There are 10 winning cards: the 4 twos, the 4 threes, and the 2 black fours. There are subsequently 42 losing cards, so the odds against a winning card are the number of losing cards, 42, to the number of winning cards, 10. This can be reduced to

or 21 to 5.

There are 4 winning cards: the 2 red fours and the 2 black nines. There are subsequently 48 losing cards, so the odds against a winning card are the number of losing cards, 48, to the number of winning cards, 4. This can be reduced to

or 12 to 1.

Since there is only one card out of 52 (ace of spades) that wins John the $25 prize, the probability of this happening is. The value of this outcome to John is $24 - the $25 prize minus the $1 he paid to play.

Since there are three cards out of 52 (ace of clubs, ace of diamonds, ace of hearts) that win John a $5 prize, the probability of this happening is. The value of this outcome to John is $4 - the $5 prize minus the $1 he paid to play.

Since there are 48 out of 52 cards that do not win John a prize, the probability of this happening is. The value of this outcome to John is, since he had to pay $1 to play.

To find an expected value of a game, multiply the probability of each outcome by its value, then add the products. The expected value of this game to John is therefore

dollars, or

The expected value of the game to Johnny rounds to.

Since there is only one card out of 53 (joker) that wins Sharon the $25 prize, the probability of this happening is. The value of this outcome to Sharon is $23 - the $25 prize minus the $2 she paid to play.

Since there is only one card out of 53 (ace of spades) that wins Sharon the $10 prize, the probability of this happening is. The value of this outcome to Sharon is $8 - the $10 prize minus the $2 she paid to play.

Since there are three cards out of 53 (ace of clubs, ace of diamonds, ace of hearts) that win Sharon a $7 prize, the probability of this happening is. The value of this outcome to Sharon is $5 - the $7 prize minus the $2 she paid to play.

Since there are 48 out of 53 cards that do not win Sharon a prize, the probability of this happening is. The value of this outcome to Sharon is, since he had to pay $2 to play.

To find an expected value of a game, multiply the probability of each outcome by its value, then add the products. The expected value of this game to Sharon is therefore

dollars, or

The expected value of the game to Sharon rounds to.

For this problem, we will multiply together the odds of each draw (assuming he draws one card at a time...the odds won't change if he draws three at once, but it's easier to visualize this way) resulting in a card that works for Jeff's goal of having one player from each team. The first draw cannot fail, as he needs one player from each team and the first card he draws must be from one of the teams. After this draw, he has 14 cards remaining, and 10 of these are players on the two teams that can still offer a player.

So the odds of a successful second draw are.

The last draw is the trickiest, as there would now be 13 cards remaining, with only 5 being players from the team that he still needs represented. When we multiply all of these odds together, we get

which is 27.5%.

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:

Set A:

Set B:

In Set A, there are five consonants out of a total of seven letters, so the probability of drawing a consonant from Set A is.

In Set B, there are three consonants out of a total of six letters, so the probability of drawing a consonant from Set B is.

The question asks for the probability of drawing two consonants, meaning the probability of drawing a constant from Set A and Set B, so probability of the intersection of the two events is the product of the two probabilities:

Set A:

Set B:

In Set A, there are five consonants out of a total of seven letters, so the probability of drawing a consonant from Set A is.

In Set B, there are three consonants out of a total of six letters, so the probability of drawing a consonant from Set B is.

The question asks for the probability of drawing at least one consonant, which can be interpreted as a union of events. To calculate the probability of a union, sum the probability of each event and subtract the intersection:

The interesection is:

So, we can find the probability of drawing at least one consonant:

Set A:

Set B:

Between Set A and Set B, there are two potential matching pairs of letters: AA and XX. The amount of possible combinations is the number of values in Set A, multiplied by the number of values in Set B,.

Therefore, the probability of drawing a matching set is:

First calculate the number of students:

The probability of drawing an honors student will then be the total number of honors students divided by the total number of students attending the school: