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Statistics and Probability Problems with Solutions

Problems on statistics and probability are presented. The solutions to these problems are at the bottom of the page.

  • Given the data set 4 , 10 , 7 , 7 , 6 , 9 , 3 , 8 , 9 Find a) the mode, b) the median, c) the mean, d) the sample standard deviation. e) If we replace the data value 6 in the data set above by 24, will the standard deviation increase, decrease or stay the same?
  • Find x and y so that the ordered data set has a mean of 42 and a median of 35. 17 , 22 , 26 , 29 , 34 , x , 42 , 67 , 70 , y
  • Given the data set 62 , 65 , 68 , 70 , 72 , 74 , 76 , 78 , 80 , 82 , 96 , 101, find a) the median, b) the first quartile, c) the third quartile, c) the interquartile range (IQR).
  • The exam grades of 7 students are given below. 70 , 66 , 72 , 96 , 46 , 90 , 50 Find a) the mean b) the sample standard deviation
  • Twenty four people had a blood test and the results are shown below. A , B , B , AB , AB , B , O , O , AB , O , B , A AB , A , O , O , AB , B , O , A , AB , O , B , A a) Construct a frequency distribution for the data. b) If a person is selected randomly from the group of twenty four people, what is the probability that his/her blood type is not O?
  • When a die is rolled and a coin (with Heads and Tails) is tossed, find the probability of obtaining a) Tails and an even number, b) a number greater 3, c) Heads or an odd number,
  • A box contains red and green balls. The number of green balls is 1/3 the number of red balls. If a ball is taken randomly from the box, what is the probability that the ball is red?
  • A committee of 6 people is to be formed from a group of 20 people. The committee has to have the number of women double that of the men. In how many ways can this committee be formed if there are 12 men?
  • A student's marks in five tests are 36%, 78%, 67%, 88% and 98%. The weights for the five tests are 1, 2, 2, 3, 3 respectively. Find the weighted mean μ of the five tests.
  • In a group of 40 people, 10 are healthy and every person the of the remaining 30 has either high blood pressure, a high level of cholesterol or both. If 15 have high blood pressure and 25 have high level of cholesterol, a) how many people have high blood pressure and a high level of cholesterol? If a person is selected randomly from this group, what is the probability that he/she b) has high blood pressure (event A)? c) has high level of cholesterol(event B)? d) has high blood pressure and high level of cholesterol (event A and B)? e) has either high blood pressure or high level of cholesterol (event A or B)? f) Use the above to check the probability formula: P(A or B) = P(A) + P(B) - P(A and B).
  • A committee of 5 people is to be formed randomly from a group of 10 women and 6 men. Find the probability that the committee has a) 3 women and 2 men. a) 4 women and 1 men. b) 5 women. c) at least 3 women.
  • In a school, 60% of pupils have access to the internet at home. A group of 8 students is chosen at random. Find the probability that a) exactly 5 have access to the internet. b) at least 6 students have access to the internet.
  • The grades of a group of 1000 students in an exam are normally distributed with a mean of 70 and a standard deviation of 10. A student from this group is selected randomly. a) Find the probability that his/her grade is greater than 80. b) Find the probability that his/her grade is less than 50. c) Find the probability that his/her grade is between 50 and 80. d) Approximately, how many students have grades greater than 80?

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Solutions to the above Problems

  • The given data set has 2 modes: 7 and 9
  • order data : 3 , 4 , 6 , 7 , 7 , 8 , 9 , 9 , 10 : median = 7
  • (mean) : m = (3+4+6+7+7+8+9+9+10) / 9 = 7
  • The standard deviation will increase since 24 is further from away from the other data values than 6.
  • x = 36 , y = 77
  • median = 75
  • first quartile = 69
  • third quartile = 81
  • interquartile range = 81 - 69 = 12
  • sample standard deviation ≈ 18.6 (rounded to 1 decimal place)
  • 1 - (7/24) = 17/24 ≈ 0.71 (rounded to 2 decimal places)
  • μ = Σ x P(X = x) = 0×0.24 + 1×0.38 + 2×0.20 + 3×0.13 + 4×0.05 = 1.37
  • 1) using definition σ = √[ Σ (x - μ) 2 P(X = x) ] = √[ (0-1.37) 2 ×0.24 + (1-1.37) 2 ×0.38 + (2-1.37) 2 ×0.2 + (3-1.37) 2 ×0.13 + (4-1.37) 2 ×0.05 ] ≈ 1.13 (rounded to 2 decimal places) 2) using computing formula σ = √[ Σ x 2 P(X = x) - μ 2 ] = √[ 0 2 ×0.24 + 1 2 ×0.38 + 2 2 ×0.2 + 3 2 ×0.13 + 4 2 ×0.05 - 1.37 2 ] ≈ 1.13 (rounded to 2 decimal places)
  • If there 12 men, then there are 20 - 12 = 8 women. The committee has six people with the number of women double that of the men, hence the committee has 4 women and 2 men. The number of ways of selecting 4 women from 8 is given by: 8 C 4 = 70. The number of ways of selecting 2 men from 12 is given by: 12 C 2 = 66. The number of selecting 4 women and 2 men to form the committee is given by : 8 C 4 × 12 C 2 = 70 × 66 = 4620
  • μ = Σ x i × f i / Σ f i Σ x i × f i = 1×2 + 2×6 + 3×10 + 4×6 + 5×2 + 6×2 = 90 Σ f i = 2 + 6 + 10 + 6 + 2 + 2 = 28 μ = 90 / 28 ≈ 3.21 (rounded to 2 decimal places)
  • Let the marks be: x 1 = 36%, x 2 = 78%, x 3 = 67%, x 4 = 88%, x 5 = 98% and the respective weights be: w 1 = 1, w 2 = 2, w 3 = 2, w 4 = 3, w 5 = 3. The weighted mean = Σ x i ×w i / Σ w i Σ x i ×w i = 36% × 1 + 78%×2 + 67%×2 + 88%×3 + 98%×3 = 884% Σ w i = 1 + 2 + 2 + 3 + 3 = 11 weighted mean = 884% / 11 = 80%
  • a) Let x be the number of people with both high blood pressure and high level of cholesterol. Hence (15 - x) will be the number of people with high blood pressure ONLY and (25 - x) will be the number of people with high level of cholesterol ONLY. We now express the fact that the total number of people with high blood pressure only, with high level of cholesterol only and with both is equal to 30. (15 - x) + (25 - x) + x = 30 solve for x: x = 10 b) 15 have high blood pressure,hence P(A) = 15/40 = 0.375 c) 25 have high level of cholesterol, hence P(B) = 25/40 = 0.625 d) 10 have both,hence P(A and B) = 10/40 = 0.25 e) 30 have either, hence P(A or B) = 30/40 = 0.75 f) P(A) + P(B) - P(A and B) = 0.375 + 0.625 - 0.25 = 0.75 = P(A or B)
  • a) In what follows n C r = n! / [ (n - r)!r! ] and is the number of combinations of n objects taken r at the time and P(A) is the probability that even A happens. There are 16 C 5 ways to select 5 people (committee members) out of a total of 16 people (men and women) There are 10 C 3 ways to select 3 women out of 10. There are 6 C 2 ways to select 2 men out of 6. There are 10 C 3 * 6 C 2 ways to select 3 women out of 10 AND 2 men out of 6. P(3 women AND 2 men) = 10 C 3 * 6 C 2 / 16 C 5 = 0.412087 b) similarly: P(4 women AND 1 men) = 10 C 4 * 6 C 1 / 16 C 5 = 0.288461 c) similarly: P(5 women ) = 10 C 5 * 6 C 0 / 16 C 5 = 0.0576923 (in 6 C 0 the 0 is for no men) d) P(at least 3 women) = P(3 women or 4 women or 5 women) since the events "3 women" , "4 women" and "5 women" are all mutually exclusive, then P(at least 3 women) = P(3 women or 4 women or 5 women) = P(3 women) + P(4 women) + P(5 women) ≈ 0.412087 + 0.288461 + 0.0576923 ≈ 0.758240
  • a) If a pupil is selected at random and asked if he/she has an internet connection at home, the answer would be yes or no and therefore it is a binomial experiment. The probability of the student answering yes is 60% = 0.6. Let X be the number of students answering yes when 8 students are selected at random and asked the same question. The probability that X = 5 is given by the binomial probability formula as follows: P(X = 5) = 8 C 5 (0.6) 5 (1-0.6) 3 = 0.278691 b) P(X ≥ 6) = P(X = 6 or X = 7 or X = 8) Since all the events X = 6, X = 7 and X = 8 are mutually exclusive, then P(X ≥ 6) = P(X = 6) + P(x = 7) + P(X = 8) = 8 C 6 (0.6) 6 (1-0.6) 2 + 8 C 7 (0.6) 7 (1-0.6) 1 + 8 C 8 (0.6) 8 (1-0.6) 0 ≈ 0.315394
  • a) x = 80 , z = (80 - 70)/10 = 1 Probablity for grade to be greater than 80 = 1 - 0.8413 = 0.1587 b) x = 50 , z = (50 - 70)/10 = -2 Probablity for grade to be less than 50 = 0.0228 c) The z-scores for x = 50 and x = 80 have already been calculated above. Probablity for grade to be between 50 and 80 = 0.8413 - 0.0228 = 0.8185 d) 0.1587 * 1000 ≈ 159 (rounded to the nearest unit)
  • a) 500 - (170+90+60+50) = 130 tons of steel/iron was recycled. b) 60/500 = 0.12 = 12% of the total recycled was glass.

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If I throw 2 standard 5-sided dice, what is the probability that the sum of their top faces equals to 10? Assume both throws are independent to each other. Solution : The only way to obtain a sum of 10 from two 5-sided dice is that both die shows 5 face up. Therefore, the probability is simply \( \frac15 \times \frac15 = \frac1{25} = .04\)

If from each of the three boxes containing \(3\) white and \(1\) black, \(2\) white and \(2\) black, \(1\) white and \(3\) black balls, one ball is drawn at random. Then the probability that \(2\) white and \(1\) black balls will be drawn is?

2 fair 6-sided dice are rolled. What is the probability that the sum of these dice is \(10\)? Solution : The event for which I obtain a sum of 10 is \(\{(4,6),(6,4),(5,5) \}\). And there is a total of \(6^2 = 36\) possible outcomes. Thus the probability is simply \( \frac3{36} = \frac1{12} \approx 0.0833\)

If a fair 6-sided dice is rolled 3 times, what is the probability that we will get at least 1 even number and at least 1 odd number?

Three fair cubical dice are thrown. If the probability that the product of the scores on the three dice is \(90\) is \(\dfrac{a}{b}\), where \(a,b\) are positive coprime integers, then find the value of \((b-a)\).

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Suppose a jar contains 15 red marbles, 20 blue marbles, 5 green marbles, and 16 yellow marbles. If you randomly select one marble from the jar, what is the probability that you will have a red or green marble? First, we can solve this by thinking in terms of outcomes. You could draw a red, blue, green, or yellow marble. The probability that you will draw a green or a red marble is \(\frac{5 + 15}{5+15+16+20}\). We can also solve this problem by thinking in terms of probability by complement. We know that the marble we draw must be blue, red, green, or yellow. In other words, there is a probability of 1 that we will draw a blue, red, green, or yellow marble. We want to know the probability that we will draw a green or red marble. The probability that the marble is blue or yellow is \(\frac{16 + 20}{5+15+16+20}\). , Using the following formula \(P(\text{red or green}) = 1 - P(\text{blue or yellow})\), we can determine that \(P(\text{red or green}) = 1 - \frac{16 + 20}{5+15+16+20} = \frac{5 + 15}{5+15+16+20}\).

Two players, Nihar and I, are playing a game in which we alternate tossing a fair coin and the first player to get a head wins. Given that I toss first, the probability that Nihar wins the game is \(\dfrac{\alpha}{\beta}\), where \(\alpha\) and \(\beta\) are coprime positive integers.

Find \(\alpha + \beta\).

If I throw 3 fair 5-sided dice, what is the probability that the sum of their top faces equals 10? Solution : We want to find the total integer solution for which \(a +b+c=10 \) with integers \(1\leq a,b,c \leq5 \). Without loss of generality, let \(a\leq b \leq c\). We list out the integer solutions: \[ (1,4,5),(2,3,5), (2,4,4), (3,3,4) \] When relaxing the constraint of \(a\leq b \leq c\), we have a total of \(3! + 3! + \frac{3!}{2!} + \frac{3!}{2!} = 18 \) solutions. Because there's a total of \(5^3 = 125\) possible combinations, the probability is \( \frac{18}{125} = 14.4\%. \ \square\)

Suppose you and 5 of your friends each brought a hat to a party. The hats are then put into a large box for a random-hat-draw. What is the probability that nobody selects his or her own hat?

How many ways are there to choose exactly two pets from a store with 8 dogs and 12 cats? Since we haven't specified what kind of pets we pick, we can choose any animal for our first pick, which gives us \( 8+12=20\) options. For our second choice, we have 19 animals left to choose from. Thus, by the rule of product, there are \( 20 \times 19 = 380 \) possible ways to choose exactly two pets. However, we have counted every pet combination twice. For example, (A,B) and (B,A) are counted as two different choices even when we have selected the same two pets. Therefore, the correct number of possible ways are \( {380 \over 2} = 190 \)

A bag contains blue and green marbles. If 5 green marbles are removed from the bag, the probability of drawing a green marble from the remaining marbles would be 75/83 . If instead 7 blue marbles are added to the bag, the probability of drawing a blue marble would be 3/19 . What was the number of blue marbles in the bag before any changes were made?

Bob wants to keep a good-streak on Brilliant, so he logs in each day to Brilliant in the month of June. But he doesn't have much time, so he selects the first problem he sees, answers it randomly and logs out, despite whether it is correct or incorrect.

Assume that Bob answers all problems with \(\frac{7}{13}\) probability of being correct. He gets only 10 problems correct, surprisingly in a row, out of the 30 he solves. If the probability that happens is \(\frac{p}{q}\), where \(p\) and \(q\) are coprime positive integers, find the last \(3\) digits of \(p+q\).

Out of 10001 tickets numbered consecutively, 3 are drawn at random .

Find the chance that the numbers on them are in Arithmetic Progression .

The answer is of the form \( \frac{l}{k} \) .

Find \( k - l \) where \(k\) and \(l\) are co-prime integers.

HINT : You might consider solving for \(2n + 1\) tickets .

You can try more of my Questions here .

A bag contains a blue ball, some red balls, and some green balls. You reach into​ the bag and pull out three balls at random. The probability you pull out one of each color is exactly 3%. How many balls were initially in the bag?

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Amanda decides to practice shooting hoops from the free throw line. She decides to take 100 shots before dinner.

Her first shot has a 50% chance of going in.

But for Amanda, every time she makes a shot, it builds her confidence, so the probability of making the next shot goes up, But every time she misses, she gets discouraged so the probability of her making her next shot goes down.

In fact, after \(n\) shots, the probability of her making her next shot is given by \(P = \dfrac{b+1}{n+2}\), where \(b\) is the number of shots she has made so far (as opposed to ones she has missed).

So, after she has completed 100 shots, if the probability she has made exactly 83 of them is \(\dfrac ab\), where \(a\) and \(b\) are coprime positive integers, what is \(a+b\)?

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Students spend their lives collecting, organizing, and analyzing data, so why not teach them a few skills to help them on their way. Data management is probably best done on authentic tasks that will engage students in their own learning. They can collect their own data on topics that interest them. For example, have you ever wondered if everyone shares the same taste in music as you? Perhaps a survey, a couple of graphs and a few analysis sentences will give you an idea.

Statistics has applications in many different fields of study. Budding scientists, stock market brokers, marketing geniuses, and many other pursuits will involve managing data on a daily basis. Teaching students critical thinking skills related to analyzing data they are presented will enable them to make crucial and informed decisions throughout their lives.

Probability is a topic in math that crosses over to several other skills such as decimals, percents, multiplication, division, fractions, etc. Probability worksheets will help students to practice all of these skills with a chance of success!

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Mean, Median, Mode and Range -- Sorted Sets (Sets of 5 from 10 to 99)

Mean, Median, Mode and Range Worksheets

statistics and probability problem solving

Calculating the mean, median, mode and range are staples of the upper elementary math curriculum. Here you will find worksheets for practicing the calculation of mean, median, mode and range. In case you're not familiar with these concepts, here is how to calculate each one. To calculate the mean, add all of the numbers in the set together and divide that sum by the number of numbers in the set. To calculate the median, first arrange the numbers in order, then locate the middle number. In sets where there are an even number of numbers, calculate the mean of the two middle numbers. To calculate the mode, look for numbers that repeat. If there is only one of each number, the set has no mode. If there are doubles of two different numbers and there are more numbers in the set, the set has two modes. If there are triples of three different numbers and there are more numbers in the set, the set has three modes, and so on. The range is calculated by subtracting the least number from the greatest number.

Note that all of the measures of central tendency are included on each page, but you don't need to assign them all if you aren't working on them all. If you're only working on mean, only assign students to calculate the mean.

In order to determine the median, it is necessary to have your numbers sorted. It is also helpful in determining the mode and range. To expedite the process, these first worksheets include the lists of numbers already sorted.

  • Calculating Mean, Median, Mode and Range from Sorted Lists Sets of 5 Numbers from 1 to 10 Sets of 5 Numbers from 10 to 99 Sets of 5 Numbers from 100 to 999 Sets of 10 Numbers from 1 to 10 Sets of 10 Numbers from 10 to 99 Sets of 10 Numbers from 100 to 999 Sets of 20 Numbers from 10 to 99 Sets of 15 Numbers from 100 to 999

Normally, data does not come in a sorted list, so these worksheets are a little more realistic. To find some of the statistics, it will be easier for students to put the numbers in order first.

  • Calculating Mean, Median, Mode and Range from Unsorted Lists Sets of 5 Numbers from 1 to 10 Sets of 5 Numbers from 10 to 99 Sets of 5 Numbers from 100 to 999 Sets of 10 Numbers from 1 to 10 Sets of 10 Numbers from 10 to 99 Sets of 10 Numbers from 100 to 999 Sets of 20 Numbers from 10 to 99 Sets of 15 Numbers from 100 to 999

Collecting and Organizing Data

statistics and probability problem solving

Teaching students how to collect and organize data enables them to develop skills that will enable them to study topics in statistics with more confidence and deeper understanding.

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  • Constructing Line Plots from Larger Data Sets Construct Line Plots with Smaller Numbers and Lines with Ticks Provided Construct Line Plots with Smaller Numbers and Lines Only Provided Construct Line Plots with Smaller Numbers Construct Line Plots with Larger Numbers and Lines with Ticks Provided Construct Line Plots with Larger Numbers and Lines Only Provided Construct Line Plots with Larger Numbers

Interpreting and Analyzing Data

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Answering questions about graphs and other data helps students build critical thinking skills. Standard questions include determining the minimum, maximum, range, count, median, mode, and mean.

  • Answering Questions About Stem-and-Leaf Plots Stem-and-Leaf Plots with about 25 data points Stem-and-Leaf Plots with about 50 data points Stem-and-Leaf Plots with about 100 data points
  • Answering Questions About Line Plots Line Plots with Smaller Data Sets and Smaller Numbers Line Plots with Smaller Data Sets and Larger Numbers Line Plots with Larger Data Sets and Smaller Numbers Line Plots with Larger Data Sets and Larger Numbers
  • Answering Questions About Broken-Line Graphs Answer Questions About Broken-Line Graphs
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  • Answering Questions About Pictographs Answer Questions About Pictographs

Probability Worksheets

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  • Calculating Probabilities with Dice Sum of Two Dice Probabilities Sum of Two Dice Probabilities (with table)

Spinners can be used for probability experiments or for theoretical probability. Students should intuitively know that a number that is more common on a spinner will come up more often. Spinning 100 or more times and tallying the results should get them close to the theoretical probability. The more sections there are, the more spins will be needed.

  • Calculating Probabilities with Number Spinners Number Spinner Probability (4 Sections) Number Spinner Probability (5 Sections) Number Spinner Probability (6 Sections) Number Spinner Probability (7 Sections) Number Spinner Probability (8 Sections) Number Spinner Probability (9 Sections) Number Spinner Probability (10 Sections) Number Spinner Probability (11 Sections) Number Spinner Probability (12 Sections)

Non-numerical spinners can be used for experimental or theoretical probability. There are basic questions on every version with a couple extra questions on the A and B versions. Teachers and students can make up other questions to ask and conduct experiments or calculate the theoretical probability. Print copies for everyone or display on an interactive white board.

  • Probability with Single-Event Spinners Animal Spinner Probability ( 4 Sections) Animal Spinner Probability ( 5 Sections) Animal Spinner Probability ( 10 Sections) Letter Spinner Probability ( 4 Sections) Letter Spinner Probability ( 5 Sections) Letter Spinner Probability ( 10 Sections) Color Spinner Probability ( 4 Sections) Color Spinner Probability ( 5 Sections) Color Spinner Probability ( 10 Sections)
  • Probability with Multi-Event Spinners Animal/Letter Combined Spinner Probability ( 4 Sections) Animal/Letter Combined Spinner Probability ( 5 Sections) Animal/Letter Combined Spinner Probability ( 10 Sections) Animal/Letter/Color Combined Spinner Probability ( 4 Sections) Animal/Letter/Color Combined Spinner Probability ( 5 Sections) Animal/Letter/Color Combined Spinner Probability ( 10 Sections)

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Probability

How likely something is to happen.

Many events can't be predicted with total certainty. The best we can say is how likely they are to happen, using the idea of probability.

Tossing a Coin

When a coin is tossed, there are two possible outcomes:

Heads (H) or Tails (T)

  • the probability of the coin landing H is ½
  • the probability of the coin landing T is ½

Throwing Dice

When a single die is thrown, there are six possible outcomes: 1, 2, 3, 4, 5, 6 .

The probability of any one of them is 1 6

In general:

Probability of an event happening = Number of ways it can happen Total number of outcomes

Example: the chances of rolling a "4" with a die

Number of ways it can happen: 1 (there is only 1 face with a "4" on it)

Total number of outcomes: 6 (there are 6 faces altogether)

So the probability = 1 6

Example: there are 5 marbles in a bag: 4 are blue, and 1 is red. What is the probability that a blue marble gets picked?

Number of ways it can happen: 4 (there are 4 blues)

Total number of outcomes: 5 (there are 5 marbles in total)

So the probability = 4 5 = 0.8

Probability Line

We can show probability on a Probability Line :

Probability is always between 0 and 1

Probability is Just a Guide

Probability does not tell us exactly what will happen, it is just a guide

Example: toss a coin 100 times, how many Heads will come up?

Probability says that heads have a ½ chance, so we can expect 50 Heads .

But when we actually try it we might get 48 heads, or 55 heads ... or anything really, but in most cases it will be a number near 50.

Learn more at Probability Index .

Some words have special meaning in Probability:

Experiment : a repeatable procedure with a set of possible results.

Example: Throwing dice

We can throw the dice again and again, so it is repeatable.

The set of possible results from any single throw is {1, 2, 3, 4, 5, 6}

Outcome: A possible result.

Example: "6" is one of the outcomes of a throw of a die.

Trial: A single performance of an experiment.

Example: I conducted a coin toss experiment. After 4 trials I got these results:

Three trials had the outcome "Head", and one trial had the outcome "Tail"

Sample Space: all the possible outcomes of an experiment.

Example: choosing a card from a deck

There are 52 cards in a deck (not including Jokers)

So the Sample Space is all 52 possible cards : {Ace of Hearts, 2 of Hearts, etc... }

The Sample Space is made up of Sample Points:

Sample Point: just one of the possible outcomes

Example: Deck of Cards

  • the 5 of Clubs is a sample point
  • the King of Hearts is a sample point

"King" is not a sample point. There are 4 Kings, so that is 4 different sample points.

There are 6 different sample points in that sample space.

Event: one or more outcomes of an experiment

Example Events:

An event can be just one outcome:

  • Getting a Tail when tossing a coin
  • Rolling a "5"

An event can include more than one outcome:

  • Choosing a "King" from a deck of cards (any of the 4 Kings)
  • Rolling an "even number" (2, 4 or 6)

Hey, let's use those words, so you get used to them:

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COMMENTS

  1. Statistics and Probability

    Unit 7: Probability. 0/1600 Mastery points. Basic theoretical probability Probability using sample spaces Basic set operations Experimental probability. Randomness, probability, and simulation Addition rule Multiplication rule for independent events Multiplication rule for dependent events Conditional probability and independence.

  2. Statistics Problems

    Problem 1. In one state, 52% of the voters are Republicans, and 48% are Democrats. In a second state, 47% of the voters are Republicans, and 53% are Democrats. Suppose a simple random sample of 100 voters are surveyed from each state. What is the probability that the survey will show a greater percentage of Republican voters in the second state ...

  3. Probability Questions with Solutions

    Probability Questions with Solutions. Tutorial on finding the probability of an event. In what follows, S is the sample space of the experiment in question and E is the event of interest. n(S) is the number of elements in the sample space S and n(E) is the number of elements in the event E. . Questions and their Solutions Question 1 A die is rolled, find the probability that an even number is ...

  4. Probability: the basics (article)

    You might intuitively know that the likelihood is half/half, or 50%. But how do we work that out? Probability =. In this case: Probability of an event = (# of ways it can happen) / (total number of outcomes) P (A) = (# of ways A can happen) / (Total number of outcomes) Example 1. There are six different outcomes.

  5. Statistics and Probability Problems with Solutions

    The probability of the student answering yes is 60% = 0.6. Let X be the number of students answering yes when 8 students are selected at random and asked the same question. The probability that X = 5 is given by the binomial probability formula as follows: P (X = 5) = 8 C 5 (0.6) 5 (1-0.6) 3 = 0.278691.

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    18.05 Introduction to Probability and Statistics (S22), Problem Set 10 Solutions. pdf. 119 kB 18.05 Introduction to Probability and Statistics (S22), Problem Set 11 Solutions. Course Info Instructors Dr. Jeremy Orloff; Dr. Jennifer French Kamrin; Departments Mathematics; As Taught In ...

  7. Probability

    The probability that you will draw a green or a red marble is \frac {5 + 15} {5+15+16+20} 5+15+16+205+15. We can also solve this problem by thinking in terms of probability by complement. We know that the marble we draw must be blue, red, green, or yellow. In other words, there is a probability of 1 that we will draw a blue, red, green, or ...

  8. How to Solve Probability Problems

    Solution: The sum of probabilities of all the sample points must equal 1. And the probability of getting a head is equal to the probability of getting a tail. Therefore, the probability of each sample point (heads or tails) must be equal to 1/2. Example 2 Let's repeat the experiment of Example 1, with a die instead of a coin.

  9. Probability Problems: Solve them the easy way!

    Finding the probability of a simple event happening is fairly straightforward: add the probabilities together. For example, if you have a 10% chance of winning $10 and a 25% chance of winning $20 then your overall odds of winning something is 10% + 25% = 35%. This only works for mutually exclusive events (events that cannot happen at the same ...

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    Having di erent sample spaces for the same probability experiment is not immediately a problem. The problem comes in with the way many textbooks de ne probability: Na ve De nition of Probability: The probability of event A happening is: P(A) = number of outcomes in event A total number of outcomes in the sample space

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    Welcome to the statistics and probability page at Math-Drills.com where there is a 100% chance of learning something! This page includes Statistics worksheets including collecting and organizing data, measures of central tendency (mean, median, mode and range) and probability.. Students spend their lives collecting, organizing, and analyzing data, so why not teach them a few skills to help ...

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    This introduction to probability and statistics explores probability models, sample spaces, compound events, random samples, and a whole lot more. ... Math; 7th grade; Unit 7: Statistics and probability. 1,100 possible mastery points. Mastered. Proficient. ... (example problem) (Opens a modal) Practice.

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    Consider statistics as a problem-solving process and examine its four components: asking questions, collecting appropriate data, analyzing the data, and interpreting the results. This session investigates the nature of data and its potential sources of variation. Variables, bias, and random sampling are introduced. View Transcript.

  16. Statistics As Problem Solving Part A: A Problem-Solving Process (15

    Part A: A Problem-Solving Process (15 minutes) The word statistics may bring to mind polls and surveys, or facts and figures in a newspaper article. But statistics is more than just a bunch of numbers: Statistics is a problem-solving process that seeks answers to questions through data. By asking and answering statistical questions, we can ...

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  18. Probability

    Example: there are 5 marbles in a bag: 4 are blue, and 1 is red. What is the probability that a blue marble gets picked? Number of ways it can happen: 4 (there are 4 blues). Total number of outcomes: 5 (there are 5 marbles in total). So the probability = 4 5 = 0.8

  19. Statistics

    General. Statistics is the distinct branch of mathematical science that deals with obtaining, analyzing, and drawing conclusions about a data set. "Applied statistics" is a subset of statistics that deals primarily with statistical analysis on information gathered from an experiment. Most data sets from statistics are from samples from a much ...

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  21. Part A: Statistics as a Problem-Solving Process (25 minutes)

    Classroom Case Studies, Grades K-2 Part A: Statistics as a Problem-Solving Process (25 minutes) Part A: Statistics as a Problem-Solving Process (25 minutes) A data investigation should begin with a question about a real-world phenomenon that can be answered by collecting data. After the children have gathered and organized their data, they ...