The probability that a card, randomly drawn from a standard deck of 52 cards without the joker, is an ace (or any other given rank) is ; if the joker is added, the probability decreases to . The decrease in probability is

The probability that a card, randomly drawn from a standard deck of 52 cards without the joker, is a diamond (or any other given suit) is ; if the joker is added, the probability decreases to . The decrease in probability is

.

The probability of rolling a 6 with a fair die is .

Since a roll of 6 is 5% more likely than any one of the other rolls, and the other rolls are equally probable, then, theoretically, there should be 105 rolls of 6 for every 100 rolls of 5, 100 rolls of 4, etc. - that is, 105 out of 605, rolls should theoretically be a 6. This makes the probability of rolling a 6 on the altered die

.

Therefore, Judy's alteration has increased the probability of rolling a 6 by

The probability of rolling a 5 with a fair die is .

Since a roll of 6 on the altered die is 4% more likely than any one of the other rolls, and the other rolls are equally probable, then, theoretically, there should be 104 rolls of 6 for every 100 rolls of 5, 100 rolls of 4, etc. Consequently, 100 out of 604 rolls should theoretically result in a 5. This makes the probability of rolling a 5 on the altered die

.

Therefore, Pilar's alteration has decreased the probability of rolling a 5 by

The probablilty of a fair coin coming up tails is . 

If the coin is altered as in the problem, then the odds of the coin coming up tails as opposed to heads are 100 to 104; restated, the probability of it coming up tails is 

Mick has decreased the probability of the coin coming up tails by 

The correct answer is , since there are 2 dice and we are looking for the probabilty than any pair shows up, the first die can then be any number, therefore we assign it a probability of one and the second die must be the same number as the first, and therefore has a probabilty of . Multiplying these two probabilities, we obtain  .

In this problem, we are asked for the probabilty that a specific pair shows up. The probability that the number one shows up on a single die is . Here, we want this event 'the 1 shows up' to occur twice. The result is given by calculating  which is .

Remember, the probability of two succesful events is calculated by multiplying their individual probabilities.

The best way to solve this problem is to count the possible outcomes for this event. Indeed, a result of five is given by different sums; 1+4, 2+3, 3+2, 1+4. The fact that we count twice each of these sum may seem counterintuitive but it stems from the fact that in probabilty we consider events to occur successively, whereas in reality the dice have been thrown at the same time. But if they were thrown one after the other, we would note that obtaining a 2 after 3 is different that obtaining a 3 after a 2.

We then have 4 possible outcomes for the success of the event 'the sum is 5'. The total number of possible outcomes is given by the multiplication of all the possible values on each die or , which is equal to 36.

The final answer is obtained by dividing the number of possible outcomes by the total number of outcomes, in other words by calculating .

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