A problem on probability (so what are the odds...?)?

Question

if we were to randomly choose any numbers from zero to nine and lets say we similarly choose 4 times. now dat would give us 4 numbers

so what is the probability for 2 numbers among them to be the same.?

thanks a lot.

Answer 1

Probability of choosing a number between 0 and 9 is 1/10.

The Binomial distribution has the probability

function P(x=r)= nCr p^r (1-p)^(n-r)

r=0,1,2,.....,n

where nCr = n! / r! (n-r)!

Let us assume that the number is 2 that you want to appear twice. It can be any number. I just chose 2.

n=4 (choose 4 times)

p=1/10 (probability of any number)

r=2 (number of times a number appears in 4 draws)

(4C2)(1/10)^2(9/10)^2

=6(1/100)(81/100) = 486/1000 =0.0486

Multiply this by 10 for each number, 0,1,2,3,4,5,6,7,8,9

0.0486 x 10 = 0.486 is the probability that two of them are the same out of 4.

Answer 2

Choosing 4 digits among ten is like randomly picking a number from 0 to 9999. You have 10^4 possibilities. Now within these 10 000 numbers from 0000 to 9999 you have to count those who have at least 2 equal digits.

To do so, first consider the digit which appears at least two times is given, determine where these two digits lay in the 4 digits sequence.

The first digit is in position 0, the second must be in position 1, 2 or 3.

This make 3 possibilities.

The first digit is in position 1, the second must be in position 2 or 3.

This make 2 possibilities.

The first digit is in position 2, the second must be in 3.

This make 1 possibilities.

So we have 3+2+1=6 possibilities.

Note that this number is C42=4!/[2!*(4-2)!]=4!/(2!*2!)=4!/4=3!=6.

Finally multiply by 10, the number of digits. So we get 60.

The searched probability is 60/10000=6/1000.

Cnp appears in the Newton Formula: (a+b)^n=SIGMA(Cnp*a^p*b^(n-p)) where p takes all integer values from 0 to n.

Cnp is called the number of combinations of p elements within n elements and  equals n!/(p!*(n-p)!) p<=n.

Furthermore given a finite set X with m elements and a finite set Y with n elements, the set of all applications from X to Y is finite ans has n^m elements.

This is used to prove that:

1) Given an alphabet of 26 letters you can form 26^n words whose length is n.

2) Given a set of 10 digits you can form 10^n numbers whose length is n.

I Hope it helps.

Bye.

Answer 3

Suppose the same numbers are the 1st and 2nd numbers picked.

There are 10 choices for the 1st number. The is only 1 choice of the 1nd number because it must be the same as the first. There are now 9 choices for the 3rd because 1 of then already used by 1st and 2nd number. Similarly, the 4th has only 8 choices.

Probability of this happening is (10*1*9*8)/10^4=0.072 .

But the same numbers need not be the 1st and 2nd numbers. In fact, there are 6 possibilities that have 2 numbers the same:

6*0.072=0.432