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Let $S$ be the set of all 4-digit numbers greater than 6999.

An element of $S$ is randomly chosen. Consider the following events:

  • A: The chosen number is even;
  • B: The digits of the chosen number are all different.

Find out if these events are independent.

About this exercise my book gives me the hint that independent events have the following property:

$$p(A \cap B) = p(A)\cdot p(B)$$

I know that

  • $p(A) = \frac{3\cdot10\cdot10\cdot5}{3000} = \frac{1}{2}$
  • $p(B) = \frac{3\cdot9\cdot8\cdot7}{3000} = \frac{63}{125}$
  • $p(A)\cdot p(B) = \frac{63}{250}$

My problem is that I am not sure how to calculate $p(A \cap B)$.

I think I could calculate all probabilities for the positions of odd or even numbers, so that the last digit is always even, but that would take an eternity (or maybe not?).

However, these events are obviously not independent because if the digits all have to be different then the digits of the leftmost 3 numbers are always going to influence the 4th digit.

Is this correct? Or do I really have to calculate all posibilities? If so, yes there any quicker way of doing that?

2 Answers 2

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Numbers with $4$ distinct digits : $\frac{10!}{4!}=5040$ (Including numbers beginning with a $0$)

The first digit must be $7,8,9$ : $\frac{3}{10}\cdot 5040=1512$ numbers are remaining , $504$ with each starting digit.

If the first digit is $8$, we have $\frac{4}{9}\cdot 504=224$ even numbers If it is $7$ or $9$, we have twice $\frac{5}{9}\cdot 504=280$ numbers, so in total $2\cdot 280+224=784$ numbers.

The rest should be easy.

  • 1
    Try to find out the meaning of the fractions $\frac{4}{9}$ and $\frac{5}{9}$ – 2017-02-13
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HINT

It may be easier to work with this:

$A$ and $B$ are independent iff $P(A) = P(A|B)$

So, given that you have a number with all different digits, what is the chance it is even? Or: Out of all the numbers with 4 different digits, what is the proportion whose last digit is even?

Also, you write:

However, these events are obviously not independent because if the digits all have to be different then the digits of the leftmost 3 numbers are always going to influence the 4th digit.

Careful!

'Influence' is not the same as independence!

For example, suppose I have 4 objects numbered 1 through 4. Suppose 1 and 2 are red, and 3 and 4 are blue. I now pick a random object. Let $A$ be the event that I picked an even-numbered object, and let $B$ be the event that I picked a red object.

Now, we can certainly make sense of event $B$ 'influencing' the outcome of $A$, in the sense that if I pick a red object, I can no longer pick 4.

However, given that I pick a red object, the probability of it being even-numbered is $\frac{1}{2}$ ... which is the same as picking an even-numbered object period. So in this case $P(A|B) = P(A) = \frac{1}{2}$, so $A$ and $B$ are in fact independent.

The moral is: you have to be really careful with using your intuitive notion of 'influence' in thinking about 'independence'!

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    I think $P(A | B) = \frac{ 3 \cdot 9 \cdot 8 \cdot 5}{3000} = \frac{1}{25}$, $\frac{1}{25} \not = \frac{1}{2}$, so they are not independent. Is this correct? – 2017-02-13
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    No, that's not the correct calculation for $P(A|B)$. Also, intuitively, it should still be close to half ... but is it exactly half? The fact that for the first digit you have a choice of 2 odd digits and 1 even digit may well introduce an asymmetry ... – 2017-02-13