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

Prove or disprove via proof that events X and Y can be independent and mutually exclusive if both of their probabilities are greater than 0.

Chooste the best definition of tree diagram.
a) the ratio of the number of times an event occurs to the total number of trials or times the activiti is perfomed.
b) the set of all possible outcomes of that experiment.
c) a diagram that shows all the possible outcomes of an event.
d) probability based on reasoning written as a ratio of the number of favorable outcomes to the number of possible outcomes.

There are two fundamental principles of counting; Fundamental principle of addition and fundamental principle of multiplication.

I often got confused applying them. I know that if there are two jobs, say $m$ and $n$, such that they can be performed independently in $m$ and $n$ ways respectively, then either of the two jobs can be performed in $m+n$ ways and when two jobs are performed in succession, they can be performed in $m\times <!--\; \times \; -->n$ ways.

My question is how to identify whether jobs are independent or in succession?

Is there any simple way to identify this? Are there any keywords?

A four digit no. Is to be formed by numbers(0,2,4,6,8).Find number of such numbers which are divisible by 2 or 5 when repetition is allowed ?

Satish picks a card at random from an ordinary pack. If the card is ace, he stops; if not, he continues to pick cards at random, without replacement, until either an ace is picked, or four cards have been drawn. The random variable C is the total number of cards drawn. Construct a tree diagram to illustrate the possible outcomes of the experiment, and use it to calculate the probability distribution of C.

License plates are made using 3 letters followed by 2 digits. How many plates can be made if repetition of letters and digits is allowed?

What is the formula to find the number of permutations to 4 sets of 6 numbers?

Five people are going to be seated in a row of seven chairs. How many different ways can they be seated?

In order to conduct an experiment, five students are randomly selected from a class of 20. How many different groups of five students are possible?

How do you find the number of distinguishable permutations of the group of letters: M, I, S, S, I, S, S, I, P, P, I?

How many ways can you make a sandwich by choosing 4 out of 10 ingredients?

In how many ways can you arrange six out of eight books on a shelf?

How do you find the number of distinguishable permutations using the letters in FOOTBALL?

How do you evaluate 9P3?

What is the difference between events that are mutually exclusive and those that are not mutually exclusive?

How do you determine the sample space for the experiment where two county supervisors are selected from five supervisors, A, B, C, D, E to study a recycling plan?

There are 7 red and 5 yellow fish in an aquarium. Three fish are randomly caught in a net. Find the probability that the fish were:
a) All red
b) Not all of the same color.

I have solved this question using the method for dependent events as follows:
a) $\frac{7}{12}\cdot <!--\; \cdot \; -->\frac{6}{11}\cdot <!--\; \cdot \; -->\frac{5}{10}=\frac{210}{1320}$
b) $1-<!--\; -\; -->(\frac{7}{12}\cdot <!--\; \cdot \; -->\frac{6}{11}\cdot <!--\; \cdot \; -->\frac{5}{10}+\frac{5}{12}\cdot <!--\; \cdot \; -->\frac{4}{11}\cdot <!--\; \cdot \; -->\frac{3}{10})=\frac{1170}{1320}$
Is it possible to use combinations to solve this question? If yes, how?

The letters of the word CONSTANTINOPLE are written on 14 cards, one of each card. The cards are shuffled and then arranged in a straight line. How many arrangements are there where no two vowels are next to each other?

We throw a die a number of times. We let $Y_k$ be the number of different faces that came up after $k$ throws. Now we want $P[Y_k\le <!--\; \le \; -->5]$.

According to the answer sheet, we use the principle of inclusion-exclusion here to arrive at the following answer:
$P(Y_k\le <!--\; \le \; -->5)=(\genfrac{}{}{0ex}{}{6}{5})(\frac{5}{6}{)}^k-<!--\; -\; -->(\genfrac{}{}{0ex}{}{6}{4})(\frac{4}{6}{)}^k+(\genfrac{}{}{0ex}{}{6}{3})(\frac{3}{6}{)}^k-<!--\; -\; -->(\genfrac{}{}{0ex}{}{6}{2})(\frac{2}{6}{)}^k+(\genfrac{}{}{0ex}{}{6}{1})(\frac{1}{6}{)}^k-<!--\; -\; -->(\genfrac{}{}{0ex}{}{6}{0})(\frac{0}{6}{)}^k$ (which I believe to be equal to) $P(Y_k\le <!--\; \le \; -->5)=P(Y_k=5)-<!--\; -\; -->P(Y_k=4)+P(Y_k=3)-<!--\; -\; -->P(Y_k=2)+P(Y_k=1)-<!--\; -\; -->P(Y_k=0)$

But I just can't wrap my head around how the principle of inclusion-exclusion was exactly used here. My first thought was that it has something to do with the fact that if we see 5 faces, we've also seen at least 4 faces, etc. so that $[Y_k\le <!--\; \le \; -->4]$ is a subset of $[Y_k\le <!--\; \le \; -->5]$. I've tried to manually calculate (using the principle of inclusion-exclusion) $P(Y_k=1\cup <!--\; \cup \; -->Y_k=2\cup <!--\; \cup \; -->Y_k=3\cup <!--\; \cup \; -->Y_k=4\cup <!--\; \cup \; -->Y_k=5)$ but I don't arrive at the same answer. So far I haven't come to a good intuitive understanding yet, it would be great if someone could help me grasp this.