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

How do you find the value of the permutation P(4,2)?

I've been studying stats, and currently taking my first ever engineering based stats course in college. It covers Probability extensively and other stats topics. Currently, I'm stuck on recognizing key points in a problem involving permutations / combinations vs. fundamental counting principle. I have 2 example problems and what would help the most is key things to look to recognize using the counting principle vs permutations / combinations formulas. Here's one that uses the permutations / combinations according to my student solutions manual

A friend of mine is giving a dinner party. His current wine supply includes 8 bottles of zinfandel, 10 of merlot, and 12 of cabernet (he only drinks red wine), all from different wineries.

If he wants to serve 3 bottles of zinfandel and serving order is important, how many ways are there to do this?

If 6 bottles of wine are to be randomly selected from the 30 for serving, how many ways are there to do this?

If 6 bottles are randomly selected, how many ways are there to obtain two bottles of each variety?

If 6 bottles are randomly selected, what is the probability that this results in two bottles of each variety being chosen?

If 6 bottles are randomly selected, what is the probability that all of them are the same variety?

Here's one that uses counting principle

The composer Beethoven wrote 9 symphonies, 5 piano concertos (music for piano and orchestra), and 32 piano sonatas (music for solo piano).

a) How many ways are there to play first a Beethoven symphony and then a Beethoven piano concerto?

b) The manager of a radio station decides that on each successive evening (7 days per week), a Beethoven symphony will be played followed by a Beethoven piano concerto followed by a Beethoven piano sonata. For how many years could this policy be continued before exactly the same program would have to be repeated?

Any ideas would be helpful to recognize the clues in the problems.

A game is played by rolling a six sided die which has four red faces and two blue faces. One turn consists of throwing the die repeatedly until a blue face is on top or the die has been thrown 4 times
Adnan and Beryl each have one turn. Find the probability that Adnan throws the die more turns than Beryl
I tried : Adnan throws two times and Beryl throws once = $\frac{2}{3}$ x $\frac{1}{3}$
Adnan throws three times and Beryl throws once = $\frac{4}{9}$ x $\frac{1}{2}$
Adnan throws three times and Beryl throws twice = $\frac{4}{9}$ x $\frac{2}{3}$
Adnan throws four times and Beryl throws once = $\frac{8}{27}$ x $\frac{1}{2}$
Adnan throws four times and Beryl throws twice = $\frac{8}{27}$ x $\frac{2}{3}$
Adnan throws four times and Beryl throws three times = $\frac{8}{27}$ x $\frac{4}{9}$
The answer says 0.365

If the probability that Joe will buy a pizza is 0.5 and the probability that Elaine will buy a pizza is 0.35, then what is the probability that at least one of the two will buy a pizza on their next visit to the pizza place?
I know that this is a very basic problem. However, I am a little confused about the "at least" statement in the question. I would really like to understand why the correct answer is correct, so please go in depth for how you got your answer. I am thinking that you might use the addition rule for probability, but again, the "at least" confuses me.

For the experiment of tossing a coin three times, how do you find the probability of getting at least two heads?

I was thinking today that if something with a probability of occurring of 1% happened 100 times, then the probability of that something happening is 100%, I believe that according to the addition rule for probabilities the probabilities for each event should be added up to get the total probability thus 1/100 + 1/100 + 1/100 ... up to 100 = 100/100 = 1 = 100%.

Now, there's still the possibility that the event didn't occur any one of those 100 times when it could have, because each time is independent. If such is the case, then obviously the probability is not 100%.

I believe I'm wrong and that I'm doing something wrong. So I would very much appreciate any guidance as to how to go about calculating the probability of something that happens 100 times that has a chance of occurring of 1% every time. For example, let's say there's a probability of 1% of dying from eating too much Cap'n Crunch, if I ate too much Cap'n Crunch 100 times, what is the probability that I will die?

How many natural odd numbers are between 100 and 999 that have all different digits?

There are two conditions in this question: (1)the number must be odd and (2)must have all different digits. I thought that condition 1 is more important. So, because tehere is five odd numbers, $\{1,3,5,7,9\}$, the units position has 5 choices. The tens position has 9 choices, since it must be different from the units. The hundreds position has 7 choices, because can't be 0 and must be different from the previous positions.
One gets $7\cdot <!--\; \cdot \; -->9\cdot <!--\; \cdot \; -->5=315$. But using Excel I found that there are 320 odd numbers with all different digits between 100 and 999. Where I'm wrong? Thanks

A group of pre-school children is drawing pictures ( one child is making one picture ) using 12-colours pencil set. Given that
(i) each pupil employed 5 or more different colours to make his drawing; (ii) there was no identical combination of colours in the different drawings; (iii) the same colour appeared in no more than 20 drawings,
find the maximum number of children who have taken part in this drawing activity.
( As each child can be identified with his/her unique combination of colours, the number of children can not exceed
C(5,12) + C(6,12) + C(7,12) +...+ C(12,12)
But how to NARROW it using the condition (iii) ? )

How do you find the permutation or combination of 7P4?

Suppose we perform 2 experiments and this first experiment can result in any one of $m$ possible outcomes. Suppose the first experiment results in outcome $i$. Then, the second experiment can result in any of $n_i$ possible outcomes $i=1,..,m$. $\mathbf{Q}\mathbf{s}\mathbf{:}$ what is the number of possible outcomes of the two experiments?

Attempt:
If we have $1,...,m$ possible outcomes and say this experiment results in one of them say $i$. and so second experiment can result in $n_i$ for each $i$ so by multiplication principle we have $n_1\cdot <!--\; \cdot \; -->n_2\cdot <!--\; \cdot \; -->..\cdot <!--\; \cdot \; -->n_m$ possible outcomes. Is this correct?

For the experiment of tossing a six sided die twice, what is the probability that the sum is less than 11?

At a party 66 handshakes occurred. Each person shook hands exactly once with each of the other people present. How many people were present?

Elijah forgot his pass code for a certain website. He knows it is made up of 5 digits and no digit can be repeated. If he randomly chooses 5 different digits, what is the probability that he chooses his pass code?

A coin is tossed three times: The probability of zero heads is 1/8 and the probability of zero tails is 1/8.

And my question is: What is the probability that all three tosses result in the same outcome?

So, if P(zero heads)= 1/8 , then that should be the same of p(all tails)?

If so, we could use the Addition Rule which is $P(A\cup <!--\; \cup \; -->B)=P(A)+P(B)$

where A and B are disjoint events, i.e. A$A\cap <!--\; \cap \; -->B=\mathrm{\varnothing <!--\; \varnothing \; -->}$, A is the event of tossing all heads and B is the event of tossing all tails.

I'm not sure how to continue after that... would the complement be used?

How many different permutations can you make with the letters in the word seventeen?

How do you determine the number of ways a computer can randomly generate an prime integer from 1 through 12?

How do you calculate permutations of a word?

How do you evaluate 20P2?