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:

Since this problem deals with a probability with two potential outcomes, it is a binomial distribution, and so the probability of an event is given as:

Where  is the number of events,  is the number of "successes" (in this case, a "heads" outcome), and  is the probability of success (in this case, fifty percent).

Per the question, we're looking for the probability of at least three heads; three head flips or four head flips would satisfy this:

Thus the probability of three or more flips is:

Since this problem deals with a probability with two potential outcomes, it is a binomial distribution, and so the probability of an event is given as:

Where  is the number of events,  is the number of "successes" (in this case rolling a four), and  is the probability of success (one in four).