With this probability problem, it is important to note that there are two paths to the desired outcome: two red marbles, and two blue marbles. So you will want to calculate the probability of each path and then add them together.

To draw two red marbles, Evelyn would need to draw one of 6 red marbles from the jar of 15, and then draw one of the remaining 5 red marbles from the remaining 14. So that probability looks like:

This fraction reduces to .

To draw two blue marbles, Evelyn would need to draw one of the 9 blues from the jar of 15, and then one of the remaining 8 blues from the remaining 14. That probability looks like:

This fraction reduces to .

When you add the two probabilities together (adding because red, red and blue, blue are two distinct probabilities that each satisfy the question), you'll get .

This probability problem begins with an incomplete table, but provides you with the information necessary to fill it in. You know that there are 60 total girls, and that twice as many girls are in the band as the choir.  So you can set up a system of equations for the girls:

B + C = 60

B = 2C

So 2C + C = 60, meaning that, among the girls column, C = 20 and B = 40

You then know that in the choir, there are twice as many girls as boys. Since there are 20 girls in the choir, you know that there are 10 boys in the choir. When you fill that into the table, you can then see that there must be 30 total choir members:

And you can then fill in the rest of the table. With 40 total boys and 10 in the choir, there must be 30 boys in the band, and that then means that, in the band, 30 boys + 40 girls equals 70 total members.

The question asks for the probability that any one student at random is a boy in the band. There are 30 such boy band members out of 100 total students, so the answer is 30%.

Remember that the probability of an outcome occurring is calculated as the number of favorable outcomes (in this case, bullseye targets) divided by the number of total outcomes. And here we know that the number of total outcomes in the provided situation has "changed" from the provided probabilities. If Emma were to throw 100 darts, 10 would hit the bullseye, 30 would land on an odd number, 30 would land on an even number, 10 would hit the unscored outer edge of the board, and 20 would miss completely. But since we're told that she did not miss the board completely, our calculation is that the number of favorable outcomes is still 10, but the number of total outcomes omits those 20 total misses. So the total number of outcomes is 10 bullseyes + 30 odds + 30 evens + 10 outer edges = 80. 

Therefore the probability is 10 favorable outcomes divided by 80 total outcomes, for a probability of 12.5%.

In approaching this conditional probability problem it is important to recognize that you do not need it to snow BOTH days in order for the event to run: as long as it snows on one or the other the event will be held. So this problem is asking for the probability of snow on either Saturday or Sunday. There are a few ways to determine this probability.

One is to note that if it snows Saturday (70%), then the event will run, but that if it does NOT snow Saturday (30%) then there's a 70% probability that it will still snow Sunday. So the two sequences of "Snows Saturday" or "Doesn't Snow Saturday, but then Snows Sunday" will combine for the total probability of the event running:

Snows Saturday = 70%
Does Not Snow Saturday, then Snows Sunday = (30%)(70%) = 21%

Snows either Saturday or Sunday = 70% + 21% = 91%. 

You could also note that the only outcome that does NOT work for you is if it fails to snow both days. Everything else does work, so you could subtract the probability of "No Snow, then No Snow" from 100% to get to the total:

100% - Neither
100% - (30%)(30%) = 100% - 9% = 91%.

With dependent probability - probability for which the outcome of one event impacts the probability of the next event - it is important to consider the sequence of events. Here, Jenny needs a particular sequence: Popcorn, then another Popcorn.

For her first selection, there are 5 bags of popcorn out of 13 total bags, so the probability is .

For her second draw, there will be 4 bags of popcorn left out of a total of the 12 remaining bags, for a probability of , which reduces nicely to .

Because she needs a bag of popcorn AND another bag of popcorn, you'll multiply the two individual probabilities. When you multiply the two fractions, you have .

When you are asked to find the median, your first step should always to be to check to see if the numbers are in ascending order. If they aren’t, your next step should be to put them in order. If you have variables, remember that without any additional information about the set, you don’t know where in the set the variable should be placed, so just put it to one side or the other until you integrate additional information. For this set, you should rearrange them to get:

{5, 9, 10, x}

You are also told that the median to this set is 7. Since the set has an even number of terms, that means that 7 must be the average of the two middle numbers. And since 7 is halfway between 5 and 9, those two numbers must be the two middle numbers. You can therefore rewrite the set in ascending order as

{x, 5, 9, 10}

It should be obvious that x could be 1, 3, or 4. It could also be 5, since that would not change the median at all since the two middle numbers would still be 5 and 9. However, it could not be 7, since the average of 7 and 9 is 8, not 7.

Since the five values are given in ascending order, you know that the median - which is defined as the middle value for an odd-numbered set - is the third value, xx. For the median to then equal the mean, you'll calculate the mean. The mean is the sum of the values divided by the number of values, or here:  . So you know that: 

Median = Mean

Now you can solve for xx. To eliminate the denominator, multiply both sides by 5 to get: 

Then combine like terms:

Then subtract  from both sides: 

Then you can solve for .

Remember that Average = Sum of Terms divided by Number of Terms. Using that equation, you can determine:

How many points Martin scored total in the first  games:  games times 32 points per game = 

How many points Martin scored total in the whole season: That  total from the first  games plus the 14 points he scored in the last game: .

Now using that, and knowing that there were  games in the whole season, you can use his known average of 30 points per game and set that equal to the total number of points  divided by the total number of games .

So your calculation is:

Multiply both sides by the denominator of  to get:

Distribute the multiplication to get:

And then subtract 30x30x and 1414 from each side to combine like terms and you'll have: 

So 

But remember that the total number of games is , so the answer is .

Conceptually, you could also realize that the difference between his average before the last game (32) and his points in the last game (14) is 18 points. For that 18 points to result in a decrease of 2 in the average (from 32 to 30), it must have been distributed over 9 games (18 divided by 9 = 2).

This problem tests the definition of “median.” Remember: the median is the middle value in a set with an odd number of elements (like List B). However, with sets containing an even number of elements (like List A), there can be no single “middle” element. Therefore, the median for sets containing an even number of elements is the average between the two numbers equally “in the middle.”

One of the first things we should notice with this problem is that the only difference between List A and List B is the addition of element y. The sets are otherwise identical. The problem tells us that the medians of the two sets are the same. Since List B contains an unknown element, we cannot initially calculate its median. However, the median of List A is fairly easy to calculate: it is the average between the two middle values (in this case, the average of 20 and 32 = 26. Since the medians of the two sets are the same, y must be equal to this new median.

This problem can also be solved by avoiding any math. Conceptually speaking, we know that the median for List A is going to be somewhere between 20 and 32. Since all of the elements of List A are contained in List B and because the problem states that the two lists have the same median, the extra element in List B, y, must represent the median of both sets. Looking down at the answer choices, there is only one answer between 20 and 32.

Since the average equals the sum of the terms divided by the number of terms, you can set this calculation up as:

Where you have 6 terms (the five knowns, plus xx) and you can quickly sum the given terms by pairing them off to find zeros. 50 + 40 = 90. 47 + 43 = 90. And then there's 55. So 90 + 90 + 55 = 180 + 55 = 235.

Now that you have your equation, multiply both sides by 6: 

Then subtract  from both sides: 

And you can solve for .