Rom

I don't think the last answer in #1 should be checked.  We're given no information about a probability of vaccinating a cat.

One would think the vet would know how many cats they vaccinated exactly.  After all someone has to pay the bill!

The rest is correct.

In #2 the first two are indeed likely to be normally distributed, if for no other reason than that there are so many people entering the store the law of large numbers kicks in.

The rest of #2 you have correct.

\(\text{The math books can be either to the left or right of the english books. That's 2 choices}\\ \text{Within the math books there are $3!$ arrangements}\\ \text{Within the english books there are $5!$ arrangements}\\ \text{The total number of arrangements is thus $2 \cdot 3! \cdot 5! = 1440$}\)

\(\text{What you need is $n \ni \dbinom{n}{4} \geq 365$}\\ \text{Whipping out our pocket Pascal's Triangle we see that}\\ \dbinom{11}{4} = 330\\ \dbinom{12}{4} = 495\\ \text{So a group of 12 people and you're good to go}\)

\(\text{let the number of cards left in the deck be $N$}\\ N \pmod{4} = 3\\ N \pmod{3} = 2\\ N \pmod{5} = 2\\ N \approx 52,~N < 52\)

\(N =4k+3\\ 4k+3 \pmod{3} = 2\\ k +(3k+3) \pmod 3 = 2\\ k \pmod{3} = 2\\ k = 3j + 2\\ N = 12j + 11\\ 12j + 11 \pmod{5} = 2\\ 2j+1 + (10j + 10) \pmod{5} = 2\\ 2j+1 \pmod{5} = 2\\ 2j \pmod{5}=1\\ 2j \pmod{5}=6\\ j \pmod{5} = 3\\ j = 5i+3\\ N = 60i+36+11 = 60i + 47\\ N=47 \)

\(\text{Give an apple to each child. You now have to distribute 3 apples and 6 oranges w/o any restrictions}\\ \text{This is a simple 2D stars and bars problem. You can independently distribute the apples and oranges}\\ N_A = \dbinom{3+4-1}{4-1} = \dbinom{6}{3} = 20\\ N_O = \dbinom{6+4-1}{4-1} = \dbinom{9}{3} = 84\\ N=N_A N_O = 1680\)

\(P[\text{first was blue | second was blue}] = \\ \dfrac{P[\text{second was blue | first was blue}]P[\text{first was blue}]}{P[\text{second was blue}]} = \\ \dfrac{\dfrac 1 5 \dfrac 1 3}{\dfrac 1 5 \dfrac 1 3 + \dfrac 1 3 \dfrac 2 3} = \dfrac{\dfrac{1}{15}}{\dfrac{13}{45} }= \dfrac {3}{13}\)

\(T= \dfrac{80}{64} + \dfrac{200-80}{50} = \dfrac{5}{4}+\dfrac{12}{5} = \dfrac{73}{20}\)

\(\bar{v} = \dfrac{200}{\frac{73}{20}} = \dfrac{4000}{73} \approx 54.8 ~mph\)

more expensive is closer to the mark

A choc frog is 11 cents dearer than a green frog --> A choc frog costs 11 cents more than a green frog

\(p= \dfrac{2}{\dbinom{31}{2}}\cdot \dfrac{1}{58}\)

No there's no solution to this.