If Hector has a shelf of books. He wants to choose a random book to read. If there are 24 books on the shelf and he has an equal chance of choosing any particular one, what is the probability that he picks the third book from the left?

This is a wordy problem, but if we cut through to what it's asking it is fairly simple.

We need to find the probability of an event.

The event is randomly choosing a book from a book shelf.

There are 24 books total, and we need to pick 1.

So, our probability will be 

\(\displaystyle P= \frac{\#Desired}{Total\#}=\frac{1}{24}\)

So our answer is:

\(\displaystyle \frac{1}{24}\)

Each event is independent of another.

A fair coin has either heads or tails, which means the probability of her rolling heads or tails will be \(\displaystyle \frac{1}{2}\).

The probability of her rolling two tails is:  \(\displaystyle (\frac{1}{2})^2 = \frac{1}{4}\)

In a six sided dice, the odds of rolling any particular number is one-sixth.

If Julie was to roll a number other than four, this means that she has five out of six possible rolls that she can land on. 

The probability of rolling a number other than four is:  \(\displaystyle \frac{5}{6}\)

Multiply the probabilities.

\(\displaystyle \frac{1}{4}\times \frac{5}{6} = \frac{5}{24}\)

The answer is:  \(\displaystyle \frac{5}{24}\)

\(\displaystyle P (7, 5)\)

\(\displaystyle \frac{7!}{(7 - 5)!} = \frac{7!}{2!} = \frac{7 \times 6 \times 5 \times 4 \times 3 \times 2!}{2!} = \frac{5,040}{2}=2520\) 

There are five ways to choose the first scoop, then four ways to choose the second scoop, and finally three ways to choose the third scoop:

5 * 4 * 3 = 60

2 men can be selected:

\(\displaystyle C\binom{5}{2 } = 10\)

2 women can be selected out of 4 women:

\(\displaystyle C\binom{4}{2} = 6\)

Finally, after the selection process, these men and women can fill the executive body in \(\displaystyle 4!=24\) ways.

This gives us a total of \(\displaystyle 10\times 6\times 24 = 1440\)

In this problem, order is important because once someone is chosen as a position they can not be chosen again, and once a position is filled, no one else can fill that in mind.

The presidential spot has a possibility of 15 choices, then 14 choices for vice president and 13 for the treasurer.

So:

\(\displaystyle 15*14*13=2730\)

There are 7 letters in the word jubilee, so initially we can calculate that there are \(\displaystyle 7! = 5,040\) ways to re-arrange those letters. However, The letter e appears twice, so we're double counting. Divide by 2 factorial (2) to get \(\displaystyle 2,520\).

At first, it makes sense that there are \(\displaystyle 6! = 720\) ways to re-arrange these letters. However, the letter A appears 3 times and the letter N appears twice, so divide first by 3 factorial and then 2 factorial:

\(\displaystyle \frac{720}{3! \cdot 2! } = \frac{720}{6 \cdot 2 } = \frac{720 } { 12} = 60\)

To solve, evaluate \(\displaystyle \binom{24}{4} = \frac{24! }{4! \cdot 20!} = 10,626\)

There are 3 winners out of the total set of 13. That means we're calculating \(\displaystyle 13P3 = \frac{13! }{(13-3)!} = \frac{13!}{10!} = 1,716\)