The answer is 0.65 because Pr(~Rain) is the complement of Pr(Rain) and both events are mutually exclusive.

When two events are mutually exclusive, $\displaystyle Pr(A\ and\ B)=Pr(A)+Pr(B)$. Since probabilities must sum up to 1, this implies that $\displaystyle Pr(B)=1-Pr(A)$. 

Since all of the events are mutually exclusive (one of the parties must win), you can get the probability of either D or R winning by adding their probabilities. 

Since the probability of D winning is .11 and R winning is .78, the probability of D or R winning is .89. 

In a deck of cards, there are 52 total cards, 13 hearts, 4 kings, and 1 king that is a heart.

So, $\displaystyle 13/52 + 4/52 - 1/52 = 16/52 = 4/13.$

There are $\displaystyle 52$ cards in a standard deck:  $\displaystyle 13$ cards in $\displaystyle 4$ suits.  There are $\displaystyle 3$ face cards (King, Queen, Jack) in each suit, so there are $\displaystyle 12$ total face cards.

Thus the probability of choosing a single face card is $\displaystyle \frac{12}{52}$ or $\displaystyle \frac{3}{13}$.

In this case, we want to know the probability of multiple, mutually exclusive possible outcomes. To determine the probability of the two possible outcomes, simply add them together.  This is called the addition rule.

$\displaystyle 1/5+1/5=2/5$

First find the number of ways to get 0, 1, or 2 heads.

Remember, $\displaystyle \binom{n}{k}=\frac{n!}{k!(n-k)!}$.Also, $\displaystyle \binom{n}{0}=1$ and $\displaystyle \binom{n}{1}=n$.

$\displaystyle 0$ heads: $\displaystyle \binom{10}{0} = 1$

$\displaystyle 1$ head: $\displaystyle \binom{10}{1} = 10$

$\displaystyle 2$ heads: $\displaystyle \binom{10}{2} = 45$

$\displaystyle \textup{number of possible outcomes of 10 tosses}= 2^{10} = 1024$

$\displaystyle p\left ( x\right )=\frac{\textup{number\ of\ desired\ outcomes}}{\textup{number\ of\ all\ possible\ outcomes}}$

$\displaystyle p(\textup{0 heads in 10 coin tosses}) = \frac{1}{1024}$

$\displaystyle p(\textup{1 head in 10 coin tosses}) = \frac{10}{1024}$

$\displaystyle p(\textup{2 heads in 10 coin tosses}) = \frac{45}{1024}$

Now combine these to find the probability of seeing at most 2 heads in 10 coin tosses:

$\displaystyle p(\textup{at most 2 heads in 10 coin tosses})$

$\displaystyle =p(\textup{0 heads})+p(\textup{1 head})+p(\textup{2 heads})$

$\displaystyle =\frac{1}{1024} +\frac{10}{1024} +\frac{45}{1024}$

$\displaystyle =\frac{56}{1024}$  

We want to know the probability of multiple possible outcomes that are not mutually exclusive. To do this, we use the addition rule with one step that we would not use if the possible outcomes were mutually exclusive.  Add the probabilities of each possible outcome, subtract from that sum the number counted twice, then reduce the answer to the least common denominator.

$\displaystyle 65/150 + 15/150 - 10/150= 70/150 = 7/15$

We want to know the probability of multiple possible outcomes that are not mutually exclusive. To do this, we use the addition rule with one step that we would not use if the possible outcomes were mutually exclusive. 

Add the probabilities of each possible outcome, subtract from that sum the number counted twice, then reduce the answer to the least common denominator. In our case, we are told that 2 of those who have a business degree are also entrepreneurs therefore we need to subtract to from the total to get the correct probability.

$\displaystyle 7/15+5/15=12/15$

$\displaystyle 12/15-2/15=10/15$

$\displaystyle 10/15=2/3$

Bear in mind how the addition rule applies here. This will be a lengthy application, but applied properly, it will see you through this problem. It helps to think of this as four events that are not mutually exclusive - drawing a queen, drawing a jack, drawing a king or drawing a diamond. We then realize that we have to avoid counting the king, queen and jack of diamonds twice.

$\displaystyle P(face \bigcup diamond)$

$\displaystyle P(king \bigcup queen \bigcup jack \bigcup diamond)$

$\displaystyle P(king) + P(queen) + P(jack) + P(diamond) - P(diamond \bigcap face)$

$\displaystyle \frac{1}{13} + \frac{1}{13} + \frac{1}{13} + \frac{1}{4} - \frac{3}{52}$

$\displaystyle \frac{4}{52} + \frac{4}{52} + \frac{4}{52} + \frac{13}{52} - \frac{3}{52} = \frac{22}{52} = \frac{11}{26}$

In this case, we want to know the probability of multiple, mutually exclusive possible outcomes. The possible outcomes are mutually exclusive because one can of food could not be both beans and tomato sauce. To determine the probability of the two possible outcomes, add them together and then find the least common denominator.

$\displaystyle 12/150 + 48/150 = 60/150 = 2/5$