The chance of choosing a white marble is \(\displaystyle 4\) out of \(\displaystyle 16\), or \(\displaystyle \dpi{100} \frac{1}{4}\)

 \(\displaystyle \dpi{100} \frac{1}{4}\) expressed as a percentage is \(\displaystyle 25\%\)

To find the probability, we use a fraction:

\(\displaystyle \frac{part}{total}\)

There are 12 total shapes, and 2 black circles:

\(\displaystyle \frac{2}{12}\)

Reduce:

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

Thus, the probability is 1 out of 6.

Probability involves part over whole. First, add up how many items are in the bag. There are 10 items. Then, count how many pens there are (4). Put 4 over 10 to get \(\displaystyle \frac{4}{10}\). Then, you must simplify it to find the right answer. 2 goes into both the numerator and denominator, so it can be reduced to \(\displaystyle \frac{2}{}5\).

Given that there are 5 black marbles, 3 blue marbles, and 2 green marbles, there are a total of 10 marbles. 

\(\displaystyle 5+3+2=10\)

If Ana picks a marble, there are 5 marbles that are black. Thus, there would a 5 out of 10 chance that she would get a black marble.

\(\displaystyle \frac{5\ \text{black}}{10\ \text{total}}=\frac{5}{10}\)

We can convert to a percent by multiplying to get 100 in the denominator.

\(\displaystyle \frac{5}{10}=\frac{5\times10}{10\times10}=\frac{50}{100}\)

Fifty over one hundred is equal to \(\displaystyle 50\%\). There is a \(\displaystyle 50\%\) chance that she will get a black marble because half of the marbles in the bag are black.

Initially the discs in the box were: \(\displaystyle 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11.\)

6 of these numbers are odd and 5 are even.

When Beth picked disc number 9, there remained 5 odd discs and 5 even discs. 

Odd discs remaining: \(\displaystyle 1,3,5,7,11\)

There are ten total discs remaining. To find the probability of an odd disc:

\(\displaystyle \frac{\text{odd}}{\text{total}}=\frac{5}{10}\)

This fraction is equal to one half, or \(\displaystyle 50\%\).

When flipping a coin, the possibility that it will be heads or tails will always be 50/50. Even though Andrew flips the coin twice before the third flip, the prior flips do not affect the outcome of the third flip. 

The Molly baked a pie with 6 slices. 3 slices have a sugar topping, 1 slice has a fruit topping, and 2 slices have no topping. Therefore, the proportion of sugar slices to the rest of the pie is:

\(\displaystyle \frac{\text{sugar}}{\text{total}}=\frac{3}{6}=\frac{1}{2}\)

Given that 50% is the percent equivalent of \(\displaystyle \frac{1}{2}\), that is the correct answer.

If there are 5 turtles racing, there is a 2 out of 5 chance that a certain turtle will be first or second place.

\(\displaystyle \frac{2\ \text{options}}{5\ \text{total}}\)

Convert this value to a decimal.

\(\displaystyle \frac{2}{5}=\frac{4}{10}=0.40\)

Convert the decimal to a percentage.

\(\displaystyle 0.40=40\%\)

If there are 5 answer choices, and Gregory eliminates 2 of the answer choices, then there are 3 answer options left.

\(\displaystyle 5-2=3\)

Of the 3 total options left, 1 of them is correct.

\(\displaystyle \frac{1\ \text{correct}}{3\ \text{total}}=\frac{1}{3}\)

If he guesses, there is a \(\displaystyle \frac{1}{3}\) chance that he will answer the question correctly because he is guessing among 3 options. Therefore, the correct answer is \(\displaystyle \frac{1}{3}\).

To find the chance, we need to set up a fraction that looks like this:

\(\displaystyle \frac{\text{puppies with spots}}{\text{total puppies}}\)

Using the information from the question, we know that this fraction is equal to:

\(\displaystyle \frac{\text{puppies with spots}}{\text{total puppies}}=\frac{3}{4}\)

There is a \(\displaystyle \frac{3}{4}\) chance that the puppy picked will have spots.