Pick a sequence of 10 balls from a sack containing red, blue, green, yellow, white and black balls. Each time a ball is picked, it is replaced in the sack before the next ball is picked.
a) How many sequences have at least 3 red balls?
b) At least 4 blue balls?
c) exactly 2 black balls AND at least 4 yellow balls?
I don't know where to begin. I believe the total number of sequences is $6^{10}$. For part a) would we choose the first 3 slots out of 5 rather than 6? Overall just very confused.
Best Answer
To make notation easier, I will reword the problem to be about strings of length 10 using the characters A,B,C,D,E,F. (Just replace "A" with red, "B" with blue, etc... to go back to the original wording of the question)
The wording for part (a) is a bit confusing. In my initial interpretation, we are counting the number of ways of picking balls such that in the process of picking them, we see exactly three colors total in any order and any amount (except zero). For example, ABACCCCCAC or DAEEEEEEEE. For part (a), we try to count how many length-10 strings have exactly three letters appearing.
Actually counting $(\star)$ will require a bit of thought. We approach via Inclusion-Exclusion.
Let $\mathcal{A},\mathcal{B},\mathcal{C}$ denote the events where the character $A,B,C$ are not present respectively. Let $U$ denote the universal event.
We have by inclusion-exclusion $|\mathcal{A}^c\cap\mathcal{B}^c\cap\mathcal{C}^c|=|U\setminus(\mathcal{A}\cup\mathcal{B}\cup\mathcal{C})|=|U|-|\mathcal{A}|-|\mathcal{B}|-|\mathcal{C}|+|\mathcal{A}\cap\mathcal{B}|+|\mathcal{A}\cap\mathcal{C}|+|\mathcal{B}\cap\mathcal{C}|-|\mathcal{A}\cap\mathcal{B}\cap\mathcal{C}|$
$=3^{10}-3\cdot 2^{10}+3\cdot 1^{10}-0=55980$
So, the number of length-10 strings whose characters come from the alphabet A,B,C,D,E,F where exactly three characters appear in the string are $\binom{6}{3}(3^{10}-3\cdot 2^{10}+3)=1119600$.
Another possible interpretation of part (a) is that you forgot a crucial word. Perhaps the original question was "How many strings have exactly three RED balls." (or some other color)
This question is much easier.
For a total of $\binom{10}{3}\cdot 5^7$ possible sequences.
(I just noticed the title of question reads "at least" instead of "exactly" as it does in the body of the question. For the question of "at least" approach as in part (b) below)
Part (b) is approached similarly to the second interpretation of part (a) with a bit of inclusion-exclusion thrown in.
Let us count how many ways violate this condition. Count how many have less than 4 blue balls. The number with exactly three blue balls will be $\binom{10}{3}\cdot 5^7$ as calculated previously. With exactly two blue balls will be $\binom{10}{2}\cdot 5^8$. Similarly for exactly one and exactly zero blue balls.
There are then a total of $6^{10}-\binom{10}{3}5^7-\binom{10}{2}5^8-\binom{10}{1}5^9-5^{10}$ sequences with at least four blue balls.
Note, we could have approached directly. By approaching indirectly, we save ourselves some arithmetic. Approaching directly, we have $\binom{10}{4}5^6+\binom{10}{5}5^5+\dots+\binom{10}{9}5^1+\binom{10}{10}5^0$. You will see these equal the same thing.
For part (c), approach similarly to part (b). Break it into cases based on the number of yellow balls and add (or subtract from the total). In the case of picking exactly 2 black balls and 5 yellow balls for example,