High school probability Uncovered: Discover the Secrets with Plainmath's Expert Help

How do you find the probability of obtaining at least one tail when a coin is tossed five times?

In how many ways can six people sit in a six passenger car?

How do you write all possible selections of three letters that can be formed from the letters A, B, C, D, E, and F?

The question is from a standardized test. It is based around a type of probability. I have not taken a class that involves probability, as it is not taught in our standard curriculum. I am not positive on if my partial efforts are correct, but I and many others I know would appreciate some clarification. The problem goes (keep in mind this is from memory, I do not have answer choices memorized. So if you are unsure I cannot provide any choices to go from.)

"You have a jar of 10 cookies. In this jar are 5 chocolate chip cookies, 3 oatmeal cookies, and 2 raisin cookies. You reach in at random and take out a cookie. Without replacing the first cookie you pulled, you pull out a second. What is the probability that the first and second cookies will be the same type?"

As stated, this is from memory. I have spoken with friends to confirm and all agree this is the exact problem. We may be wrong, so if this is mathematically impossible to solve, be sure to let me (us) know.

From my very limited knowledge of probability, I cannot see a way to form one probability from three. You would take the probability of the first pick (for each type of cookie), 5/10, 3/10, and 2/10, and multiply them respectively by 4/9, 2/9, and 1/9. That would give you the probability of picking the same for each type.

How would you form one single probability from those three (if possible)?

As I said. I could be wrong, as if this is just not possible, then I obviously am not recalling it clearly enough. Thanks in advance.

How do you evaluate 9P4?

In how many ways can a 12- question true and false exam be answered?

This example is from the book first course in probability Example 4a.
Let $X$ denote a random variable that takes on any of the values −1, 0, and 1 with respective probabilities $P\{X=-1\}=.2P\{X=0\}=.5>P\{X=1\}=.3$.
Solution.Let $Y=X^2$. Then the probability mass function of $Y$ is given by
$P\{Y=1\}=P\{X=-1\}+P\{X=1\}=.5$
$P\{Y=0\}=P\{X=0\}=.5$
What rule is used to compute $P\{Y=1\}$? When $Y=X^2$ Why addition is used ?

You can answer any 10 questions from a total of 12 questions on an exam. In how many different ways can you select the questions?

How do you evaluate 10P7 using a calculator?

How do you find out how many ways the integers 1,2,3,4,5,6,7,8 can be placed in a circle if: at least three odd numbers are together?

A professor gives his students 7 essay questions to prepare for an exam. Only 5 of the questions will actually appear on the exam. How many different exams are possible?

How do you evaluate ${\displaystyle x{=}_6{P}_3}$?

For the experiment of tossing a coin three times, how do you find the probability of getting a head on the first toss?

You randomly select a card from a 52 card deck. What is the probability that the card is not a club?

Having been reprimanded for posting a question on the wrong site, I hope I'm not transgressing this time. In the addition rule for non mutually exclusive events $P(A\cup <!--\; \cup \; -->B)=P(A)+P(B)-<!--\; -\; -->P(A\cap <!--\; \cap \; -->B)$, while $P(A\cap <!--\; \cap \; -->B)$ can be determined by observation, why cannot it always be calculated from $P(A)\cdot <!--\; \cdot \; -->P(B)$?

Example: A lottery box contains 50 lottery tickets numbered 1 to 50. If a lottery ticket is drawn at random, what is the probability that the number drawn is a multiple of 3 or 5? $P(X\cup Y)=P(X)+P(Y)-<!--\; -\; -->P(X\cap Y)$ Therefore,
$P(XUY)=8/25+1/5-<!--\; -\; -->3/50=(16+10-<!--\; -\; -->3)/50$
$=23/50$
But using $P(X\cap Y)=P(X)\cdot <!--\; \cdot \; -->P(Y)=8/25\cdot <!--\; \cdot \; -->1/5=8/(25\cdot <!--\; \cdot \; -->5)=8/125$ which is NOT 3/50?
There are situations where $P(X\cap Y)=P(X)xP(Y)$ works perfectly, but not in others.

Note : This is not homework, it is self-study.

An employer interviews eight people for four openings in the company. Three of the eight people are women. If all eight are qualified, in how many ways could the employer fill the four positions if a)the selection is random and b)exactly two are women?

Part a) is already taken care of, now for part b). My reasoning goes like this:
There must be 2 women chosen, so imagine that for the first position we choose a woman. There's 3 ways to choose. For the second position we choose another woman, there's 2 ways to choose. For the third position we choose a man, there's 5 ways to choose. For the fourth position we choose another man, there's 4 ways to choose. Then we multiply: $3\cdot <!--\; \cdot \; -->2\cdot <!--\; \cdot \; -->5\cdot <!--\; \cdot \; -->4=120$. But the answer in the book is 30.

How do you determine the number of ways a computer can randomly generate an integer that is divisible by 4 from 1 through 12?