What is the probability of getting a spade or a face card?

A = number of black cards B = number of picture cards (or face cards) C = number of black picture cards

There are 26 black cards (spades and clubs), so A = 26.

There are 3 picture cards (Jack, Queen, King) in each suit. There are 4 suits (clubs, hearts, spades, diamonds). So there are 3*4 = 12 picture cards. This means B = 12

There are 2 suits which are black (spades and clubs) with 3 face cards per suit, so 2*3 = 6 cards which are both black cards and face cards. So C = 6

The number of cards that are either black cards or a face card, or both, is...

D = A+B-C D = 26+12-6

D = 32

So there are 32 cards that are either black cards or a face card, or both

This is out of 52 cards total, so

probability of selecting a black card or a picture card = D/52

probability of selecting a black card or a picture card = 32/52

probability of selecting a black card or a picture card = 8/13

The final answer, as a fraction, is 8/13

Sadie C.

asked • 08/05/20

find the odds against selecting a face card or a spade when a card is drawn from a standard deck of cards

3 Answers By Expert Tutors

William W. answered • 08/06/20

Math and science made easy - learn from a retired engineer

"Odds in favor" is defined as the number of successes divided by the number of failures while "odds against" is defined as the number of failures divided by the number of successes. Obviously, we want the latter.

In this case, since a face card is Jack, Queen, or King, and there are 4 different suits, there are 12 face cards. We also know that there are 13 spades, HOWEVER, 3 of those spades are already included in the face cards we have counted. So the number of cards that are face cards OR spades is 12 + (13 - 3) = 22. That makes 30 cards that are NOT face cards or spades (52 total cards - 22 that are face cards or spades)

The number of failures is 30 and the number of successes is 22 the the odds against is 30:22 or 15:11

Assuming a standard deck of cards includes 52 cards with no Jokers.

There are 13 spade cards from A to K

Three of the spades are face cards (J Q K)

There are 12 face cards

Three for each suit.

Since we want to avoid double counting the J Q K of spades there are

13 spades and 9 face cards that aren't spades.

This is 22 total cards.

The probability is 22/52 or 11/26 of drawing one of the cards so the odds that this doesn't happen is one minus that.

1-11/26 = 15/26 chance to not draw a face or spades.

This is also a 57.7% chance if percentage is needed.

The probability of selecting a face card or a spade = P(face card) + P(spade) - P(face card and space) = 12/52 + 13/52 - 3/52 = 22/52 = 11/26. So probability of NOT selecting a face card or a space = 15/26, so odds are 15:11 against selecting a face card or a spade.

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Consider the experiment of selecting a card from an ordinary deck of 52 playing cards. Determine the probability of each outcome.

1.

A face card is drawn.

12/52 = 3/13

2.

A red card or a card showing a 5 is drawn.

P(Red OR 5) = P(Red) + P(5) - P(Red AND 5) = 26/52 + 4/52 - 2/52 = 28/52 = 7/13

3.

A non-face card or a 7 is drawn.

P(non-face OR 7) = P(non-face) + P(7) - P(non-face AND 7) = 40/52 + 4/52 - 4/52 = 40/52 = 10/13

4.

A card drawn is neither a king nor a spade.

There are 4 kings and there are 13 spades, with one that is both a king and a spade, so there are 4 + 13 - 1 = 16 cards that are either kings or spades. That leaves 52-16=36 cards that are neither a king nor a spade. Our desired probability is 36/52 = 9/13.

5.

A card that is a black face card is drawn.

There are 6 black face cards, so the probability is 6/52 = 3/26.

6.

A card that is not a face card is drawn.

We know from question (1) that P(face card) = 3/13. Because the event "non-face card" is the complement to "face card," P(non-face card) = 1 - P(face card) = 1 - 3/13 = 10/13.

A box contains three red balls and two white balls. A ball is selected at random from the box, its color is recorded, and the ball is replaced. A second ball is then selected at random and its color is recorded. The outcome associated with this type of selection is an ordered pair (first drawn second draw). An example outcome is (red,white).

7.

List the sample space for this experiment.

There are four ordered pairs in the sample space, where R stands for a red ball drawn and W represents a white ball drawn: {(R,R),(R,W),(W,R),(W,W)}.

8.

Determine the probability that both balls are red.

Because we are drawing with replacement, the outcome of the second draw does not depend on the outcome of the first. Therefore, the events are independent and we can mutliply to determine the desired probability: P(R,R) = P(R)*P(R) = 3/5 * 3/5 = 9/25.

9.

Determine the probability that both balls are white.

Using the same justification as in question (8), P(W,W) = P(W)*P(W) = 2/5 * 2/5 = 4/25.

10.

Explain why the probabilities determined for (8) and (9) do not sum to 1.

The events in questions (8) and (9) do not represent all possible outcomes of the experiment, for we could get (R,W) or (W,R) as well.

The table of information was recently collected at the Jamestown School dining hall.

Use this information to calculate the probabilities indicated for someone picked at random from the school's student population.

For the questions here, probabilities are determined by creating a ratio of "favorable cases" to "total cases."

 Class Males Females 9th grade 91 101 10th grade 95 105 11th grade 103 98 12th grade 97 101

11.

P(male)

386/791

12.

P(10th grader)

200/791

13.

P(male | 9th grader)

91/192

14.

P(9th grader | male)

91/386

15.

P(junior or senior)

399/791

16.

P(11th grader and female)

98/791

17.

Suppose we know that for event A, P(A) = 0.65. If ~A represents the event complementary to A, what is P(~A)?

P(~A) = 1 - P(A) = 1 - 0.65 = 0.35

18.

An experiment has three possible outcomes, X, Y, and Z. If P(Z) = P(X) and P(Y) = 3·P(X), determine P(X), P(Y), and P(Z).

Let P(X) = a. Then P(Z) = a and P(Y) = 3a. We must have P(X) + P(Y) + P(Z) = 1, because X, Y, and Z comprise the entire sample space. Then P(X) + P(Y) + P(Z) = 1 implies that a + 3a + a = 1. Solving for a yields a = 1/5. This gives us P(X) = P(Z) = 1/5 and P(Y) = 3/5.

A family has three children, ages 5, 7, and 10. If we assume that the probability of giving birth to a boy is the same as the probability of giving birth to a girl, determine the following probabilities.

For these problems, we use the following information, where B represents a boy and G represents a girl. The set of all outcomes is a set of 8 equally likely outcomes, shown in birth order from youngest to oldest:

 BBB BBG BGB BGG GBB GBG GGB GGG

19.

P(three girls)

1/8

20.

P(at least two boys)

4/8 = 1/2

21.

P(the oldest is a girl)

4/8 = 1/2

22.

P(two children are of the same gender)

6/8 = 3/4 (interpreted as "exactly two")

23.

P(there are an equal number of boys and girls)

0: This cannot happen in a family of three.

Three different gaming machines were on exhibit at a local casino. A sign on each machine showed its sample space (in dollars) and the probability of each guaranteed output in the sample space.

24.

If you wanted to select the machine that produced the largest average (mean) output over a long time period, which machine would you select? Explain.

Select Machine A. Over the long haul, here's what each machine will do, expressed as a weighted average using the sum of the product of each outcome with its probability. This is called expected value:

• Machine A: (2/7)*1 + (3/7)*3 + (2/7)*5 = 21/7 = 3
• Machine B: (5/9)*0 + (4/9)*6 = 24/9 = 8/3
• Machine C: (1)*2 = 2

25.

Suppose that you and two friends are beginning to play the three machines. If you have first choice on which machine you'll play, which of the three would you choose? Explain.

I interpret this to mean I want to choose the machine whose most likely outcome is greatest. The most likely outcome of Machine A is 3, of B is 6, and of C is 2. I'd choose Machine B.

26.

A fair die is rolled three consecutive times. What is the probability that the digit sum of the three rolls is 15 or larger?

There are 6*6*6 = 216 possible outcomes when a fair die is rolled three times. We now need to determine how many of those result in a sum of 15 or more.

The possible sums are 18, 17, 16, and 15. We list below the ways to get each of those. Each entry is an ordered triple showing the result on the first, second, and third roll, respectively. For example, (6,5,6) represents getting a 6 on the first roll, a 5 on the second, and a 6 on the third.

 sum of 15 sum of 15 sum of 17 sum of 18 (6,6,3) (6,3,6) (3,6,6) (6,5,4) (6,4,5) (5,6,4) (5,4,6) (4,6,5) (4,5,6) (5,5,5) (6,6,4) (6,4,6) (4,6,6) (6,5,5) (5,6,5) (5,5,6) (6,6,5) (6,5,6) (5,6,6) (6,6,6)

This list shows 20 ways to get the desired sums. The probability we seek is 20/216 = 5/54.

27.

A fair die is rolled six times. Determine the probability that each of the six equally likely outcomes appears exactly once in those six rolls.

Again we rely on the classic probability ratio, comparing favorable cases to the total number of cases.

There are 6*6*6*6*6*6=6^6 (6 raised to the sixth power) different outcomes when a fair die is rolled 6 times. The favorable outcomes are just the ways to arrange the digits {1,2,3,4,5,6}, because each must appear once. This can be done in P(6,6) = 6! ways.

Therefore, the desired probability is 6!/(6^6) = 5/324 or approximately 0.0154.

A production line is equipped with two quality-control check points that tests all items on the line. At check point #1, 10% of all items failed the test. At check point #2, 12% of all items failed the test. We also know that 3% of all items failed both tests.

28.

If an item failed at check point #1, what is the probability that it also failed at check point #2?

In symbolic form, we are asking for P(fail #2 | fail #1). Using the formula for determining conditional probability, P(fail #2 | fail #1) = P(fail #2 AND fail #1)/P(fail #1) = (0.03)/(0.10) = 3/10.

29.

If an item failed at check point #2, what is the probability that it also failed at check point #1?

Here we are asking for P(fail #1 | fail #2). Using the formula, P(fail #1 | fail #2) = P(fail #1 AND fail #2)/P(fail #2) = (0.03)/(0.12) = 3/12 = 1/4.

30.

What is the probability that an item failed at check point #1 or at check point #2?

P(fail #1 OR fail #2) = P(fail #1) + P(fail #2) - P(fail #1 AND fail #2) = 0.10 + 0.12 - 0.03 = 0.19.

31.

What is the probability that an item failed at neither of the check points?

We are asking for P(no failure). The event "no failure" is the complement of the event "failure in at least one test." We know from question (30) that P(failure in at least one test) = 0.19, so P(no failure) = 1 - 0.19 = 0.81.

George knows that a rare disease, D, within his family can be passed on to his children, and that the probability is 0.10 that the inherited disease will be passed on to a child. We signify this at P(D) = 0.10.

32.

Determine the probability that none of George's three children inherit the disease from George.

Use ~D to represent the complement of event D. Then ~D represents "did not pass on the disease" and because P(D) = 0.10, P(~D) = 0.90.

We want P(~D child #1 AND ~D child #2 AND ~D child #3). We assume that these are independent events, that is, that whether the disease has passed to one child doesn't change the probability of it being passed to another. Assuming this, we have

P(~D child #1 AND ~D child #2 AND ~D child #3) = P(~D child #1)*P(~D child #2)*P(~D child #3) = (0.9)*(0.9)*(0.9) = 0.729, the desired probability.

33.

If P(D) = x, determine the largest value of x possible so that the solution to (32) would be greater than or equal to 90%.

Using the information, symbolism, and strategy from question (32), we seek x so that x*x*x is greater than or equal to 0.9. By guess-and-check or by using the cube-root function on a calculator, we determine that x = 0.9655, rounded to the nearest ten-thousandth.

Suppose the Atlanta Braves and the New York Yankees meet in the next Major League Baseball World Series, where the two teams play until one team has won four games. Let B represent the event that the Braves win a game and let Y represent the event that the Yankees win a game, with P(B) = 0.4 and P(Y) = 0.6). Assume that these probabilities do not change throughout the series. Determine the following probabilities for this World Series.

For the questions below, we use two strategies: (1) Use multiplication of consecutive probabilities to determine the probability of a specific sequence of games. (2) Count in some way the number of ways a certain event can occur. Question (34) illustrates use of strategy #1 and question (35) uses both strategies.

We use the events B and Y above, writing them in sequence to show a string of results. for instance, BYBYY represents the Braves winning game 1 and game 3 and the Yankees winning games 2, 4, and 5.

We also interpret the statement above,"Assume that these probabilities do not change throughout the series," to indicate that the outcome of one game does not influence the outcome of another.

34.

P(Atlanta wins no games)

This will only happen if the Yankees win the first 4 games. Thus we seek P(YYYY) and by the independence of game results, P(YYYY)=P(Y)*P(Y)*P(Y)*P(Y)=0.6^4 = 0.1296.

35.

P(series is tied 2 games each after the first 4 games)

There are six ways this can happen, each with the same probability:

 BBYY BYBY BYYB YYBB YBYB YBBY

The probability of each is (0.6)^2*(0.4)^2 = 0.0576. Multiply this by 6 to get the desired probability: P(series is tied 2 games each after the first 4 games) = 0.3456

36.

P(the series ends in 5 games)

To end in 5 games, the series record must be 4 wins and 1 loss for one of the teams. The probability of each will change depending on which team has 4 wins:

 Atlanta wins the series 4-1 BBBYB BBYBB BYBBB YBBBB P(Atlanta wins series 4-1) = (4ways)(probability of each way) = 4(0.4)^4(0.6) = 0.06144
 Yankees wins the series 4-1 YYYBY YYBYY YBYYY BYYYY P(Yankees wins series 4-1) = (4ways)(probability of each way) = 4(0.6)^4(0.4) = 0.20736

Because the two events (Braves win 4-1, Yankees win 4-1) are mutually exclusive, we add the probabilities of each event. Therefore, P(the series ends in 5 games) = 0.06144 + 0.20736 = 0.2688.

37.

P(if the series lasts 7 games, Atlanta will win).

We seek P(Atlanta wins | series last 7 games). This is a conditional probability statement and we apply the conditional formula here.

P(Atlanta wins | series last 7 games) = P(Atlanta wins AND series last 7 games)/P(series lasts 7 games)

There are 40 different win-lose sequences that lead to a 7-game series. In 20 of these sequences, Atlanta wins 4 games to 3 and in the other 20, the Yankees win 4 games to 3.

Each of the 20 ways that Atlanta wins the series in 7 games has the same probability: (0.4)^4(0.6)^3 = 0.0055296. So P(Atlanta wins AND series last 7 games) = 20*(0.4)^4(0.6)^3 = 20*0.0055296 = 0.110592.

Each of the 20 ways that the Yankees win the series in 7 games has the same probability: (0.6)^4(0.4)^3 = 0.0082944. So P(Yankees win AND series last 7 games) = 20*(0.6)^4(0.4)^3 = 20*0.0082944 = 0.165888.

Because the event "Yankees win in 7" and the event "Braves win in 7" are mutually exclusive, we add the two results to determine P(series last 7 games). This is 0.165888 + 0.110592 = 0.27648 = P(series last 7 games). This says, by the way, that under these conditions and assumptions, about 25% of the time the series will last 7 games.

We can now compute the desired conditional probability:

P(Atlanta wins | series last 7 games) = P(Atlanta wins AND series last 7 games)/P(series lasts 7 games) = (0.110592)/(0.165888 + 0.110592) = 0.4. Does this seem like a surprising result? Given that we get to game 7, we can think of the series boiling down to one game, and the probability that atlanta wins that one game is just P(B) = 0.4.

Another way to think about (0.110592)/(0.165888 + 0.110592) is top think of it as a weighted outcome: P(result A and condition #1) compared to the sum of the probabilities of all possible results under condition #1. Here, there were just two results possible under the condition that we get to game 7: Either the Braves win or the Yankees win.

Other situations may extend that. Suppose under a certain condition, 10 different things - - all mutually exclusive - - could happen. Then the probability that the first of those different things does happen under the given condition is just the probability that that first thing happens and the condition happens compared to the sum of the 10 probabilities for the 10 different things that could happen with the condition in place.

What is the probability of getting a card of spade or an ace?

Chance of getting a Spade or an ace=16/52=4/13.

How many spade face cards are in a deck of 52 cards?

- There are 4 suits (Clubs, Hearts, Diamonds, and Spades) and there are 13 cards in each suit (Clubs/Spades are black, Hearts/Diamonds are red) - Without replacement means the card IS NOT put back into the deck.

What is the probability of getting a face card?

Hence, the probability of drawing a red face card from a deck of cards is 3/26. Q.

How many face cards are a spade?

A standard 52-card deck comprises 13 ranks in each of the four French suits: clubs (♣), diamonds (♦), hearts (♥) and spades (♠). Each suit includes three court cards (face cards), King, Queen and Jack, with reversible (double-headed) images.

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