Two independent events that each have a probability of 1/2 of occuring.

1/2 * 1/2 = 1/4

Probability = # of Tickets You Bought / # of Tickets Sold

Probability = 8/526 = 0.0152

Probability of each event = 1 side of die / # of sides = 1/6

Probability for multiple events = P1 * P2 * P3

1/6 x 1/6 x 1/6 = 1/216

The sample space for rolling two six-sided dice is 36.  We can get 7 six different ways:  1,6  2,5  3,4  4,3  5,2  6,1 so the probability of getting a 7 is \(\displaystyle \frac{6}{36}\) or \(\displaystyle \frac{1}{6}\).  The probability of NOT getting a 7 is \(\displaystyle 1 - \frac{1}{6} = \frac{5}{6}\).  We can add up all the things we want or we can subtract from 1 what we don't want. 

The probability of choosing 1 white marble is \(\displaystyle \frac{1}{2}\).

The probability of choosing a red marble is \(\displaystyle \frac{1}{8}\).

To find the chances of both events happneing you must multiply the probabilities together. 

To find the probability of a marble being chosen, we divide the number of the certain marble by the total number of marbles. In this case, we are finding the probability of NOT choosing a blue marble, or the probability of choosing a purple or red marble. The probability of choosing a purple marble is \(\displaystyle \frac{12}{25}\), and that for a red marble is \(\displaystyle \frac{3}{25}\). Together, this adds up to \(\displaystyle \frac{15}{25}\), which is reduced down to \(\displaystyle \frac{3}{5}\).

If 30% of owners have both a cat and a dog and 10% have neither, that leaves 60% of owners who either have only a cat or only a dog. \(\displaystyle 1 - 0.3 - 0.1 = 0.6\)

There are twice as many owners that only own a dog as owners that only own a cat. Thus, \(\displaystyle 2c = d\).

The total number of cat-only and dog-only owners make up 60% of those surveyed. Thus, \(\displaystyle c + d = 0.6\)

Substitute \(\displaystyle 2c\) for \(\displaystyle d\) in the second equation. \(\displaystyle c + 2c = 0.6\)

Solve for \(\displaystyle c\), \(\displaystyle c= 0.2\)

Plug answer into first equation to get value of \(\displaystyle d\), the probability of selecting an owner that only has a dog.

The sample space for rolling two six-sided dice is \(\displaystyle 36\).

Even numbers less than 6 are 4 and 2:

\(\displaystyle 2: 1,1\)

\(\displaystyle 4: 1,3 \; \; \; 2,2 \; \; \; 3,1\)

So the probability of getting an even number less than six is \(\displaystyle \frac{4}{36} = \frac{1}{9}\)

So the probabilty of not getting an even number less than six is

 \(\displaystyle 1 - \frac{1}{9}\) or \(\displaystyle \frac{8}{9}\)

This is a probability question. All you have to account for is that one of the twenty people at this party brought a cake last time so they will not need to bring it this time. Therefore we take one person away from the twenty leaving nineteen people. Leslie has a \(\displaystyle \frac{1}{19}\) probabiity of being chosen.

\(\displaystyle 26\; black \; cards + 26 \; red \; cards = 52 \; total \; cards\)

Probability that first card is black:  \(\displaystyle \frac{26}{52} = \frac{1}{2}\)

Now there are 25 black cards remaining in the deck.

Probability that second card is black: \(\displaystyle \frac{25}{51}\)

Now there are 24 black cards remaining in the deck.

Probability that third card is black:  \(\displaystyle \frac{24}{50} = \frac{12}{25}\)

The probability of getting three black cards in a row is

\(\displaystyle \frac{1}{2} * \frac{25}{51} * \frac{12}{25} = \frac{300}{2550}=\frac{2}{17}\)