Probability - Finite Mathematics

Card 0 of 68

Question

Jack and Jill agree to a game. A card is drawn at random from a standard deck of 52. If the card is a spade, Jack plays Jill $75. If the card is anything else, Jill pays Jack $25.

True or false: This is an example of a fair game.

Answer

The fairness or unfairness of the game is a function of the expected value, which can be calculated by multiplying the probability of each outcome by its value, and adding the products.

We will examine the value of the game to Jack; a positive value indicates a gain to Jack, and a negative value indicates a gain to Jill.

Since only suits matter in this game, there are four equiprobable outcomes. One is favorable to Jill, and the other three are favorable to Jack. There are two events, a win for Jill and a win for Jack, which we will call and , respectively. Their probabilities and their values to Jack are:

: A spade is drawn, and Jill wins.

Probability: $P(W) = \frac{1}{4}$

Value to Jack:

: A card of any of the other three suits is drawn, and Jack wins:

Probability: $P(W') = \frac{3}{4}$

Value to Jack:

The expected value of one play of the game to Jack is

$=\frac{1}{4} (-$75)+ \frac{3}{4}($25)$

The game is fair.

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Question

From a standard deck of 52 cards (the joker is not included) a card is drawn. The person who draws it reveals only that the card is black.

A second card is drawn without replacement. What is the probability that the card will be a spade?

Answer

This is a conditional probability problem.

There are thirteen spades and thirteen clubs; since the first card is known to be black, the probability of the first card being a spade (or club) is

$P (S_{1}) = P (C_{1}) = \frac{1}{2}$ ,

The probability of the second card being a spade depends on the suit of the first card. The probability of the second card being a spade, given that the first card was also a spade, is

$P (S_{2}| S_{1}) = \frac{12}{51}$

The probability of the second card being a spade, given that the first card was a club, is

$P (S_{2}| C_{1}) = \frac{13}{51}$

The overall probability that the second card will be a spade is

$P (S_{2}| S_{1}) P (S_{1}) +P (S_{2}| C_{1})P(C_{1})$

$= \frac{12}{51} \cdot \frac{1}{2}+ \frac{13}{51} \cdot \frac{1}{2}$

$= \frac{25 }{102}$

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Question

A penny is flipped ten times; each flip results in heads.

True or false: The coin must be loaded so that it comes up heads more often than tails.

Answer

Suppose the coin is indeed fair. The probability of the coin coming up heads in one trial is $\frac{1}{2}$ ; by the multiplication principle, the probability that the coin will come up heads each of ten consecutive times is $\left (\frac{1}{2} \right )^{10} = \frac{1}{1, 024}$ . While this makes this event extremely unlikely, it is not impossible. Thus, the fact that the coin comes up ten heads in ten flips is not conclusive evidence that the coin is unfair.

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Question

A card is drawn at random from a standard deck of 52 cards (the joker is not included), and its rank and suit are recorded.

Which of the following changes both the probability of drawing a black card and that of drawing an ace?

Replacing the ace of spades with the joker

Adding the joker

Removing the ace of spades

Answer

If a card is drawn at random from a standard deck of 52, the probability of drawing one of the 26 black cards is $\frac{1}{2}$ ; the probability of drawing an ace is $\frac{1}{13}$ . Now, examine each of the three scenarios.

(1) If the ace of spades is replaced with the joker, this leaves 25 black cards and 3 aces out of a total of 52 cards. The probability of drawing one of the 25 black cards is $\frac{25}{52}$ ; the probability of drawing an ace is $\frac{3}{52}$ .

(2) If the joker is added, there are still 26 black cards and 4 aces, but the deck now has 53 cards. The probability of drawing one of the 25 black cards is $\frac{26}{53}$ ; the probability of drawing an ace is $\frac{4}{53}$ .

(3) If the ace of spades is removed, this leaves 25 black cards and 3 aces out of a total of 51 cards. The probability of drawing one of the 25 black cards is $\frac{25}{51}$ ; the probability of drawing an ace is $\frac{3}{51} = \frac{1}{17}$ .

In all three cases, both probabilities have changed.

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Question

The twelve face cards (kings, queens, jacks) are separated from a standard deck of 52 cards. Two cards are selected at random from the twelve, without replacement. What is the probability that both cards will be kings?

Answer

Two cards are drawn from the deck without regard to order, so the sample space is the set of all combinations of two cards from a set of twelve. The size of this sample space is

The event is the set of all combinations of two cards from the set of four kings. The size of this event space is

The probability of the event is

$P(E) = \frac{n(E)}{n(S)} = \frac{6}{66}= \frac{1}{11}$

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Question

The odds in favor of an event occurring are 17 to 4. To the nearest hundredth, what is the probability of the event happening?

Answer

If the odds in favor of an event are to , the probability of that event is $\frac{a}{a+b}$ . Setting and , the probability of the event is

$\frac{17}{17+4} = \frac{17}{21} \approx 0.81$ .

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Question

The probability of an event is $\frac{17}{25}$ . What are the odds in favor of the event?

Answer

The odds in favor of an event are equal to the ratio of the probability of the event to that of the opposite event. Therefore, we want to determine $\frac{P(E)}{P(E')}$ :

$P(E)= \frac{17}{25}$

$P(E')=1- \frac{17}{25}= \frac{8}{25}$

$\frac{P(E)}{P(E')} = \frac{\frac{17}{25}}{\frac{8}{25}}= \frac{17}{8}$ ,

or 17 to 8 in favor.

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Question

Jack and Jill agree to a game. A card is drawn at random from a standard deck of 52 (no joker). If the card is a face card (king, jack, queen) or a ten, Jack plays Jill $50. If the card is anything else, Jill pays Jack $25.

Which of the following is true of the game?

Answer

The fairness or unfairness of the game is a function of the expected value, which can be calculated by multiplying the probability of each outcome by its value, and adding the products.

We will examine the value of the game to Jack; a positive value indicates a gain to Jack, and a negative value indicates a gain to Jill.

Since only ranks matter in this game, there are thirteen equiprobable outcomes. Four are favorable to Jill, and the other nine are favorable to Jack. There are two events, a win for Jill and a win for Jack, which we will call and , respectively. Their probabilities and their values to Jack are:

: A king, queen, jack, or ten is drawn, and Jill wins.

Probability: $P(W) = \frac{4}{13}$

Value to Jack:

: A card of any of the other nine ranks is drawn, and Jack wins:

Probability: $P(W') = \frac{9}{13}$

Value to Jack:

The expected value of one play of the game to Jack is

$=\frac{4}{13} (-$50)+ \frac{9}{13}($25)$

$\approx $1.92$ in Jack's favor.

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Question

Jack and Jill agree to a game, the rules of which are as follows:

A card will be drawn from a standard 53-card deck (including the joker). If a club is drawn, Jack will pay Jill $100. If a card of any other suit is drawn, Jill will pay Jack $20.

For the game to be fair, what must happen in the event the joker is drawn?

(Note: The joker is not considered to be of any of the four suits.)

Answer

We will consider the value of each event to Jill; positive values refer to Jack paying Jill, and negative values refer to Jill paying Jack.

There are 53 equiprobable outcomes. Along with their probabilities and their values to Jill, they are:

: The joker is drawn. 1 out of 53 outcomes is favorable to this event, so

$P(J) = \frac{1}{53}$

The value to Jill, , is what is to be determned.

: A club is drawn. 13 out of 53 outcomes are favorable to this event, so

$P(C)= \frac{13}{53}$

Since Jill wins $100 in this event, the value to her is

: A card other than a club or a joker is drawn. 39 out of 53 outcomes are favorable to this event, so

$P(O)= \frac{39}{53}$

Since Jill loses $20, the value to her is

.

The expected value of any game can be calculated by multiplying the probability of each event by its value, and adding the products. That is,

The game is to be fair, meaning that . Making this and the other substitutions, we get

$$ 0= \frac{1}{53} A(J)+ \frac{13}{53} ($ 100)+ \frac{39}{53} (-$20)$

Solve for

$$ 53(0)=53 \left [\frac{1}{53} A(J)+ \frac{13}{53} ($ 100)+ \frac{39}{53} (-$20) \right ]$

Jill must agree to pay Jack $520 in the event of the draw of the joker.

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Question

Two fair dices are rolled. If the first die is an odd number what is the probability that the sum of the two dice equal 8?

Answer

First, recall that each die is rolled independently.

Now, it is known that the first die is an odd number therefore the possible options are:

$\text{Die}_1=1,3,5$

From here, find the number of combinations each one will have to make a sum of 8 with the other die.

$\text{Die}_2=1,2,3,4,5,6$

By investigation, die number one cannot be one because it along with any value from die two will not equal eight.

Therefore, investigate when die one is three.

$\text{Sum}=3+5=8$

Next, if die one is five then the possible sum is,

$\text{Sum}=5+3=8$

Therefore there are a total of two possibilities of rolling a sum equaled to eight.

From here, identify the total number of options the sum could be when it does not equal eight.

$\\text{Possibilities for Die}_1\times \text{Possibilities for Die}_2 \=3\times 6 \=18$

$\text{Probability of Sum of 8}=\frac{\text{Combinations equalling 8}}{\text{Total Possibilities}}$

$\text{Probability of Sum of 8}=\frac{2}{18}=\frac{1}{9}$

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Question

A coin is loaded so that it comes up heads 60% of the time. If it is tossed six times, what is the probability that it will come up heads at least five times?

Choose the closest response.

Answer

Since each trial can have one of two outcomes, heads or tails, this is an example of a Bernoulli process.

The probability that trials will result in successes in a Bernoulli process is

$C(n, k) p^{k}q^{n-k}$ ,

where is the probability of a success in any given trial and is the probability of a failure.

Since there are six tosses, set . We can call heads the success with probability , and tails the failure with probability . Accordingly, the probability that exactly heads will come up is

$P(k) = C(6, k) 0.6^{k}0.4^{6-k}$

Since we are looking for the probability of heads coming up at least five times, evaluate as follows:

$P(5) = C(6, 5) 0.6^{5}0.4^{6-5} = 6 (0.07776) (0.4) = 0.186624$

$P(6) = C(6,6) 0.6^{6}0.4^{6-6} = 1 (0.046656)(1) = 0.046656$

Of the five choices, 0.25 comes closest.

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Question

Jan and Jeff are playing a game of poker dice, in which a turn comprises the following process:

Roll all five dice - note: these are standard six-sided dice, which you may assume to be fair.

Keep the dice you want, then roll the remaining dice in hope of improving your score.

Repeat Step 2, after which whatever you have stands.

Jan is down to her last turn. She has rolled the dice once and has the following: 1-2-3-4-6.

She must score a "large straight", which comprises five dice in sequence. (1-2-3-4-5, or 2-3-4-5-6). If she re-rolls the "6", what is the probability that she will get her large straight on her second or third roll?

Answer

Let be the event that Jan gets a straight - that is, that she rolls a "5" on her second or third turn.

The easiest way to find the probability that Jan will complete her large straight - by rolling a "5" - is to find the probability that she will not complete it. This happens if she rolls any number other than "5" on both rolls, which, by the multiplication principle, will happen with probability

$P(\overline{W}) = \frac{5}{6} \times \frac{5}{6} = \frac{25}{36}$

The probability that she will get the straight is this probability subtracted from 1, or

$P(W)= 1- P(\overline{W}) = 1- \frac{25}{36} = \frac{11}{36}$

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Question

Mike and Spike are playing a game of poker dice, in which a turn comprises the following process:

Roll all five dice - note: these are standard six-sided dice, which you may assume to be fair.

Keep the dice you want, then roll the remaining dice in hope of improving your score.

Repeat Step 2, after which whatever you have stands.

Mike is down to his last turn. He has rolled the dice once and has the following: 1-3-3-3-5.

He must roll a five of a kind in order to score. If he re-rolls the "1" and the "5", what is the probability that he will get five "3's"?

Answer

In order for Mike to get his five-of-a-kind, one of three things has to happen:

He can roll two "3's" on his next turn. There is only one way out of thirty-six for this to happen, so the probability of rolling two 3's is $\frac{1}{36}$ .

He can roll one "3" on his next turn, then one "3" on his third turn. There are ten ways out of thirty-six to roll exactly one "3" with a pair of dice (1-3, 2-3, 4, 3, 5-3, 6-3, 3-1, 3-2, 3-4, 3-5, 3-6), making this probability $\frac{10}{36} = \frac{5}{18}$ . After this, he will roll one die; the probability of rolling a "3" is $\frac{1}{6}$ . By the multiplication principle, the probability of this scenario happening is $\frac{5}{18} \times \frac{1}{6} = \frac{5}{108}$ .

He can roll no "3's" on his next turn, then two "3's" on his third turn. There are twenty-five ways out of thirty-six to roll two dice and not get a 3, so the probability of this is $\frac{25}{36}$ ; the probability of rolling two "3's" on the third turn is $\frac{1}{36}$ . By the multiplication principle, the probability of this scenario happening is $\frac{25}{36} \times \frac{1}{36}= \frac{25}{1,296}$ .

These three scenarios are mutually exclusive and together comprise the event that Mike gets his two 3's. Add the probabilities:

$\frac{1}{36}+ \frac{5}{108} + \frac{25}{1,296}$

$= \frac{36}{1,296}+ \frac{60}{1,296} + \frac{25}{1,296}$

$= \frac{121}{1,296}$

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Question

A number of black and white balls are placed in a box. A ball is drawn and replaced, and its color is noted; this is repeated 100 times. At the end of the experiment, it is noted that a black ball was drawn 23 times; it is concluded that the probability of drawing a black ball from the box is $\frac{23}{100}$ .

This is an example of:

Answer

Empirical probability is based on repeated experimentation and observation; theoretical probability is calculated based on equally probable events and the number of such events favorable to a given outcome.

The probability that a black ball will be drawn is hypothesized based on an experiment, in which the number of times a black ball was drawn compared to the number of total draws was observed. The probability was derived based on this observation. This is therefore an example of empirical probability.

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Question

A fair coin comes up heads with probability $\frac{1}{2}$ . This probability is an example of:

Answer

Empirical probability is based on repeated experimentation and observation; theoretical probability is calculated based on equally probable events and the number of such events favorable to a given outcome.

The two outcomes for the flip of a fair coin are two, which are equally probable. The probability of each outcome is $\frac{1}{2}$ based on theoretical probability.

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Question

The twelve face cards (kings, queens, jacks) are separated from a standard deck of 52 cards. Two cards are selected at random from the twelve, without replacement. What is the probability that the two cards will be of different color?

Answer

Two cards are drawn from the deck without regard to order, so the sample space is the set of all combinations of two cards from a set of twelve. The size of this sample space is

The event is the set of all selections of one card from a set of six red cards and one card from a set of six black cards. The selections are independent, so by the multiplication principle,

$n(E ) = 6 \times 6 = 36$

The probability of the event is

$P(E) = \frac{n(E)}{n(S)} = \frac{36}{66} = \frac{6}{11 }$ .

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Question

The four aces, the four deuces, and the joker are separated from a standard deck of 53 cards. Two cards are selected at random from the nine, without replacement. What is the probability that both cards will be red?

Note: The joker is considered to be neither red nor black.

Answer

Two cards are drawn from the deck without regard to order, so the sample space is the set of all combinations of two cards from a set of nine. The size of this sample space is

The event is the set of all combinations of two cards from the set of four red cards. The size of this event space is

The probability of the event is

$P(E) = \frac{n(E)}{n(S)} = \frac{6}{36} = \frac{1}{6 }$

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Question

The probability of an event is $\frac{17}{25}$ . What are the odds in favor of the event?

Answer

The odds in favor of an event are equal to the ratio of the probability of the event to that of the opposite event. Therefore, we want to determine $\frac{P(E)}{P(E')}$ :

$P(E)= \frac{17}{25}$

$P(E')=1- \frac{17}{25}= \frac{8}{25}$

$\frac{P(E)}{P(E')} = \frac{\frac{17}{25}}{\frac{8}{25}}= \frac{17}{8}$ ,

or 17 to 8 in favor.

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Question

From a standard deck of 52 cards (the joker is not included) a card is drawn. The person who draws it reveals only that the card is black.

A second card is drawn without replacement. What is the probability that the card will be a spade?

Answer

This is a conditional probability problem.

There are thirteen spades and thirteen clubs; since the first card is known to be black, the probability of the first card being a spade (or club) is

$P (S_{1}) = P (C_{1}) = \frac{1}{2}$ ,

The probability of the second card being a spade depends on the suit of the first card. The probability of the second card being a spade, given that the first card was also a spade, is

$P (S_{2}| S_{1}) = \frac{12}{51}$

The probability of the second card being a spade, given that the first card was a club, is

$P (S_{2}| C_{1}) = \frac{13}{51}$

The overall probability that the second card will be a spade is

$P (S_{2}| S_{1}) P (S_{1}) +P (S_{2}| C_{1})P(C_{1})$

$= \frac{12}{51} \cdot \frac{1}{2}+ \frac{13}{51} \cdot \frac{1}{2}$

$= \frac{25 }{102}$

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Question

A card is drawn at random from a standard deck of 52 cards (the joker is not included), and its rank and suit are recorded.

Which of the following changes both the probability of drawing a black card and that of drawing an ace?

Replacing the ace of spades with the joker

Adding the joker

Removing the ace of spades

Answer

If a card is drawn at random from a standard deck of 52, the probability of drawing one of the 26 black cards is $\frac{1}{2}$ ; the probability of drawing an ace is $\frac{1}{13}$ . Now, examine each of the three scenarios.

(1) If the ace of spades is replaced with the joker, this leaves 25 black cards and 3 aces out of a total of 52 cards. The probability of drawing one of the 25 black cards is $\frac{25}{52}$ ; the probability of drawing an ace is $\frac{3}{52}$ .

(2) If the joker is added, there are still 26 black cards and 4 aces, but the deck now has 53 cards. The probability of drawing one of the 25 black cards is $\frac{26}{53}$ ; the probability of drawing an ace is $\frac{4}{53}$ .

(3) If the ace of spades is removed, this leaves 25 black cards and 3 aces out of a total of 51 cards. The probability of drawing one of the 25 black cards is $\frac{25}{51}$ ; the probability of drawing an ace is $\frac{3}{51} = \frac{1}{17}$ .

In all three cases, both probabilities have changed.

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