What Are the Calculations Behind Probabilities in Combinatorial Problems?

[lesdavies123]

Hi, there are two problems I can't really solve yet they don't seem that difficult. The two of them seem pretty related to me, I think there's something I'm not getting I'll detail my attempts at solving but any help especially with the steps to the solution would be really appreciated as I have the answers, these are not homework or anything like that just me practicing.

The first one is: Consider 4 digits, what are the probabilities of getting 1 different digit, 2 of them, 3 of them or 4.
I have no problem with 1 digit, it is 10/1000. The third I can do also and the 4th, it is especially with two different digits that I struggle.

So I thought the two digits which can repeat twice can be 0,1;0,2;0,3;0,4;0,5,etc...1,2;1,3;1,4;1,5..etc,2,3;2,4;2,5...etc,2,9; until 8,9 so a total of 45 possibilities.. but out of 10000? because 10^4? Anyways the answer is supposed to be 630... I don't know how that's possible.. Thanks

2nd problem

From the population of five symbols a, b, c, d, e, a sample of size 25 is taken. Find the probability that the sample will contain five symbols of each kind.

The only thing I could think of was (4/5)^5 x (3/5)^5 x (2/5)^5 x (1/5)^5 which is obviously wrong. Apparently the answer is 25!/(5!^5 X 5^25 ) so if anyone could explain the logic behind this answer would be great! Thank you guys! :)

 

[HallsofIvy]

What, exactly, do you mean by "two different digits"? Different from what? If you number were "1101" would that count as "one different digit" (since the single digit "0" is different from the other digits) or would it count as "two different digits" (since it has "1" and "0" as digits)?

 

[lesdavies123]

I had interpreted it as counting as 2 different digits! Or else 4 different would be impossible.

 

[mathman]

For your second question, there is an unstated assumption that all the letters are equally probable.
1/5^25 is the probability for one particular sequence of 25 letters.
25!/ 5!^5 is the number of sequences with 5 of each letter.

 

[blue_raver22]

Hi there,

First, let's address the first problem. You are correct in your thinking that there are 45 possible combinations of two digits that can repeat twice out of the 10000 possible combinations. However, the reason the answer is 630 is because you need to consider the order of the digits as well.

For example, the combination 0,1,0,1 can be arranged in 4!/2!2! = 6 ways (you can think of it as 4 choices for the first digit, 3 choices for the second digit, 2 choices for the third digit, and 1 choice for the fourth digit, but since there are two 0's and two 1's, we need to divide by 2! twice to account for the repeated digits). Similarly, the combination 1,0,1,0 can also be arranged in 6 ways. So, out of the 45 possible combinations, there are actually 6+6=12 ways to arrange the digits. This means that the total number of combinations with two different digits is 45 x 12 = 540.

Now, for the second problem, let's break it down step by step. We have a population of 5 symbols (a, b, c, d, e) and we are taking a sample of size 25. This means that we will have a total of 25 symbols in our sample, and we want to know the probability that we will have exactly 5 of each symbol.

First, let's consider the probability of getting 5 a's in our sample. This would be (1/5)^5, since there is a 1 in 5 chance of getting an a for each of the 5 symbols in our sample.

Similarly, the probability of getting 5 b's in our sample would be (1/5)^5, and the same goes for c, d, and e.

Since we want to have 5 of each symbol, we need to multiply these probabilities together. This gives us (1/5)^5 x (1/5)^5 x (1/5)^5 x (1/5)^5 x (1/5)^5 = (1/5)^25.

However, we also need to consider the number of ways we can arrange these 25 symbols in our sample. This is where the 25! comes in. This represents all the possible ways