In this problem, the best way to calculate the answer is to count the possible outcomes. We can see that the sum can be obtained with the following sums: 2+6,3+5, 4+4, 5+3, 6+2 .

We have a total of 5 outcomes. As previously discussed, some sums are counted twice because in probabilty we study events as if they were successive events even when they are not. For this same reason, the 4+4 is counted only once, because a dice yielding a 4 followed by a dice yielding a 4  is the same as a dice yielding a 4 followed by a dice yielding a 4. 

There are a total of  possible outcomes and the final answer is .

Here we are asked for the probability of drawing any pair. Therefore, the first card can be any card and we can assign this event a probability of one.

Furthermore, the second card drawn must be the same card to form a pair. Therefore, its probability is .

We use 51 because we already have drawn a card. You will notice that to form a pair there are 3 other cards in the deck that can be drawn, for example if an ace is drawn first, there must be three other aces in the deck.

 is the final answer

In order to draw a pair of ace, the first card drawn must be an ace, which has a probability of being drawn of .

Since there are 4 aces in the deck. The second ace has a probability of being drawn of , since there are only three aces left in the deck. 

Therefore to obtain the probability of these two successive events, we have to multiply each probability.

The final answer is given by , or 

First of all, the first card can be any card, so the probabilty of drawing any card is 1, since it is necessarily a success. The second card must belong to the same suit (don't be confused by the fact that the cards can be red or black, whenever color is mentioned it is said to reference the suits), in total there are 13 cards by colors. Since we have already drawn one card, there are only 12 left out of a total of 51 cards, after we have drawn two card of the same suit, there will be only 11 cards of this suit left out of 50.

The probability is given by the following calculation :

The answer can be obtained by firstly calculating the probabilty of her missing twice and hitting the target once, in any order.

This probability is obtained by calculating  .

 is the  probability of her failing, which is calculated by substracting  .

The answer can be obtained by finding the opposite of the event 'she hits at least one target'. This event is 'the woman misses all her 4 shots'.

This probabilty is calculated by raising  to the power of 4, since she misses all her 4 shots.

The result is then substracted from 1, 1 being the total of all the possible probabilities for all events in a given problem.

, which is the final answer.

This simple problem asks us to find the probability of an event occuring 3 times.

A head shows up with a probability of , since we throw the coin thrice, the final answer is given by  or   .

This trick question is in fact very simple. Since the throws are independent (in other words, what happened before an event has no influence on the future outcomes), then the probabilty is , indeed there are no reasons for any of the head or tail to have more chance of showing up at any point.

Again since we are asked to look for the probability of an event occuring at least once, we can take a shortcut and calculate the probability of the opposite event; 'no tail shows up'. The probability of this event is given by  or .

We substract this  to 1, the sum of all the probabilities, to obtain . 

Logically, we can see that the probability of having a tail show up at least once in 5 flips should be pretty high since its probability is , therefore common sense supports our choice.

A face card constitutes a Jack, Queen, or King, and there are twelve in a deck, so the probability of drawing a face card is .

There are thirteen spades in the deck, so the probability of drawing a spade is .

Keep in mind that there are also three cards that fit into both categories: the Jack, Queen, and King of Spades; the probability of drawing one is 

Thus the probability of drawing a face card or a spade is: