90

To solve this problem we need to figure out two thigs. First, we need to get the number of groups of three colors that include purple, and then we need to calculate how many different color schemes can we make with three given colors.

For the first part since we know we are using purple the only thing we need to do then is choose two colors from the remaining 6, and we know there’s {6 \choose 2} way’s to do that. Then given any set of three colors, How many color schemes can we create? To answer this question you can think of the three possible meanings for a color as placesments on a line, thus if you have your three colors on a line, the first one means correct, the second one means incorrect placement for the letter, and the last one means the letter is not on the string, then the question becomes how many lines can you make with three elements, and it should be clear to you thet this is 3!. Therefore the total number of schemes that include purple would be

{6 \choose 2} \cdot 3! = 90

\dfrac{100}{26^6} = 100\cdot \dfrac{P(26^6-1, 99)}{P(26^6, 100)} = 1 - \dfrac{P(26^6- 1, 100)}{P(26^6, 100)}

Whenever you are thrying to calculated the probability of something happening at least once over a series of repeated trials, it is a very good idea to consider the complement rule because in most cases calculating the probability of something never happening is easier.

Knowing that, let’s calculate the probability of not getting the string DSCTEN in any of the days. Well, the first part would be figuiring out how many possible strings exists, since there’s 26 letters on the English alphabet, it should be clear that the total number of strings is 26^6, now the number of ways we can choose 100 of them withouth chossing DSCTEN is {26^6-1 \choose 100}, and therefore the probability of DSCTEN not being in our selection of 100 strings is \frac{26^6-1 \choose 100}{26^6 \choose 100}. Notice that in this I’m thinking in terms of combinations, but it would be equally valid to think in therms of permutations, and in that case we would do the number of permutations wihtout the string DSCTEN over the total number of permutations (\dfrac{P(26^6- 1, 100)}{P(26^6, 100)}).

Now we only need to use the complement rule, and we’ll get that the answer is 1 - \frac{26^6-1 \choose 100}{26^6 \choose 100} = 1 - \dfrac{P(26^6- 1, 100)}{P(26^6, 100)}

Note: P stands for permutation in this problem

Bubble 1: A and B are independent. A and B are conditionally dependent given C.

We can test if A and B are independent by using the equation: \mathbb{P}(A \cap B) = \mathbb{P}(A) \cdot \mathbb{P}(B).

• We know \mathbb{P}(A) = \frac{6}{26} because there are six vowels the Stringle string can start with out of 26 letters in the alphabet.
• We know \mathbb{P}(B) = \frac{13}{26} = \frac{1}{2} because the letters A through M is numbered 1 to 13, so there are 13 possible letters out of 26.
• We know \mathbb{P}(A \cap B) = \frac{3}{26} because the letters A, E, and I are between A and M and are also vowels out of all 26 letters.

When plugging these values into the equation we get: \begin{align*} \mathbb{P}(A \cap B) &= \mathbb{P}(A) \cdot \mathbb{P}(B) \\ \frac{3}{26} &= \frac{6}{26} \cdot \frac{1}{2} \\ \frac{3}{26} &= \frac{3}{26} \end{align*}

This proves that A and B are independent.

Recall we can test if something is conditionally independent by doing \mathbb{P}(A \cap B|C) = \mathbb{P}(A|C) \cdot \mathbb{P}(B|C).

• We know there are 25 letters in the alphabet that are not equal to Z.
• We can calculate \mathbb{P}(A|C) = \frac{6}{25} because there are six vowels out of the 25 non-Z letters.
• We can calculate \mathbb{P}(B|C) = \frac{13}{25} because there are 13 letters out of the 25 non-Z letters.
• We can calculate \mathbb{P}(A \cap B|C) = \frac{3}{25} because the letters A, E, and I are between A and M and are also vowels out of 25 non-Z letters.

When plugging these values into the equation we get: \begin{align*} \mathbb{P}(A \cap B|C) &= \mathbb{P}(A|C) \cdot \mathbb{P}(B|C) \\ \frac{3}{25} &= \frac{6}{25} \cdot \frac{13}{25} \\ \frac{3}{25} &\neq \frac{78}{625} \end{align*}

Since \mathbb{P}(A \cap B|C) \neq \mathbb{P}(A|C) \cdot \mathbb{P}(B|C) this means that A and B are conditionally dependent given C.