There are 18 people in a room, and each of the 18 writes down an integer at random from 1 to 100 inclusive.?

Question

Find the probability that at least two people wrote down the same number. Express your answer as a decimal to 4 signifigant digits.

Answer 1

0.8037 or 80.37%

This is why:

P(E)=1-P(not E); so the probability of at least 2 people having the same number is 1- the probability that each person has a unique number.

The probability that each person has a unique number:

First person has 100 choices out of 100 numbers, so 100/100.
Second person can't choose the first person's number, so only 99 choices out of 100, so 99/100;
Third only 98/100;
. . .
18th person, has (100-17)/100=84/100

Therefore the probability that none of these people have the same number is (100/100)•(99/100)•(98/100)•(97/100)•. . .•(84/100) ≈19.63%

So the probability that at least two people have the same number is 100%-19.63%=80.37%.

Answer 2

I'm going to assume you're familiar with the binomial distribution, which deals with the number of successes in a fixed number of Bornoulli trials. Here, a success "p" is defined as picking a specific number between 1 and 100, and a failure "q" is equal to 1 - p.

The probability that one person writes down one specific number is 1/100, so we'll say p = 0.01 and q = 0.99. Where X is the number of people who pick that specific number, here is the distribution function for the random variable X:

f(X) = (18 choose X) * .01^X * .99 ^(18-X)

We basicly want to know f(2) + f(3) + f(4) + ... + f(18). That would take a long time to calculate, so instead, I'll use this formula:

f(X>=2) = f(X<2) = 1 - f(1) - f(0)
= 1 - [(18) * .01 * .99^17] - [(1) * 1 * .99^18]
= 1 - .1517 - .8345
= .0138

Answer 3

It's probable that three people wrote the same number.

The chance of a person NOT repeating:
1st: 0/100
2nd: 1/100
3rd: 2/99
4th: 3/98
[...]
18th: 17/84

Sum them up, and the number, approximately 2 (plus the first person, so 3), is how many will have chosen the same number.

The other variant of this problem is 30 people in a class and two having the same birthday (approximately a 50% chance, barring twins).

Answer 4

This looks like a homework problem so I'm not going to answer it directly, but I'll give you a hint. It's easier to figure out the probability that NONE of the people wrote down the same number. Do that first, and then it's easy to figure out the answer to the real question.

Answer 5

homework?