Practice Problems
Problems With Solutions
1. Aaron selects a card from a standard deck. What is the probability that it is...
- a face card?
- less than a 4 or a heart?
- an ace and a face card?
There are 12 cards which have a face on them (4 Jacks, 4 Queens, 4 Kings); the probability of getting a face card is P(face) = 12 / 52.
The events {card less than 4} and {card is a heart} are not mutually exclusive. Thus, P(less than 4 or heart) = P(less than 4) + P(heart) - P(hearts less than 4) =
8 / 52 + 13 / 52 - 2 / 52 = 19 / 52
There are no Aces that have a face card on them! (see below)
Thus P(ace and face card) = 12 / 52 = 0
2. Julia picks 3 cards from a deck of 52.
- What is the probability that all 3 cards are diamonds?
- What is the proability that all 3 cards are 7's?
There are C(13, 3) = 286 ways to choose three cards that are all diamonds out of C(52, 3) = 22,100 ways to select three cards, so the probability of selecting 3 diamonds is 286 / 22,100 = .013.
There are C(4, 3) = 4 ways to choose 3 7's out of C(52, 3) = 22,100 ways to select three cards. Thus P(3 7's) = 4 / 22,100 = .00018.
3. You are dealt 4 clubs from an ordinary deck of 52. What is the probability that your next card is a club? If you are dealt two cards instead of one, what are your chances that exactly one of them is a club?
There are 13 clubs in a deck of cards. If you are dealt 4 of them, there are only 9 clubs left in the deck. In all, there are 52 - 4 = 48 cards left in the deck. The probability of the next card being a club is 9 / 48.
Two cards out of 48 left in the deck may be dealt C(48,2) = 1128 ways. One club may be dealt in 9 * 39 = ways (there are 9 ways to choose a club, and 39 ways to choose the last card so that it is NOT a club). Therefore, the probability of receiving exactly one club from the next two cards is 351 / 1128 = .311
4. A gambler is playing a game of poker has the following hand:
The gambler is permitted to discard two cards, and be dealt two new cards from the dealer. Should the gambler
- Discard the 4 and the Queen in hopes for a full house? (a full house requires that the gambler is dealt either a 3 and a Queen or two Queens)
- Or should she discard a 3 and a Queen, in hopes for a straight? (a straight requires that she be dealt a 2 and a 6 or a 5 and a 6)
If the gambler will win more money from a full house than a straight, would you reccomend that she attempt to get a full house?
Case 1: Attempting a Full House
To find the probability of obtaining a straight, we must find (1) the number of ways to get two Queens OR a Queen and a three, and (2) the number of ways possible to get two more cards.
There are 3 * 2 = 6 ways to be given two Queens, and 3 * 2 = 6 ways to be given a Queen and a 3 (there are three Queens left that may be picked; to each one, there are two 3's that may be picked). Thus there are 6 + 6 = 12 card combinations that will give a full house.
47 cards remain in the deck; there are C(47,2) = 1081 ways to select two cards from those remaining. Thus the probability of selecting the cards that will give a full house is 12 / 1081 = .011
Case 2: Attempting a Straight
To find the probability of obtaining a straight, we must find (1) the number of ways to get a five and a six OR a two and a six, and (2) the number of ways possible to get two more cards.
There are 3 * 4 = 12 ways to get a five and a six, and 4 * 4 = 16 ways to get a two and a six. It follows that there are 12 + 16 = 28 card combinations that will give a straight. The find the probability that those combinations are drawn is 28 / 1081 = .026
The probability of getting a straight (.026) is larger than that of getting a full house (.011) by a factor of 2 and a third. The chances are in favor of getting a straight. However, since the probabilites are relatively small, the added incentive of receiving more money from winning on a full house may influence the gambler to go against the odds!
5. The following cards are laid out on a table:
If you randomly select 2 cards, what is the probability that you add the values of the cards selected and get an even number? that the cards you select are both prime?
There are C(6, 2) = 15 ways to select 2 cards from a total of six.
To get an even number, one must have either (a) two even cards or (b) two odd cards. There are C(3, 2) = 3 ways to select 2 even cards, and C(3, 2) = 3 ways to select 2 odd cards. It follows that there are 6 ways to select two cards whose digits add to an even number.
The probability of getting two cards whose values add to an even number is 6 / 15.
A prime number has the property that the only two numbers that evenly divide it are 1 and itself. Thus 5, 7, and 2 are the prime numbers above. They may be selected in C(3, 2) = 3 ways out of a total of 15 possible ways to select two cards; thus the probability of obtaining it is 3 / 15.
6. A standard deck of cards is divided into two piles: the red cards (hearts and diamonds) and the black cards (clubs and spades). A cube numbered 1-6 is rolled; if an even number appears, a card is selected from the red pile. If an odd number is rolled, a card is drawn from the black pile.
- what is the probability that a red card is selected?
- if a two is rolled, what is the probability that of selecting a diamond?
- given that a 5 of spades is selected, what is the probability that a 4 was rolled?
- what is the probabilty that a face card is drawn given that a 3 was rolled?
The probability of a red card being selected depends on the number rolled on the cube. Since 3 numbers are even on the cube, out of a total of 6, the probability of selecting a red card is 3 / 6 = 1 / 2.
If a two was rolled, then a card is drawn from the red pile. The probability of drawing a diamond from the red pile is 1 / 2.
If one selected a Five of Spades, then an odd number must have been rolled. If an odd number was rolled, then 4 could not have been rolled. Thus P( 4 | 5 of spades) = 0.
If a three is rolled, one must select from the black pile. There are 6 face cards in that pile, so P(face) = 6 / 26.
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