanation:

The problem statement tells you half the total number of seats are in section A, so you already know that there are 25000 A seats. The revenue from those seats is

... 25000×$25 = $625,000

so the revenue from B and C seats must total

... $1,070,500 - 625,000 = $445,500

If all 25000 of the B/C seats were C seats, the revenue would be

... 25000×$15 = $375,000

The actual revenue from those seats is $445,500 -375,000 = $70,500 more than that. We know each B seat generates $5 more revenue, so there must be ...

... $70,500/$5 = 14,100 . . . . B seats

Then the balance of the 25000 B and C seats are C seats:

... 25,000 - 14,100 = 10,900 . . . . C seats

_____

<em>Alternate Solution Method</em>

The new Brainly answer format requires the answer be supplied before the working. In order to find the answer quickly so that I can fill in that section, I used a matrix method for solving the problem. The problem equations can be written ...

  • a + b + c = 50000
  • a - b - c = 0
  • 25a + 20b + 15c = 1070500

so the augmented matrix is ...

\left[\begin{array}{cccc}1&1&1&50000\\1&-1&-1&0\\25&20&15&1070500\end{array}\right]

A graphing calculator can be used to find the solution to this, generally using a function that produces the reduced row-echelon form. The attachment shows the solution using a TI-84 calculator.

___

<em>Comment on the Working</em>

Since the number of A seats is equal to the total of B and C seats, the number of A seats must be half the total number of stadium seats. Having figured that out, the problem is reduced to one of finding the mix of B and C seats that will produce the remaining revenue.

As with many mixture problems, it is convenient to look at differences. Start with the assumption that all of the desired revenue comes from the least contributor. Here, that is C seats. Then figure the difference that using a B seat makes ($20 -15 = $5) and the difference of the actual revenue and the amount that you got by assuming all C seats: 445,500 -375,000 = 70,500. Since replacing a C seat by a B seat adds $5 to the revenue, it is easy to figure the number of such replacements required in order to raise the revenue by $70,500.

If you write the equation for B seats, you find the solution to the equation mirrors this verbal description:

... 20b + 15(25000-b) = 445,500

... 5b = 445,500 - 375000 . . . . simplify, subtract 375000

... b = 70500/5 = 14100

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