The problem says "an negative or an even number," which we could interpret as meaning either a negative or an even or both — or not both. Let's work out the probabilities for each.

First, let P(E) be the probability of getting an even number and P(–) be the probability of getting a negative. It's easy to see that given our set of 9 elements,

$$P(E) = \frac{4}{9} \phantom{00} \text{and} \phantom{00} P(-) = \frac{3}{9}$$

If we mean either or both, the probability is

$$P(E \cup -) = \frac{4}{9} + \frac{3}{9} = \frac{7}{9} = 0.78$$

If we want the probability of obtaining an even or a negative, but not both, we have to subtract from the first result the probability of obtaining both:

$$ \begin{align} P &= P(E \cup -) - P(E \cap -) \\[5pt] &= \frac{7}{9} - \frac{3}{9} \cdot \frac{4}{9} \\[5pt] &= \frac{7}{9} - \frac{12}{81} \\[5pt] &= \frac{51}{81} = 0.63 \end{align}$$

Here is a Venn diagram of this scenario.

The figure shows the 52 unique possibilities for card types in a 52-card deck.

The black boxes outline our two outcomes, spades and red kings. These sets don't overlap, so the probability of drawing from either of these sets is just

$$\begin{align} \frac{13}{52} + \frac{2}{52} &= \frac{15}{52} \\[5pt] &\approx 29 \% \end{align}$$

The table shows that there are 24 total outcomes. On the left are the four outcomes for two coins, {HH, HT, TH, TT}, and above are the six outcomes for rolling a six-sided die.

Numerically, the result is $n = n_c \cdot n_c \cdot n_d,$ where $n_c$ and $n_d$ are the number of possible results for the coin flip and die roll, respectively. So:

$$ \begin{align} n &= n_c^2 \cdot n_d = 2^2 \cdot 6 \\[5pt] &= 24 \end{align}$$

There are 11 letters in the word and there are 11! ways of arranging 11 unique letters. These letters are not unique, however, so we have to divide by 2! for the number of P's, 4! for the number of ways of rearranging the S's and 4! again for the number of ways of rearranging the four I's.

$$n = \frac{11!}{2! \cdot 4! \cdot 4!} = 34,650$$

So there are 34,650 unique combinations of the letters in the name MISSISSIPPI.

This problem is similar to #4 above. There are two G's and two U's in this word, so in any configuration of the 10-letter string, these could be swapped and we wouldn't know it. The number of unique rearrangements is

$$n = \frac{10!}{2! \cdot 2!} = 907,200$$

So there are 907,200 unique arrangements of those ten letters. That's amazing.

The choices of decks, trucks and wheels are all independent, so the number of boards is just

$$ \begin{align} n &= n_d \cdot n_t \cdot n_w \\[5pt] &= 12 \cdot 3 \cdot 8 = 288 \end{align}$$

One reason to study a problem like this is that it reminds us that it's sometimes easier to solve the converse problem and then convert at the end. That is, if we're asked for P(B), sometimes it's easier to calculate P(!B), and then recognize that P(B) = 1 - P(!B).

So let's start with a single person in our room and ask what the probability is that he doesn't share a birthday with anyone else in the room. Well, of course that's the easy one. There are 365 out of 365 chances. The chance of him sharing a birthday is

$$P(B) = 1 - \left( \frac{365}{365} \right) = 0.$$

If we add a second person, the first person takes up one possible birthday, so the chances that they don't share a birthday are 364/365. Overall, the probability of sharing a birthday is then

$$P(B) = 1 - \left( \frac{365}{365} \right) \left( \frac{364}{365} \right) = 0.0027$$

Adding the third person reveals a pattern,

$$ \begin{align} P(B) &= 1 - \left( \frac{365}{365} \right) \left( \frac{364}{365} \right) \left( \frac{363}{365} \right) \\[5pt] &= 0.0082 \end{align}$$

Now we want this to be greater than or equal to 0.5, and the pattern should be fairly obvious. We can write

$$1 - \left( \frac{365}{365} \right) \left( \frac{364}{365} \right) \dots \left( \frac{365-n+1}{365} \right),$$

where $n$ is the number of people in the room.

Now this formula can be expanded using a spreadsheet, calculating probabilities for each $n.$ If you do this (it's easy), you'll find that once the number of people gets to 23 you're past the 50% probability mark. The numbers from such a spreadsheet for $n$ up to about 70 are graphed below. Notice that by 70 people or so, it is almost certain that at least two people will share the same birthday.

One way to think about why this is so is to distinguish between asking what's the probability that two people will share some pre-specified birthday, like July 30, and that they'll simply share some unspecified birthday of the 365 days available. Those are different problems.

The number of possibilities for the first number is 40. With that number gone (no replacement), the number of possibilities for the second number is 39. If we carry out this line of reasoning, we get 40·39·38·37·36·35 possibilities.

But, we have to reduce this by the number of ways of arranging those numbers, which is 6!, so we have

$$ \begin{align} n &= \frac{40 \cdot 39 \cdot 38 \cdot 37 \cdot 36 \cdot 35}{6!} \\[5pt] &= 3,838,380. \end{align}$$

So the odds of winning this lottery are 1 in 3,838,380.

We can write our pool set like this:

$$ \begin{align} P = \{&A, A, A, A, A, A, A, A \\[5pt] &A, A, A, S, S, S, S, S\}, \end{align}$$

where A is a generic astronaut and S stand for soil specialist.

For the soil specialists, we have 5 choices for the first and four for the second (no replacement), for 20 choices.

For the other astronauts, we pick two from a field of ten, so the number of groups is 10·9 = 90.

So the number of choices is 90·20 = 1800. But we have to account for the fact that some of these teams contain the same people, just arranged in a different order, and that shouldn't matter. So the number of teams is

$$n = \frac{5 \cdot 4 \cdot 10 \cdot 9}{2! \cdot 2!} = 450.$$

To solve this one, I began with the quick little python program shown here. It simply counts all of the possibilities.

For the ascending numbers, the smallest is 123 and the largest is 789, with 84 total possibilities that meet the ascending requirement. That's just the binomial $\binom{10}{3}$ which gives only one way of rearranging the numbers. That is, 789 is allowed but 798, 879, 897, 987 and 987 are not.

$$ \begin{align} \binom{9}{3} &= \frac{n!}{k!(n-k)!} \\[5pt] &= \frac{9!}{3!(9-3)!} \\[5pt] &= 84 \end{align}$$

For the descending numbers, we can use zero. The largest is 987 and the smallest is 210. We have:

$$ \begin{align} \binom{10}{3} &= \frac{n!}{k!(n-k)!} \\[5pt] &= \frac{10!}{3!(10-3)!} \\[5pt] &= 120 \end{align}$$

These are just the solutions we get from the python code.

                                
# Ascending: largest possible number
# is 789
i = 1
count = 0
while i < 8:
    j = i+1
    while j < 9:
        k = j+1
        while k < 10:
            count+= 1
            k+=1 
        j+=1
    i+= 1
    print('count (increasing) = ', count)

# Descending: smallest number is 210;
# note that zero can be used in this
# case
i = 9
count = 0
while i > 2:
    j = i-1
    while j > 1:
        k = j-1
        while k >= 0:
            count+= 1
            k-=1 
        j-=1
    i-= 1
print('count (decreasing) = ', count)
                                
                                

The results from the code are: Ascending = 84, Descending = 120

Notice that this problem is asking for a guarantee that we'll draw ten of one number, so we have to be thorough. Drawing marbles labeled 1-9 doesn't make any difference because there aren't ten of any of those. The number of those is 1 + 2 + 3 + ... + 9 = 45 marbles. We add those to our list.

Now from the remaining marbles, if we draw nine of each kind, that's is 41·9 = 369 marbles. Now if one more marble is drawn, it has to match up with one of the other groups of nine. So our minimum number is

$$45 + 369 + 1 = 415$$

Let P(Y) be the probability that the Yankees win a game, then 1-P(Y) is the probability that the RedSox win (which would, of course, be expected). Then we can list all of the possible series outcomes in a table:

Series	Probability
YRY	P(Y)[1 - P(Y)]P(Y)
RYY	[1 - P(Y)]P(Y)P(Y)
YY*	[1 - P(Y)]P(Y)P(Y)

If the Yankees win two in a row, a third game is not required.

Now if we sum these probabilities, we'll have the probability that the Yankees win the series:

$$P^2(1 - P) + P^2(1 - P) + P^2 = \frac{P}{1.6},$$

where P is an abbreviation for P(Y) just to make things easier, and the right side reflects that P(win series)·1.6 = P. Now we can solve this equation with a little bit of algebra:

$$ \begin{align} 2P^2(1 - P) + P^2 &= \frac{P}{1.6} \\[5pt] 2P^2 - 2P^3 + P^2 &= \frac{P}{1.6} \\[5pt] 3P^2 - 2P^3 &= \frac{P}{1.6} \\[5pt] 3P - 2P^2 &= \frac{1}{1.6} \\[5pt] -2P^2 + 3P &= \frac{1}{1.6} \\[5pt] P^2 - \frac{3}{2}P &= -\frac{1}{3.2} \\[5pt] \end{align}$$

$$ \begin{align} P^2 - \frac{3}{2}P + \left( \frac{3}{4} \right)^2 &= \frac{9}{16} - \frac{1}{3.2} \tag{1} \\[5pt] &= \frac{18}{32} - \frac{10}{32} \\[5pt] &= \frac{1}{4} \\[5pt] \left( P - \frac{3}{4} \right)^2 &= \frac{1}{4} \tag{2} \\[5pt] P &= \frac{3}{4} ± \frac{1}{2} \\[5pt] &= 0, \frac{5}{4}, \frac{1}{4} \tag{3} \end{align}$$

Here steps 1, 2 and 3 are the process of completing the square. In step (3) we realized that by dividing everything through by P above, we dropped a root, P = 0. Now of these three solutions, we know that 5/4 is not a probability and that P(Y) ≠ 0, so the Yankees have a probability of winning the series of ¼.