First calculate the number of students:

 \(\displaystyle 200+150+250+100=700\)

The probability of drawing an honors student will then be the total number of honors students divided by the total number of students attending the school:

\(\displaystyle \frac{200(.20)+(150)(.10)+(250)(.12)+(100)(0.30)}{700}\)

\(\displaystyle \frac{115}{700}=\frac{23}{140}\)

First calculate the number of students:

 \(\displaystyle 200+150+250+100=700\)

The percentage of seniors that do not attend honors classes is:

\(\displaystyle (1-.30)=.70\)

Therefore, the probability of selecting a student who is a senior and one who does not attend honors classes is:

\(\displaystyle \frac{100}{700}(.70)=\frac{1}{10}\)

First, consider the probability of each individual event: drawing a diamond, an even number, or a number divisible by three.

Drawing a diamond, there are four suits, so:

\(\displaystyle P(D)=\frac{1}{4}\)

Drawing an even number, there are five possible values \(\displaystyle [2,4,6,8,10]\) present in each thirteen-card suit, so:

\(\displaystyle P(E)=\frac{5}{13}\)

Drawing a number divisible by three, there are three possible values \(\displaystyle [3,6,9]\) present in each thirteen-card suit, so

\(\displaystyle P(3)=\frac{3}{13}\)

This problem deals with a union of probabilities, essentially a "this or that" option; however, since there are three events considered, the formula for the union follows the form:

\(\displaystyle P(A\bigcup B\bigcup C)=P(A)+P(B)+P(C)-P(A\bigcap B)-P(A\bigcap C)-P(B\bigcap C)+P(A\bigcap B\bigcap C)\)

The probability of drawing an even number that is also a diamond is:

\(\displaystyle \frac{5}{52}\)

The probability of drawing a diamond that is divisible by three is:

\(\displaystyle \frac{3}{52}\)

The probability of drawing an even number that is divisible by three is:

\(\displaystyle \frac{1}{13}\)

And the probability of drawing a number that is a diamond, even, and divisble by three is:

\(\displaystyle \frac{1}{52}\)

Therefore, the probability of this union is:

\(\displaystyle \frac{13+20+12-5-3-4+1}{52}=\frac{34}{52}=\frac{17}{26}\)

To solve this problem, we must first understand a basic concept of probability. In order to find the probability of two independent events happening, we multiply the probability of each of those two independent events happening together in order to find the probability of both of them occurring.

In this case, we know that Alex has a \(\displaystyle 75\%\) chance of being accepted and \(\displaystyle 25\%\) chance of being rejected.

We also know that Jeffery has a \(\displaystyle 66\%\) chance of being accepted and a \(\displaystyle 34\%\) chance of being rejected.

The question asks us the probability of the college accepting Jeffery, but not Alex, therefore we multiply the probability of the college accepting Jeffery, \(\displaystyle 66\%\), by the probability of the college rejecting Alex, \(\displaystyle 25\%\), together in order to answer this question.

\(\displaystyle 0.66*0.25=0.165\) or a \(\displaystyle 16.5\%\) chance of Alex being rejected by Jeffery being accepted.

To solve this problem, we must first understand some of the basic concepts of probability. If the probability of an outcome is \(\displaystyle 0.5\), that means that there is a \(\displaystyle 50\%\) chance of that outcome happening. Thus the probability of an outcome occurring can never exceed \(\displaystyle \textup{1, or 100\%}\) 

This question asks the probability of Martin NOT losing his car keys. It tells us that the probability of him losing his car keys is \(\displaystyle 0.0975\), or \(\displaystyle 9.75\%\). Because we know there is a \(\displaystyle 100\%\) chance he will either lose or keep his car keys and we know that probability of him losing his car keys, we simply subtract the probability of him losing his car keys from the total chance, \(\displaystyle 1-0.0975=0.9025\) or 90.25% chance that Martin will not lose his car keys next month.

To solve this problem, we must understand some basic concepts.  In the jar, there are currently \(\displaystyle 900\) candies that can be chosen. Of these \(\displaystyle \textup{900, 300}\) are skittles and \(\displaystyle 600\) are M&M's. This means that there is a \(\displaystyle \frac{300}{900}\) or \(\displaystyle \frac{1}{3}\) chance when randomly selecting a candy for the first time that it will be a skittle.

The question asks us the probability of selecting two skittles in a row, therefore we must once again calculate the probability of selecting a skittle a second time. After picking a skittle the first time, there are now \(\displaystyle 899\) candies in the jar and only \(\displaystyle 299\) of them are skittles now. This means that there is a \(\displaystyle \frac{299}{899}\) chance of randomly selecting a skittle the second time.

Now that we have our two probabilities, to find the probability of both happening, we simply multiply them together, meaning there is \(\displaystyle \frac{1}{3}*\frac{299}{899}=\frac{299}{2697}\) chance of approximately \(\displaystyle 11\%\) chance of grabbing two skittles in a row from the jar.

There are twelve face cards (king, queen, and jack for each of the four suits) in a deck, so the probability of drawing a face card on the first draw is

\(\displaystyle P(\textup{Face})=\frac{12}{52}\)

Now, there are two potential situations for the second draw given he draws a face the first time; either he drew a face from the clubs or he didn't.

The probability of a face also being a club is

\(\displaystyle P(\textup{Club}|\textup{Face})=\frac{3}{12}\)

And the probability of the complement is then:

\(\displaystyle P(\textup{NotClub}|\textup{Face})=\frac{9}{12}\)

Were a face club drawn, the probability of drawing a club next would be

\(\displaystyle P(\textup{Club}|\textup{FaceClub})=\frac{12}{51}\)

Since a club was removed from the deck.

Were a face club not drawn, the probability of drawing a club next would be

\(\displaystyle P(\textup{Club}|\textup{NotClubFace})=\frac{13}{51}\)

Therefore, the unconditional probability of drawing a club accounts for the probability of either of these scenaorios:

\(\displaystyle P(Club)=\frac{3}{12}\frac{12}{51}+\frac{9}{12}\frac{13}{51}\)

\(\displaystyle P(Club)=\frac{36+117}{612}\)

\(\displaystyle P(Club)=\frac{153}{612}\)

\(\displaystyle P(Club)=\frac{1}{4}\)

Now, for the probability of drawing the face first and the club next, it is the intersection of events:

\(\displaystyle P(Face\bigcap Club)=\frac{12}{52}\frac{1}{4}=\frac{3}{52}\)

In this problem, the order of selection for ice cream and toppings does not matter, so it's dealing with combinations.

The number of possible combinations for \(\displaystyle k\) selections out of \(\displaystyle n\) items is:

\(\displaystyle C(n,k)=\frac{n!}{(n-k)!k!}\)

So the number of possible ice cream choices in this problem is:

\(\displaystyle \frac{10!}{(10-2)!2!}\)

\(\displaystyle 45\)

And the chance of choosing a chocolate+vanilla selection is \(\displaystyle \frac{1}{45}\).

For the toppings, the number of combinations is:

\(\displaystyle \frac{4!}{(4-2)!2!}\)

\(\displaystyle 6\)

With the almond sprinkle combination having a \(\displaystyle \frac{1}{6}\) chance of being chosen.

The chance of BOTH of these selections being made is then the product of the two probalities:

\(\displaystyle \frac{1}{6*45}\)

\(\displaystyle \frac{1}{270}\)

For this problem, when selecting food items, the order of selection does not matter, so we're dealing with combinations.

For \(\displaystyle k\) choices made from \(\displaystyle n\) possible options, the number of possible combinations is

\(\displaystyle C(n,k)=\frac{n!}{(n-k)!k!}\)

So for the curry options, where two choices are made from five options

\(\displaystyle C(5,2)=\frac{5!}{(5-2)!2!}=10\)

For the rice options, with one choice made from three:

\(\displaystyle C(3,1)=\frac{3!}{(3-1)!1!}=3\)

For the lunch special, the total amount of combinations is the product of these two:

\(\displaystyle 30\)

For a twenty-sided die, there are twenty possible rolls:

\(\displaystyle [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20]\)

Of these, sixteen satisfy the problem condition:

\(\displaystyle [5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20]\) 

So the probability of rolling a five or greater is

\(\displaystyle P(>5)=\frac{16}{20}=\frac{4}{5}\)

The probability of rolling three of these rolls in a row is a type of intersection, meaning multiplication is involved:

\(\displaystyle (\frac{4}{5})^3\)

\(\displaystyle \frac{64}{125}\)