Math 1030 Chap 7

You toss a coin 200 times and get 198 tails. (a)What is the relative frequency? (b) Determine the expected frequency of the event. (c) Do you have reason to suspect the coin is unfair?

(a) .99 (198/200=.99) (b) 0.5 (how many tails would be expected 1/2) (c) Yes, because the relative frequency is not near the expected frequency.

Determine whether the following statement makes sense (or is clearly true) or does not make sense (or is clearly false). The number of different possible batting orders for 9 players on a 25-person baseball team is so large that there's no hope of trying them all out.

(25P9=25!/(25-9)!=741 Billion.) The statement makes sense because substituting the n=25 and r-9 into the forumula nPr=n!/(n-r)! results in more than 700 billion batting orders.

Properties of Probability

1. For any event, E, 0<P(E)<1. 2. For any sample space, S, P(S)=1 3. For the empty set, 0, P(0)=0

Use the theoretical method to determine the probability of the following event. Sharing a birthday with another person when you both have birthdays in June.

1/30 (June has 30 days so the probability is 1/30)

Evaluate the expression 12!/10!2!=

12!/10!2!= 12x11x10x1/10x9x8x1x2x1, Notice that the numbers in the expansion of 10! in the denominator also occur in the expansion of 12! in the numerator. So all numbers in the expansion of 10! (10.9.7.6.5.4.3.2.1) will cancel with numbers in the expansion of 12! (12x11)/(2x1)=66

Evaluate the expression 14!/10!4!

14! =14.13.12.11.10.9.8.7.6.5.4.3.2.1 _________________________________________________ 10!4! 10.9.8.7.6.5.4.3.2.1.4.3.2.1 The numbers in the expansion of 10 occur in the expansion of 14! in the numerator. So all numbers in the expansion of 10! will cancel with numbers in the expansion of 14! Which leaves a simpler expression: 14! =14.13.12.11 ______________________ =1001 10!14! 4.3.2.1

You are asked to create a 8-character password, and each character may be any of the 26 letters of the alphabet or 9 numerals 1 through 9 numerals 1 through 9. How many different passwords are possible?

35^8 because there are 26+9=35 different characters to choose from, so there are 35^8 different arrangements possible. (nxnx.....xn=n^r)

The probability of choosing four numbers that match four randomly selected balls when the balls are numbered 1 through 36.

36C4=36!/36-4!x4! 36!/32!x4!=58,905 probability 1/58,905

Answer the following question using arrangements with repetition, permutations, or combinations. How many different six-digit passwords can be formed from the numbers zero to six if repetition is not allowed?

5,040 (7x6x5x4x3x2x1=5,040)

Suppose that four 6-sided dice are tossed. How many outcomes are possible?

6x6x6x6, because each of the four dice have six possible outcomes.

What is the probability distribution?

A probability distribution represents the probabilities of all possible events of interest.

A box contains 20 chocolates, but only two of them are the dark chocolate that somebody likes. When the probability of drawing one dark chocolate at random is 0.1 what kind of probability is stated?

A theoretical probability, because both the total and the desired numbers of objects are known.

A. Answer the following question using the appropriate counting technique, which may be either arrangements with repetition, permutations, or combinations. Hos many different three-digit codes can be formed using digits 0 through 9? B. How many possible codes are there?

A. Arrangements with repetitions because there are r selections from a group of n choices can be repeated. B. There are 1000 possible codes. (10^3=1000)

A. Suppose someone gives you 10 to 3 odds that you cannot roll two even numbers with the roll of two fair dice. This means you win $10 if you succeed and you lose $3 if you fail. What is the expected value of this game to you? Should you expect to win or lose the expected value in the first game? What can you expect if you play? Rolling 2 dice potential for 3 possible event results of (2,4,6) with a possibility of 6 potential rolls of 2 dice. B. Should you expect to win(or lose) an amount equal to the exprected value in the first game? C. Should you expect to win (or lose) an amount equal to the expected value in the first game? D. What can you expect if you play 200 times? E. Averaged over 200 games you should expect to win?

A. First from the table, find the number of outcomes of event 1, rolling two even numbers. The number of outcomes is 9. From the table, the total number of outcomes is 36. Now use the theoretical probability formula to find the probability of a success, rolling two even numbers. P(success)=9/36=1/4 or 0.25, hence, P(failure)=1-P(success)=0.75. Substitute the known values into the expected value formula and evaluate. Expected value=($10x0.25)+(-$3x0.75)=$0.25 B. Should you expect to win(or lose) an amount equal to the exprected value in the first game? Win $10 or lose $3. Thus, in one game, you expect to either win $10 or lose $3. C. Should you expect to win (or lose) an amount equal to the expected value in the first game? No the outcome of one game cannot be predicted. D. What can you expect if you play 200 times? According to the law of large numbers, because there are large number of rolls, the expected value on each roll can be expected. Thus, averaged over 200 games, expect to win $0.25 in each game. If you play 200 games, you should expect to win 200x$0.25=$50.00 E. Averaged over 200 games you should expect to win? $50.00

a. What do we mean by combinations? b. What is an example of it's use?

A. The n! represents the number of permutations, which includes different orderings. The (n-r)!xr! represents the number of different orderings of n items taken r at a time. Dividing n! by (n-r)!xr! deptermines the number of permutations without counting each different ordering, which is a combination. B. The formula can be used when deciding how many ways someone can choose two hats from 10 options.

Determine the probability of the given opposite event. What is the proability that a 47% free-throw shooter will miss her next free throw?

Answer is .53 (1.-0.47=0.53) P(A)+P(notA)=1 or P(not A)=P(A)=1-P(A)

What is the probability of the given opposite event. What is the probability of rolling a fair die and not getting an outcome less than 4? (Die has 6 possible outcomes:1,2,3,4,5,6)

Answer: 1/2 P(A)=number of ways A can occur/total number of outcomes (3/6=1/2)

A DJ is preparing a playlist of 18 songs. How many different ways can the DJ arrange the first six songs on the playlist?

Answer: 13,366,080 (18!/18-6!) (18!/12!) 18!=18.17.16.15.14.13.12.11.10.9.8.7.6.5.4.3.2.1 12!=12.11.10.9.8.7.6.5.4.3.2.1 The 12.11.10.9.8.7.6.5.4.3.2.1 all cancelled out. (18x17x16x15x14x13= 13,366,080)

Pizza House offers 2 different salads, 4 different kinds of pizza, and 3 different desserts. How many different 3 course meals can be ordered.

Answer:24 (2*4*3=24)

How many different three-letter "words" can be formed from the alphabet QRS? What counting technique should be used to make this calculation?

Arrangements with repetitions because there are r selections from a group of n choices and choices can be repeated. (Identify the n in this situation = 3 letters, identify the r in this situation =3 letter words) The expression is n^r or 3^3, the number of three-letter workds that can be made is 3^3=27.

Answer the following question using the appropriate counting technique, which may be either arrangements with repetition, permutations, or combinations. How many license plates can be made of the form XXXX-YYY, where X is a letter of the alphabet and Y is a numeral 0-9? What counting technique should be used to make this calculation? (26 letters in the alphabet, 10 numbers)

Arrangements with repetitions because there are r selections from a group of n choices and choices can be repeated. How many license places can be made of the form XXXX-YYY where X is a letter of the alphabet and Y is a numeral 0-9. (nxnx...xn=n^r) Number of arrangements of letters 26^4, Number of arrangements of numbers 10^3. 26^4*10^3=456,976,000 (License plate form:XXXX-YYY)

Answer the following question using the appropriate counting technique, which may be either arrangements with repetition, permutations, or combinations. How many possible birth orders with respect to gender are possible in a family with seven children? (For example BBGGBBB and BGBBBBG are different orders.)

Arrangements with repetitions because there are r selections from a group of n choices and choices can be repeated. There are * possible birth orders for a family with seven children. (Number of outcomes of sex of child=2, the number of selections r=7) n^r=2^7, there are 128 possible birth orders for a family with seven children.

How many different triple cones can you create from 31 flavors? Assume that you specify which flavor goes on the bottom, middle and top. What counting technique will you use to make this calculation?

Arrangements with repetitions because there are r selections from a group of n choices and you can repeat choices. (31^3=29,791)

Using 31 flavors of ice cream, how many different triple cones can you create with 3 different flavors if you don't care about the order of flavors of the cone?

Combinations because the selections come from a single group of items and the order of the arrangement does not matter. Use factorial formula (nCr=n+r-1)!/r!(n-1)! (=4495)

An insurance company knows that the average cost to build a home in a new California subdivision is $97,709 and that in any particular year there is a 1 in 44 chance of a wildfire destroying all the homes in the subdivision. Based on these data and assuming the insurance company wants a positive expected value when it sells policies, what is the minimum the company must charge for fire insurance policies in this subdivision?

Consider two events, each with its own value and probability. The expected value is shown below. Expected value=(value of event 1)x(probability of event 1)+(value of event 2)x(probability of event 2) This formula can be extended to any number of events by including more terms in the sum. (97709/44=2221) The company must charge $2,221 per year because this is the expected value.

Evaluate the expression. 15!/4!(15-4)!

Evaluate the subtraction inside parenthesis first. 15-4=11. 15!/4!11!= 15.14.13.12.11!/4!.11! 15.14.13.12.11!/4!.11!=15.14.13.12/4! Expanded: 15.14.13.12/4!=15.14.13.12/4.3.2.1=32760/24=1365

What is the probability of drawing either a 3 or 5 from a standard deck of cards?

Events A and B are mutually exclusive. 1/13 + 1/13=2/13

a.You are given 3 to 1 odds against tossing three coins, meaning you win $3 if you succeed and you lose $1 you fail. Find the expected value (to you) of the game. Would you expect to win or lose money in 1 game? In 100 games? b.Would you expect to win or lose money in 1 game? c. Would you expect to win or lose money in 100 games?

Expected value =(event 1 value)x(event 1 probability)+(event 2 value)x(event 2 probability) a.(Event 1 probability (1/2x1/2x1/2=1/8) (Event 2 1-1/8=7/8) $3 (amount you win)x1/8(probability of tossing 3 heads)+-$1 (amount you lose)x7/8 (probability of tossing at least 1 tail)=-, /2, Answer: -, 0.50 b. I would expect neither because the outcome of one game cannot be predicted. c. I would expect to lose because the expected value is negative.

You are given 4 to 1 odds against tossing three heads with three coins, meaning you win $4 if you succeed and you lose $1 if you fail. Find the expected value (to you) of the game. Would you expect to win or lose money in 1 game? In 100 games?

Find the expected value (to you) for the game. Expected value=(event 1 value)x(event 1 probability)+(event 2 value)x(event 2 probability) Event 1 should be tossing 3 heads, and event 2 should be tossing at least 1 tail. (Event 1 probability (1/2x1/2x1/2=1/8) Event 1 probability=1/8 Event 2 probability(1-1/8=7/8) Probability of event 2=7/8) $4 (amount you win)x 1/8(probability of tossing 3 heads)+-$1 (amount you lose)x7/8(probability of tossing at least 1 tail)=-3/8 or -, 0.38

Addition Rule for Probability

For two events, E and F, the probability that E or F occurs is given by the following formula. P(E or F)=P(E)+P(F)-P(E and F) IE: Probaility of rolling two dice and getting less than 4 or 10. =3/36+3/36-0/3 chance of both occurring is zero 6=6/36=1/6 +0.1667

If you toss a coin 3 times and get HTH (Head, Tails, Head) and you are interested in the number of heads that appear, what different outcome would correspond to the same event?

HHT, because it has the same number of heads and tails as HTH.

When one coin is tossed, the probability of landing heads is 1/2. Assuming that the coin is fair, what does that statement mean?

If 1000 coins are tossed, there is no way to predict the precise number of heads that will be generated, though it will probably be close to 500 because probability does not guarantee certain results.

A $1 slot machine at a casino is set so that it returns 97% of all the money put into it in the form of winnings, with most of the winning in the form of huge but low-probability jackpots. What is the probability of winning when you put $1 into this slot machine?

It cannot be calculated from the given data, but it is certainly quite low. We would need to know the value of the jackpots to determine the correct probability.

Use the at least once rule to find the probability of the following event. Getting rain at least once in 11 days if the probability of rain on each singel day is 0.3

P(at least one event A in n trials=1-(P(no A in one trial)) P(no A in one trial)=1-0.3=0.7 So, P(no A in one trial)=0.7. P(at least one event A in n trials)=1-(P(no A in one trial))n=1-0.7^11= .980226733 rounded to the nearest thousandth 0.980

Use the at least once rule to find the probability of the following event. Getting rain at least once in 6 days if the probability of rain on each single day is 0.4

P(at least one event A in n trials=1-(P(no A in one trial)) P(no A in one trial)=1-0.4=.6. P(at least one event A in n trials)=1-(P(no A in one trial))n=1-0.6^6=0.953344 round to the nearest thousandth = 0.953

A teacher has 23 students, and 5 of them will be chosen to participate in a play that has 5 distinct characters. Which of the following questions requires calculating permutations to solve?

Once the 5 children have been chosen, how many different ways can their roles be arranged? TO answer this question, the total number of possible arrangements, or permutations, needs to be found. Since the roles are distinct, the order of the children matters.

Randomly meeting either a man or a Republican in a group composed of 30 Democratic men, 50 Republican men, 50 Democratic women, and 30 Republican women.

Overlapping "either/or" probability. P(A)+P(B)-P(A and B) 1.find total of group sample. 2. find the probability of meeting a man. 3. find the probability of meeting a republican. 4.deduct the probability of meeting a republican man. (30+50+50+30=160 total of group) the probability of meeting a man is 1/2 (30+50=80/160=1/2), The probability of meeting a republican (50+30=80/160=1/2) The probability of meeting a Republican man is 5/16 (50/160=5/16) 1/2+1/2-5/16=11/16

What is the probability of randomly meeting a man or a Republican in a group composed of 50 Democratic men, 20 Republican men, 30 Democratic women and 60 Republican women.

Overlapping "either/or" probability. P(A)+P(B)-P(A and B) The probability of meeting a man is 7/16. (50+20+30+60=160 all of group, men 50+20=70, 70/160=7/16) The probability of meeting a Republican is 1/2. (20+60=80, 80/160=1/2) The probability of meeting a Republican man is (20/160= 1/8). So 7/16+1/2-1/8=13/16 So the probability is 13 in 16.

Non overlapping Events formula

P(A)+P(B)

Overlapping Events formula

P(A)+P(B)-P(A and B)

Independent Events - You have a fair 8-sided die labeled 1-8. What is the probability of rolling three 1's in a row with a single fair die?

P(three 1's in a row)=P(1)xP(1)xP(1) or 1/8x1/8x1/8= 1/512 The probability of rolling three 1s in a row is 1 in 512.

Using 31 flavors of icecream for different triple cones how many different triple cones can you create if you specify which flavor goes on the bottom, middle, and top?

Permutations because the selections come from a group of items and the order of the arrangement matters. (31!/31-3!, 31!/28!=26970) cancel out 28-1 multiply 31x30x29=26970

How many different nine-digit passwords can be formed from the numbers zero to eight if repetition is not allowed?

Permutations should be used because no item may be selected more than once and the order matters. (1x(n-1)x(n-2)xx(n-r+1) Calculate how many different passwords are possible=362,880

Answer the following question using arrangements with repetition, permutations, or combinations. How many different six-character passwords can be formed from the uppercase letters A to F if repetitions are not allowed?

Permutations should be used because no item may be selected more than once and the order matters. Calculate how many different passwords are possible = (6x5x4x3x2x1=720) there are a total of 720 different passwords possible.

Answer the following question using arrangements with repetition, permutations or combinations. A city council with nine members must elect a five-person executive committee consisting of a mayor, deputy mayor, secretary, administrator, and treasurer. How many executive committees are possible?

Permutations should be used because no item may be selected more than once and the order matters.(9!/(9-5)!=(9!/4!) (9.8.7.6.5.4.3.2.1/4.3.2.1, the 4.3.2.1 all cancel each other out. Left with 9x8x7x6x5=15,120) Answer:15,120

Make a probability distribution for the given set of events. The sums that appear when two fair four-sided dice (tetrahedrons) with sides 2,3,4 and 5 are tossed.

Sum 4 Probability 1/16 Sum 5 Probability 1/8 Sum 6 Probability 3/16 Sum 7 Probability 1/4 Sum 8 Probability 3/16 Sum 9 Probability 1/8 Sum 10 Probability 1/16

Determine whether the following individual events are overlapping or non-overlapping. Then find the probability of the combined event. Drawing either a 5 or a king from a regular deck of cards. (52 cards in a deck 4 suits with 13 cards including 2,3,4,5,6,7,8,9,10,jack,king,queen,ace)

The Two events are non-overlapping. P(A) + P(B). There are 4 5's per deck and 4 kings and a total of 52 cards per deck of cards. (4/52+4/52=2/13) The probability of the combined event is 2 in 13.

The Complement

The complement of an Event E, denoted E^c, is the set of all outcomes in the sample space that are not in E. IE: choosing a red card out of a standard deck of cards; the complement is not choosing a red card.

An insurance policy sells for $200. Based on past data, and average of 1 in 100 policyholders will file a $20,000 claim, an average of 1 in 250 policyholders will file a $30,000 claim, and an average of 1 in 400 policyholders will file a $60,000 claim. Find the expected value ( to the company) per policy sold. If the company sells 10,000 policies, what is the expected profit or loss?

The expected value is the sum of the probability of each event multiplied by the event value. Event 1: The company sells an insurance policy for $1200 with probability 1. This represents a gain to the company, so it has a positive value. Event 2:The company pays out a claim of $20,000 with probability 1/100. This represents a loss to the company, so it has a negative value. Event 3: The company pays out claim of $30,000 with probability 1/250. This represents a loss to the company, so it has a negative value. Event 4: THe company pays out a claim of $60,000 with probability 1/400. This represents a loss to the company, so it has a negative value. Expected value= (value of event 1)x(probability of event1) + (value of event 2)x(probability of event 2)+(value of event 3)x(probability of event 3)+(value of event 4)x(probability of event 4) =($1200x1)+(-$20,000x1/100)+(-$30,000x1/250)+(-$60,000x1/400) (1200+-200+-120+-150)=$730 The expected value is $730. The expected profit or loss is the number of policies sold multiplied by the expected value. If the company sells 10,000 policies and the expected value is positive, the expected profit is (10,000 x $1200=12,000,000- 4,700,000=$7,300,000

Determine if the following individual events are independent or dependent. Then find the probability of the combined event. Randomly drawing and immediately eating two red pieces of candy in a row from a bag that contains 9 red pieces of candy out of 55 pieces of candy total.

The individual events are dependent because the outcome of drawing a red candy on 1st draw would impact the probability of the second draw. The probability of the combined event is P(A and B)=P(A)xP(B given A) 9/55x8/54=4/165 or .024 in decimal format rounded to the nearest thousandth.

Determine whether the following individual events are independent or dependent. Then find the probability of the combined event. Rolling two 4s followed by one 3 on three tosses on a fair die.

The individual events are independent. The probability of the combined event is 1/6x1/6x1/6=1/216. The probability of rolling two 4s followed by one 3 is 1 in 216.

Determine the following individual events are independent or dependent. Then find the probability of the combined event. Drawing either a black five or a black two on one draw from a regular deck of cards. (52 cards in standard deck half are black half are red.)

The individual events are non-overlapping. The probability of the combined event is P(A)=P(A)=probability of the combined event. (1/26+1/26=1/13) The probability is 1 in 13.

An ice cream shop offers 31 different flavors of ice cream and 12 different toppings. What counting technique should you use (multiplication principle, arrangements with repetitions, permutations, or combinations)?

The multiplication principle because that are M possible outcomes for one group and N outcomes for the other group.(31x12=372)

A soccer coach who has 15 children on her team will be playing 7 children at a time, each at a distinct position. Which of the following numbers is largest?

The number of permutations of the 7 positions that are possible with the 15 children. This number is 15!/(15-7)!, which is larger than the other two numbers.

Two fair dice are rolled. Find the odds for and the odds against getting a double 1.

The odds for getting a double 1 are 1 to 35. (1/6*1/6 -1=1/35) The odds against the event are 35 to 1. (reciprocal)

You purchase 25 lottery tickets for which the probability of winning some prize on a single ticket is 1 in 100. What is your probability of having at least one winner among your 25​ tickets?

The probability is 1-(0.99)^25 according to the "at least once" rule for independent events. ( 1/100 or 1-.01=.99)^25 tickets.

You roll two dice twice. Based on the probabilities shown in the table, what is the probability you'll get a sum of 3 on the first roll and a sum of 4 on the second roll? Event 2 3 4 5 6 7 8 9 Probabilty 1/36 2/36 3/36 4/36 5/36 6/36 5/36 4/36 Event 10 11 12 total Probability 3/36 2/36 1/36 1

The probability is 2/36x3/36 because the two events are independent.

What is the probability of drawing three fours in a row from a standard deck of cards? (4 suits with 13 cards per suit including 2,3,4,5,6,7,8,9,10,jack,king,queen,ace)

The probability is dependent on events A&B. P(A and B)=P(A)xP(B given A) first draw is 1/13 (there are 3 fours and 51 cards left for the second draw, so 3/51==1/17 ) if 4 is drawn is 1/17, the third draw if another 4 is drawn is 1/25 (2/50=1/25). Multiply all single probabilities: 1/13x1/17x1/25=1/5525

What is the probability of drawing three sevens in a row from a standard deck of cards when the drawn card is not returned to the deck each time? The probability of drawing three sevens is?

The probability is dependent on events A&B. P(A and B)=P(A)xP(B given A) if first draw is 1/13 (1/13 cards then the next draw is 3/51=1/17, the third draw will be 2/50=1/25) (1/13x1/17x1/25=1/5525)

Find the probability of being dealt 5 cards from a standard 52-card deck, and the cards are 9,10,jack,queen, and king, all of the same suit. (52 cards 4 suits, 13 cards per suit.)

The probability of being dealt this hand is * (N= number of cards 52 R=number of cards drawn at a time 5) 52!/52-5!x5! (52C5) 52x51x50x49x48/5x4x3x2x1=311875200/120= Answer possible number of hands that can be dealt from standard deck:2598960 How many ways is it possible to be dealt a 9,10,jack,queen,and king all of the same suit, where order does not matter? 4 ways; as there's 4 different suits of cards. Divide the number of ways the hand can be chosen by the total number of 5 card hands. 4/2598960=1/649740 There is a 1 in 649,740 chance of being dealth a 9,10,jack,queen, and king, all of the same suit.

The probability of rolling two dice and getting a​ double-1 (making a sum of​ 2, or​ "snake eyes") is 1 in 36. Suppose you rolled two dice twice.

The probability of getting a double-1 on at least one of the two rolls is 2/36. This was the probability that Chevalier used, and he won a lot of bets.

In a family with 7 children excluding multiple births, what is the probability of having 7 girls? Assume that a girl is as likely as a boy at each birth.

The probability of having 7 girls is 1/128. (S=2^7=128 E= 1 (how many ways can having 7 girls out 7 children occur =1) P(E)=number of elements in E/number of elements in S=n(E)/n(S)

Determine the probability of selecting a two-child family with two girls, assuming boys and girls are equally likely.

The probability of selecting a two-child family with two girls is 1/4. (BB, GG, BG, GB)

One person in a stadium filled with 100,000 people is chosen at random to win a free pair of airline tickets. What is the probability that it will not be you?

The probability that it will not be you is 0.99999, because the probability that you win is 0.00001, and the probability that you do not win is the opposite of that. (1/100,000=0.00001, 1-0.00001=0.99999)

The local weather forecast has been accurate for 19 of the past 31 days. Based on this fact, what is the relative frequency probability that the forecast for tomorrow will be accurate?

The relative frequency probability that the forecast for tomorrow will be accurate is .613 (19/31=.61290323 rounded to nearest thousandth .613)

Suppose that the probability of a hurricane striking a state in any single year is 1 in 10 and that this probability has been the same for the past 1000 years. Which of the following is implied by the law of large numbers?

The state has been hit by clost to (but not necessarily exactly) 100 hurricanes in the past 1000 years because as the years increase, the closer the proportion of hurricanes should be to 0.1.

Decide whether the following statement makes sense (or is clearly true) or does not make sense (or is clearly false). If you toss a coin four times, it's much more likely to land in the order HTHT than HHHH (H stands for heads and T for tails.)

The statement does not make sense because each outcome is equally likely since the probability of any single particular outcome is 1/2, so each set of outcomes have the same probability of (1/2)^4=1/16.

Decide whether the following statement makes sense (or is clearly true) or does not make sense (or is clearly false). I haven't won in my last 25 pulls on the slot machine, so I must be having a bad day and I'm sure to lose if I play again.

The statement does not make sense because the results of repeated trials do not depend on results of earlier trials.

The probability that Jonas will win the race is 0.6 and the probability that he will not win is 0.5

The statement does not make sense because the sum of the proabilitites of Jonas winning and not winning the race must equal to 1.

Explain if true or false. The probability of getting heads and tails when you toss a coin is 0, but the probability of getting heads or tails is 1.

The statement makes sense because heads and tails are the only possible outcomes and its impossible to get both heads and tails on a single toss.

Complement Rule of Probability

The sum of the probabilities of an event, E, and its complement, E^c, is equal to one. P(E)+P(E^c)=1 IE: The complement to the outcome of failing 35% probability your upcoming test is passing it. THus the probability is calculated as follows: P(passing)=1-P (failing) =1-.35 =0.65=65%

A group of friends are playing 5-card poker with a deck of 52 cards. For a probability distribution showing the individual probabilities of all possible hands, what would be the sum of all the individual probabilities?

The sum would equal 1 because the sum of the probabilities of all possible events in any situation is 1.

Explain why the probability is the same for any particular set of ten coin toss outcomes. How does this idea affect our thinking about streaks?

The total number of outcomes for ten coins is 2x2x2x2x2x2x2x2x2x2=1024, so every individual outcome has the same probability of 1/1024. A streak of all heads would not seem surprising since a streak of all heads is just as likely as a streak of all tails, or as likely as any other combination of outcomes.

Determine whether the following individual events are overlapping or non-overlapping. Then find the probability of the combined event. Drawing either a heart or a club from a regular deck of cards. (52 cards in a deck 4 suits with 13 cards including 2,3,4,5,6,7,8,9,10,jack,king,queen,ace)

The two events are non-overlapping. P(A)+P(B) (13/52+13/52=1/2) the probability is 1 in 2.

Which of the following correctly describes the events of being born on a Wednesday and being born in July?

The two events are overlapping. It is possible to be born on a Wednesday in July.

How many different 4-person teams (order does not matter) can be put together from a group of 23 people?

There are 23x22x21x20/4x3x2x1 different possible teams, because all players come from one group, nobody is selected twice, and order does not matter.

Explain if true or false. My chance of getting a 5 on the roll of one die is 1/6, so my change of getting at least one 5 when I roll three dice is 3/6.

This does not make sense because the real probability would be 1-(5/6)^3, which is not equal to 3/6.

Randomly selecting a three child family with either one or two girl children.

This is a non overlapping event. P(A or B)=P(A)+P(B) What is the probability that a three child family has one girl 3/8. (BBB,GGG ,BGB,GBG,BBG,GGB, BGG, GBB) What is the probability that a three child family has two girls 3/8. (3/8+3/8=3/4). The probability of randomly selecting a three-child family with either one or two girl children is 3/4.

Cameron is betting on a game in which the probability of winning is 1 in 5. He's lost five games in a row, so he decides to double his bet on the sixth game. What does this strategy show?

This shows poor logic, as he has a 80% chance of losing the double bet. The past bad luck has no bearing on the future chances. His chances of winning never change. (1/5=20% chance of winning so 80% chance of losing.)

Explain how to make a table of probability distribution.

To make a table of probability distribution, list all possible outcomes, identify the outcomes that represent the same event, and then find the probability of each event.

Consider a lottery with 100 million tickets in which each ticket has a unique number. Each ticket is sold for $1, and one ticket is drawn for a single prize of $75 million (and no other prizes). If you were to spend $1 million to purchase 1 million lottery tickets, what would the most likely result be.

You would lose the entire $1 million. No matter how many tickets are bought, the chance of winning the lottery is still 1 in 100 million.

a. Decide with method (theoretical, relative frequency, or subjective) is appropriate, and compute or estimate the following probability. A card is drawn at random out of well-shuffled deck of 52 cards. b. Find the probability of drawing a two or three.

a. The Theoretical method would be appropriate as the number of cards are known so the probability could be assessed. (52 cards in a deck 36 numbered cards) b. 2/13 (8/52=2/13 Sample =13 Event=2) P(A)=number of ways A can occur/total number of possible outcomes

Find the odds for and the odds against the event rolling a fair die and getting a 2 or a 6.

a. The odds for the event are 1 to 2. (1/6 divided by 2/6=1/2 or 1 to 2) Odds for event A=P(A)/P(notA) b. The odds against the event are 2 to 1 (reciprocal)

Decide which method (theoretical, relative frequency, or subjective) is appropriate, and compute or estimate the following probability. Randomly meeting someone born between 1:00pm and 5:00pm.

a. Theoretical b. 1/6 P(A)=number of ways A can occur/total number of outcomes (S=1+5=6, E=1)

a. Give an example in which we would be interested in an​ either/or probability. b. How do we determine whether the events are overlapping or​ non-overlapping? c.Which event is overlapping? d. Which event is non-overlapping?

a. We roll a die, hoping for a 2 or a 5. We want to know whether a person selected at random is Democrat or a man. b. Events are overlapping if they can occur together. Events are non-overlapping if they cannot occur together. c. We want to know where a person selected at random is a Democrat or a man. d. We roll a die, hoping for a 2 or a 5.

Each of the following states a probability for two events. In which case are the events​ dependent? a. the probability that an Olympic diver will score 8 or above on two consecutive dives. The first dive will affect the score of the second dive. b. the probability of choosing two pink roses from a vase with 25 roses. After the first rose is picked, there will be one less pink rose and one less out of the total in the vase.

b. the probability of choosing two pink roses from a vase with 25 roses. After the first rose is picked, there will be one less pink rose and one less out of the total in the vase.