[SOLVED] probability question(digits)

Question:

A four-digit number, in the range 0000 to 9999 inclusive, is formed. Find the probability that

(a) the number begins or ends with 0,

(b) the number contains exactly two non-zero digits.

Dear reader,

Can you help me solve this question? I have tried but didn't manage to get the real answer(but I strongly support my own answer).

Thanks (Happy)

Permutations and Combinations

Hello stpmmaths Quote:

Originally Posted by

**stpmmaths**

I thought we should consider the position(placement) of the digits? If we do it by http://www.mathhelpforum.com/math-he...1&d=1249776761 we are not considering the position. According to your statement " **imagine you have 4 empty boxes in a line in front of you. You have to (i) place a zero into two of the boxes, and then (ii) place a digit from 1 to 9 in the remaining two boxes.**

For stage (i), there are $\displaystyle \binom42 = 6$ ways of choosing which boxes shall contain the zeros.

" (Wondering)

This makes me confused between http://www.mathhelpforum.com/math-he...1&d=1249777134 and http://www.mathhelpforum.com/math-he...1&d=1249777134.

Lets say we change the question to:-

A four-digit number, in the range 0000 to 9999 **without repetitions for all digits except digit 0**, is formed. Find the probability that

(a) ....

(b) the number contains exactly two non-zero digits.

You are raising two different points here. First, the difference between $\displaystyle ^nP_r$ and $\displaystyle ^nC_r$.

- $\displaystyle ^nP_r$ gives the number of ways of choosing $\displaystyle r$ objects from $\displaystyle n$,
*and then arranging them in order*. So, for instance, if we want to choose 2 from 4, and the objects are A, B, C and D, then the choice A, C will be counted as a different choice from C, A. There are therefore $\displaystyle ^4P_2=4 \times 3 = 12$ possible choices.

- $\displaystyle ^nC_r$ gives the number of ways of choosing $\displaystyle r$ objects from $\displaystyle n$,
*where the order doesn't matter*. In the above example, A, C and C, A would be counted as the same choice, and there are therefore $\displaystyle ^4C_2=\frac{4\times 3}{2}= 6$ possible choices.

As an example, suppose that a committee chairman and secretary have to be chosen from a short-list of 4 people. Here, A, C is a different selection from C, A because A and C are fulfilling different roles. So you use $\displaystyle ^4P_2$. But if you have to choose 2 delegates to attend a conference, from 4 people, you'll use $\displaystyle ^4C_2$. Why? Because A, C and C, A are now the same choice - there's no difference between the roles of A and C.

So, in question (b), the number of different pairs of positions that can be occupied by a zero is $\displaystyle ^4C_2=6$, because one zero looks just like the other - the order in which you place them in the 'empty boxes' doesn't matter.

In each of the remaining boxes (containing non-zeros) the order in which we fill the boxes *does* matter: 0709 will be different from 0907, for instance. Each box can then be filled in 9 ways, so - using the 'r-s Principle' - the total number of ways in which these two boxes can be filled, one after the other, is $\displaystyle 9^2 =81$.

Your second question is: how is this affected if we're not allowed to choose the same non-zero digit twice? The answer is this:

- The first stage - choosing which 'boxes' are to be filled with zeros - is unaltered. It's still 6 choices.

- In the second stage, suppose we fill the left-hand of the two remaining boxes first. There are 9 ways of doing this. Since we're not allowed to choose the same digit again, there are then 8 ways of filling the last box. Total number of choices: $\displaystyle 9\times 8 = 72$. Notice that this is the number of ways of choosing 2 items from 9, where the order matters; in other words, it's $\displaystyle ^9P_2$.

So the overall answer to your modified question would be $\displaystyle 6 \times 72 = 432$ choices, or a probability of $\displaystyle 0.0432$.

Does that clear things up?

Grandad