Stats Exam 2 Review
Among respondents asked which is their favorite seat on a plane, 496 chose the window seat, 11 chose the middle seat, and 305 chose the aisle seat. What is the probability that a passenger prefers the middle seat? Is it unlikely for a passenger to prefer the middle seat? If so, why is the middle seat so unpopular? The probability that a passenger prefers the middle seat is . 014. (Round to three decimal places as needed.) Is it unlikely for a passenger to prefer the middle seat? A. Yes, because the probability that a passenger prefers the middle seat is greater than 0.05. B. No, because the probability that a passenger prefers the middle seat is greater than 0.05. C. No, because the probability that a passenger prefers the middle seat is less than 0.05. D. Yes, because the probability that a passenger prefers the middle seat is less than 0.05. Your answer is correct. If so, why is the middle seat so unpopular? A. The middle seat offers more opportunities for conversation with other passengers. B. The middle seat lacks an outside view, easy access to the aisle, and a passenger in the middle seat has passengers on both sides instead of on one side only. Your answer is correct.C. The middle seat is more comfortable than window or aisle seats. D. Tickets for the middle seat are cheaper. E. It is not true that the middle seat is unpopular.
.014 (add all values, divide by middle seat value) D B
Use the following results from a test for marijuana use, which is provided by a certain drug testing company. Among 142 subjects with positive test results, there are 23 false positive results; among 152 negative results, there are 5 false negative results. If one of the test subjects is randomly selected, find the probability of a false positive or false negative. (Hint: Construct a table.) What does the result suggest about the test's accuracy? The probability of a false positive or false negative is . 095. (Round to three decimal places as needed.) What does the result suggest about the test's accuracy? A. With an error rate of 0.905 (or 90.5%), the test appears to be highly accurate. B. With an error rate of 0.095 (or 9.5%), the test appears to be highly accurate. C. With an error rate of 0.095 (or 9.5%), the test does not appear to be highly accurate. Your answer is correct.D. With an error rate of 0.905 (or 90.5%), the test does not appear to be highly accurate.
.095 C
The data in the table below summarize results from 173 pedestrian deaths that were caused by accidents. If two different deaths are randomly selected without replacement, find the probability that they both involved intoxicated drivers. Is such an event unlikely? Pedestrian Intoxicated? Yes No Driver Yes 32 24 Intoxicated? No 98 19 The probability is . 105. (Round to three decimal places as needed.) Is such an event unlikely? A. Yes, because its probability is less than 0.05. B. No, because its probability is less than 0.05. C. Yes, because its probability is greater than 0.05. D. No, because its probability is greater than 0.05.
.105 (total drunk drivers/all deaths) D
In a region, there is a 0.8 probability chance that a randomly selected person of the population has brown eyes. Assume 10 people are randomly selected. Complete parts (a) through (d) below. a. Find the probability that all of the selected people have brown eyes. The probability that all of the 10 selected people have brown eyes is . 107. (Round to three decimal places as needed.) b. Find the probability that exactly 9 of the selected people have brown eyes. The probability that exactly 9 of the selected people have brown eyes is . 268. (Round to three decimal places as needed.) c. Find the probability that the number of selected people that have brown eyes is 8 or more. The probability that the number of selected people that have brown eyes is 8 or more is . 677. (Round to three decimal places as needed.) d. If 10 people are randomly selected, is it unusual for 8 or more to have brown eyes? A. Yes, because the probability that 8 or more of the selected people have brown eyes is less than 0.05. B. No, because the probability that 8 or more of the selected people have brown eyes is greater than 0.05. Your answer is correct.C. Yes, because the probability that 8 or more of the selected people have brown eyes is greater than 0.05. D. No, because the probability that 8 or more of the selected people have brown eyes is less than 0.05.
.107 .268 .677 B
Refer to the table below. Given that 2 of the 135 subjects are randomly selected, complete parts (a) and (b). Group O A B AB Type Rh Superscript plus 50 38 10 12 Rh Superscript minus 11 8 4 2 a. Assume that the selections are made with replacement. What is the probability that the 2 selected subjects are both group Upper O and type Rh Superscript plus? 0.1372 (Round to four decimal places as needed.) b. Assume the selections are made without replacement. What is the probability that the 2 selected subjects are both group Upper O and type Rh Superscript plus? 0.1354 (
.1372 (These are the correct answer, on review you got it wrong) .1354
The probability of a randomly selected car crashing during a year in a certain country is 0.0483. If a family has three cars, find the probability that at least one of them has a car crash during a year. Is there any reason why the probability might be wrong? The probability that at least one of them has a crash during the year is . 138. (Round to three decimal places as needed.) Is there a reason why the probability might be wrong? A. No, one outcome does not have an effect on later trials. B. Yes, the three cars are not randomly selected. Your answer is correct.C. Yes, one outcome has an effect on later trials. D. No, the three cars are representative of all cars in the country.
.138 (see ch 4) B
Multiple-choice questions each have four possible answers left parenthesis a comma b comma c comma d right parenthesis, one of which is correct. Assume that you guess the answers to three such questions. a. Use the multiplication rule to find P(WCW), where C denotes a correct answer and W denotes a wrong answer. P(WCW)equals . 140625 (Type an exact answer.) b. Beginning with WCW, make a complete list of the different possible arrangements of one correct answer and two wrong answers, then find the probability for each entry in the list. P(WCW)minussee above P(CWW)equals . 140625 P(WWC)equals . 140625 (Type exact answers.) c. Based on the preceding results, what is the probability of getting exactly one correct answer when three guesses are made? . 421875
.140625 .140625 .140625 .421875
Assume that a procedure yields a binomial distribution with a trial repeated n times. Use the binomial probability formula to find the probability of x successes given the probability p of success on a single trial. nequals17, xequals13, pequals0.65 P(13)equals . 143
.143
The table below summarizes results from a study of people who refused to answer survey questions. If one of the subjects is randomly selected, what is the probability that the selected person refused to answer? Does the probability value suggest that refusals are a problem for pollsters? Age 18-21 22-29 30-39 40-49 50-59 60 and over Responded 78 260 250 141 143 207 Refused 14 23 36 29 38 60 The probability that a randomly selected person refused to answer is . 156. (Do not round until the final answer. Then round to three decimal places as needed.) Does the probability value suggest that refusals are a problem for pollsters? A. No, the refusal rate is below 10%. The sample will likely be representative of the population. B. No, the refusal rate is below 10%. This ensures that a suitable number of people were sampled. C. Yes, the refusal rate is above 10%. This results in a sample size that is too small to be useful. D. Yes, the refusal rate is above 10%. This may suggest that the sample may not be representative of the population.
.156 (all refused/everything) D
The accompanying table describes results from eight offspring peas. The random variable x represents the number of offspring peas with green pods. Complete parts (a) through (d). LOADING... Click the icon to view the data. a. Find the probability of getting exactly 7 peas with green pods. . 282 (Type an integer or a decimal.) b. Find the probability of getting 7 or more peas with green pods. . 394 (Type an integer or a decimal.) c. Which probability is relevant for determining whether 7 is an unusually high number of peas with green pods, the result from part (a) or part (b)? The result from part (a) The result from part (b) Your answer is correct. d. Is 7 an unusually high number of peas with green pods? Why or why not? Use 0.05 as the threshold for an unusual event. A. No, since the appropriate probability is greater than 0.05, it is not an unusually high number. Your answer is correct.B. Yes, since the appropriate probability is greater than 0.05, it is an unusually high number. C. No, since the appropriate probability is less than 0.05, it is not an unusually high number. D. Yes, since the appropriate probability is less than 0.05, it is an unusually high number.
.282 (found on the chart) .394 (add 7 and 8 peas together) Result from part b A
Refer to the sample data for polygraph tests shown below. If one of the test subjects is randomly selected, what is the probability that the subject is not lying? Is the result close to the probability of 0.427 for a negative test result? Did the Subject Actually Lie? No (Did Not Lie) Yes (Lied) Positive test results 14 45 Negative test results 34 10 The probability that a randomly selected polygraph test subject was not lying is . 466. (Type an integer or decimal rounded to three decimal places as needed.) Is the result close to the probability, rounded to three decimal places, of 0.427 for a negative test result? A. No, because there is less than a 0.050 absolute difference between the probability of a true response and the probability of a negative test result. B. No, because there is more than a 0.050 absolute difference between the probability of a true response and the probability of a negative test result. C. Yes, because there is less than a 0.050 absolute difference between the probability of a true response and the probability of a negative test result. Your answer is correct.D. Yes, because there is more than a 0.050 absolute difference between the probability of a true response and the probability of a negative test result.
.466 (Total did not lie/all results) C
With one method of a procedure called acceptance sampling, a sample of items is randomly selected without replacement and the entire batch is accepted if every item in the sample is okay. A company has just manufactured 1063 CDs, and 236 are defective. If 3 of these CDs are randomly selected for testing, what is the probability that the entire batch will be accepted? Does this outcome suggest that the entire batch consists of good CDs? Why or why not? If 3 of these CDs are randomly selected for testing, what is the probability that the entire batch will be accepted? The probability that the whole batch is accepted is . 471. (Round to three decimal places as needed.) Does the result in (a) suggest that the entire batch consists of good CDs? Why or why not? A. No, because the sample will always consist of good CDs. B. Yes, because if all three CDs in the sample are good then the entire batch must be good. C. No, because only a probability of 1 would indicate the entire batch consists of good CDs. Your answer is correct.D. Yes, because it is not unlikely that the batch will be accepted.
.471 C
The accompanying table displays results from experiments with polygraph instruments. a. Find Upper P left parenthesis subject told the truth | negative test result right parenthesis . b. Find Upper P left parenthesis negative test result | subject told the truth right parenthesis. c. Compare the results from parts a. and b. Are they equal? LOADING... Click the icon to view the data table. a. Upper P left parenthesis subject told the truth | negative test result right parenthesis equals . 762 (Round to three decimal places as needed.) b. Find the probability of selecting a subject with a negative test result, given that the subject told the truth. Upper P left parenthesis negative test result | subject told the truth right parenthesisequals . 627 (Round to three decimal places as needed.) c. Compare the two values. Are they equal? Yes No
.762 (told the truth/total neg test) .627 (neg test/total truth) No
Let the random variable x represent the number of girls in a family with three children. Assume the probability of a child being a girl is 0.34. The table on the right describes the probability of having x number of girls. Determine whether the table describes a probability distribution. If it does, find the mean and standard deviation. Is it unusual for a family of three children to consist of three girls? x P(x) 0 0.287 1 0.444 2 0.229 3 0.040 Find the mean of the random variable. Select the correct choice below and, if necessary, fill in the answer box to complete your choice. A. muequals 1.02 (Round to two decimal places as needed.) Your answer is correct.B. The table is not a probability distribution. Find the standard deviation of the random variable. Select the correct choice below and, if necessary, fill in the answer box to complete your choice. A. sigmaequals . 82 (Round to two decimal places as needed.) Your answer is correct.B. The table is not a probability distribution. Is it unusual for a family with three children to have only girls? A. No, because the probability of having 3 girls is less than or equal to 0.05. B. Yes, because the probability of having 3 girls is less than or equal to 0.05. Your answer is correct.C. No, because the probability of having 3 girls is greater than 0.05. D. Yes, because the probability of having 3 girls is greater than 0.05.
1.02 find w calc .82 B
A presidential candidate plans to begin her campaign by visiting the capitals in 3 of 50 states. What is the probability that she selects the route of three specific capitals? Is it practical to list all of the different possible routes in order to select the one that is best? P(she selects the route of three specific capitals)equals StartFraction 1 Over 117600 EndFraction (Type an integer or a simplified fraction.) Is it practical to list all of the different possible routes in order to select the one that is best? A. No, it is not practical to list all of the different possible routes because the number of possible permutations is very small. B. No, it is not practical to list all of the different possible routes because the number of possible permutations is very large. Your answer is correct.C. Yes, it is practical to list all of the different possible routes because the number of possible permutations is very large. D. Yes, it is practical to list all of the different possible routes because the number of possible permutations is very small.
1/117600 (50P3) B`
Assume that the given procedure yields a binomial distribution with n trials and the probability of success for one trial is p. Use the given values of n and p to find the mean mu and standard deviation sigma. Also, use the range rule of thumb to find the minimum usual value mu minus 2 sigma and the maximum usual value mu plus 2 sigma. In an analysis of preliminary test results from a gender-selection method, 36 babies are born and it is assumed that 50% of babies are girls, so nequals36 and pequals0.5. The value of the mean is muequals 18. (Do not round.) The value of the standard deviation is sigmaequals 3. (Round to one decimal place as needed.) The minimum usual value is mu minus 2 sigmaequals 12. (Round to one decimal place as needed.) The maximum usual value is mu plus 2 sigmaequals 24.
18 (np) 3 (sq rt npq) 12 24 (find z scores
A fan of country music plans to make a custom CD with 11 of her 28 favorite songs. How many different combinations of 11 songs are possible? Is it practical to make a different CD for each possible combination? How many different combinations of 11 songs are possible? 21474180 Is it practical to make a different CD for each possible combination? A. Yes, it is practical to make a different CD for each possible combination because the number of possible combinations is very small. B. No, it is not practical to make a different CD for each possible combination because the number of possible combinations is very large. Your answer is correct.C. No, it is not practical to make a different CD for each possible combination because the number of possible combinations is very small. D. Yes, it is practical to make a different CD for each possible combination because the number of possible combinations is very large.
21474180 (28C11) B
To the right are the outcomes that are possible when a couple has three children. Refer to that list, and find the probability of each event. a. Among three children, there are exactly 2 girls. b. Among three children, there are exactly 0 boys. c. Among three children, there is exactly 1 boy. 1st 2nd 3rd boy minus boy minus boy boy minus boy minus girl boy minus girl minus boy boy minus girl minus girl girl minus boy minus boy girl minus boy minus girl girl minus girl minus boy girl minus girl minus girl a. What is the probability of exactly 2 girls out of three children? three eighths (Type an integer or a simplified fraction.) b. What is the probability of exactly 0 boys out of three children? one eighth (Type an integer or a simplified fraction.) c. What is the probability of exactly 1 boy out of three children? three eighths
3/8 1/8 3/8
A corporation must appoint a president, chief executive officer (CEO), chief operating officer (COO), and chief financial officer (CFO). It must also appoint a planning committee with three different members. There are 16 qualified candidates, and officers can also serve on the committee. Complete parts (a) through (c) below. a. How many different ways can the officers be appointed? There are 43680 different ways to appoint the officers. b. How many different ways can the committee be appointed? There are 560 different ways to appoint the committee. c. What is the probability of randomly selecting the committee members and getting the three youngest of the qualified candidates? P(getting the three youngest of the qualified candidates)equals StartFraction 1 Over 560 EndFraction
43680 (16C4) 560 (16C3) 1/560 (16C3)
In a past election, the voter turnout was 57%. In a survey, 1025 subjects were asked if they voted in the election. a. Find the mean and standard deviation for the numbers of voters in groups of 1025. b. In the survey of 1025 people, 629 said that they voted in the election. Is this result consistent with the turnout, or is this result unlikely to occur with a turnout of 57%? Why or why not? c. Based on these results, does it appear that accurate voting results can be obtained by asking voters how they acted? a. muequals 584.3 (Round to one decimal place as needed.) sigmaequals 15.9 (Round to one decimal place as needed.) b. Is the result of 629 voting in the election usual or unusual? A. This result is unusual because 629 is within the range of usual values. B. This result is usual because 629 is within the range of usual values. C. This result is unusual because 629 is below the minimum usual value. D. This result is unusual because 629 is greater than the maximum usual value. Your answer is correct. c. Does it appear that accurate voting results can be obtained by asking voters how they acted? A. Yes comma because the results indicate that 57 % is a possible turnout. nothing B. No comma because it appears that substantially more people say that they voted than nbsp the proportion of people who actually did vote. Your answer is correct.C. No comma because it appears that substantially fewer people say that they voted than nbsp the proportion of people who actually did vote.
584.3 15.9 D B
When testing for current in a cable with six color-coded wires, the author used a meter to test five wires at a time. How many different tests are required for every possible pairing of five wires? The number of tests required is 6.
6 (nCr)
Evaluate the expression. 6! A. 120 B. 720 Your answer is correct.C. 714 D. 726
720
In a clinical trial of a drug used to help subjects stop smoking, 827 subjects were treated with 1 mg doses of the drug. That group consisted of 48 subjects who experienced nausea. The probability of nausea for subjects not receiving the treatment was 0.0102. Complete parts (a) through (c). a. Assuming that the drug has no effect, so that the probability of nausea was 0.0102, find the mean and standard deviation for the numbers of people in groups of 827 that can be expected to experience nausea. The mean is 8.4 people. (Round to one decimal place as needed.) The standard deviation is 2.9 people. (Round to one decimal place as needed.) b. Based on the result from part (a), is it unusual to find that among 827 people, there are 48 who experience nausea? Why or why not? A. It is not unusual because 48 is outside the range of usual values. B. It is unusual because 48 is within the range of usual values. C. It is not unusual because 48 is within the range of usual values. D. It is unusual because 48 is outside the range of usual values. Your answer is correct. c. Based on the preceding results, does nausea appear to be an adverse reaction that should be of concern to those who use the drug? A. The drug does appear to be the cause of some nausea. Since the nausea rate is quite high (about 6%), it appears to be an adverse reaction that occurs very often. B. The drug does not appear to be the cause of any nausea. C. The drug does appear to be the cause of some nausea. Since the nausea rate is still quite low (about 6%), it appears to be an adverse reaction that does not occur very often. Your answer is correct. Question
8.4 (np) 2.9 (sq rt npq) D C
Find the probability of at least 2 girls in 8 births. Assume that male and female births are equally likely and that the births are independent events. Round to three decimal places. A. 0.965 Your answer is correct.B. 0.856 C. 0.035 D. 0.109
A
Use the given values of n and p to find the minimum usual value muminus2sigma and the maximum usual value muplus2sigma. Round to the nearest hundredth unless otherwise noted. nequals1058; pequals0.84 A. Minimum: 864.87; maximum: 912.57 Your answer is correct.B. Minimum: 876.80; maximum: 900.64 C. Minimum: 912.57; maximum: 864.87 D. Minimum: 871.86; maximum: 905.58
A (found mean and st dev then found z scores)
Determine whether the following probability experiment represents a binomial experiment and explain the reason for your answer. Four cards are selected from a standard 52-card deck without replacement. The number of aces selected is recorded. Does the probability experiment represent a binomial experiment? A. No, because there are more than two mutually exclusive outcomes for each trial. B. No, because the trials of the experiment are not independent and the probability of success differs from trial to trial. Your answer is correct.C. No, because the experiment is not performed a fixed number of times. D. Yes, because the experiment satisfies all the criteria for a binomial experiment.
B
Evaluate the expression. 12 Upper C 3 A. 4 B. 220 Your answer is correct.C. 362 comma 880 D. 440
B
In a computer instant messaging survey, respondents were asked to choose the most fun way to flirt, and it found that P(D)equals0.720, where D is directly in person. If someone is randomly selected, what does Upper P left parenthesis Upper D overbar right parenthesis represent, and what is its value? What does Upper P left parenthesis Upper D overbar right parenthesis represent? A. Upper P left parenthesis Upper D overbar right parenthesis is the probability of randomly selecting someone who did not participate in the survey. B. Upper P left parenthesis Upper D overbar right parenthesis is the probability of randomly selecting someone who does not choose a direct in-person encounter as the most fun way to flirt. Your answer is correct.C. Upper P left parenthesis Upper D overbar right parenthesis is the probability of randomly selecting someone who chooses a direct in-person encounter as the most fun way to flirt. D. Upper P left parenthesis Upper D overbar right parenthesis is the probability of randomly selecting someone who did not have a preference in regards to the most fun way to flirt. Upper P left parenthesis Upper D overbar right parenthesisequals . 280
B .280 (subtract .720 from 1)
Determine whether the following value is a continuous random variable, discrete random variable, or not a random variable. a. The square footage of a pool b. The hair color of adults in the United States c. The number of hits to a website in a week d. The time it takes to drive from City Upper A to City Upper B e. The amount of rain in City Upper A during July f. The number of statistics students now doing their homework a. Is the square footage of a pool a discrete random variable, continuous random variable, or not a random variable? A. It is a discrete random variable. B. It is a continuous random variable. Your answer is correct.C. It is not a random variable. b. Is the hair color of adults in the United States a discrete random variable, continuous random variable, or not a random variable? A. It is a discrete random variable. B. It is a continuous random variable. C. It is not a random variable. Your answer is correct. c. Is the number of hits to a website in a week a discrete random variable, continuous random variable, or not a random variable? A. It is a discrete random variable. Your answer is correct.B. It is a continuous random variable. C. It is not a random variable. d. Is the time it takes to drive from City Upper A to City Upper B a discrete random variable, continuous random variable, or not a random variable? A. It is a discrete random variable. B. It is a continuous random variable. Your answer is correct.C. It is not a random variable. e. Is the amount of rain in City Upper A during July a discrete random variable, continuous random variable, or not a random variable? A. It is a continuous random variable. Your answer is correct.B. It is a discrete random variable. C. It is not a random variable. f. Is the number of statistics students now doing their homework a discrete random variable, continuous random variable, or not a random variable? A. It is a discrete random variable. Your answer is correct.B. It is a continuous random variable. C. It is not a random variable.
B C A B A A
Evaluate the expression. 6 Upper P 4 A. 2 B. 30 C. 360 Your answer is correct.D. 24
C
Based on data from a car bumper sticker study, when a car is randomly selected, the number of bumper stickers and the corresponding probabilities are as shown below. Complete parts (a) through (d). LOADING... Click the icon to view the data table. a. Does the given information describe a probability distribution? No Yes Your answer is correct. b. Assuming that a probability distribution is described, find its mean and standard deviation. The mean is . 5. (Round to the nearest tenth as needed.) The standard deviation is 1.4. (Round to the nearest tenth as needed.) c. Use the range rule of thumb to identify the range of values for usual numbers of bumper stickers. The maximum usual value is 3.3. (Round to the nearest tenth as needed.) The minimum usual value is 0. (Round to the nearest tenth as needed.) d. Is it unusual for a car to have more than one bumper sticker? Explain. A. Yes, because the probabilities for random variable x from 2 to 9 are all less than 0.05. B. No, because the probability of having 1 bumper sticker is 0.076, which is greater than 0.05. Your answer is not correct.C. No, because the probability of more than 1 bumper sticker is 0.121, which is greater than 0.05.
Some wrong, correct answers Yes .5 1.4 (answered w 1.5, watch rounding) 3.3 (answered 3.5, ^) 0 C
The accompanying data table describes results from groups of 10 births from 10 different sets of parents. The random variable x represents the number of girls among 10 children. Complete the questions below. LOADING... Click the icon to view the data table. Use the range rule of thumb to identify a range of values containing the usual numbers of girls in 10 births. The maximum usual value is 8.5. (Round to one decimal place as needed.) The minimum usual value is 1.7. (Round to one decimal place as needed.) Based on the result, is 1 girl in 10 births an unusually low number of girls? Explain. A. No, 1 girl is not an unusually low number of girls, because 1 girl is less than the minimum usual value. B. Yes, 1 girl is an unusually low number of girls, because 1 girl is outside of the range of usual values. This is the correct answer.C. Yes, 1 girl is an unusually low number of girls, because 1 girl is less than the maximum usual value. D. Not enough information is given.
Wrong, Correct answers: 8.5 1.7 B
The brand name of a certain chain of coffee shops has a 56% recognition rate in the town of Coffleton. An executive from the company wants to verify the recognition rate as the company is interested in opening a coffee shop in the town. He selects a random sample of 10 Coffleton residents. Find the probability that the number that recognize the brand name is not 4. Round to three decimal places. A. 0.001 B. 0.067 C. 0.850 This is the correct answer.D. 0.150
Wrong, correct answer: .850 (dont use binomial prob)
Identical twins come from a single egg that split into two embryos, and fraternal twins are from separate fertilized eggs. Also, identical twins must be of the same sex and the sexes are equally likely, and sexes of fraternal twins are equally likely. Use the data table to complete parts (a) and (b) below. Sexes of Twins boy/boy boy/girl girl/boy girl/girl Identical Twins 7 0 0 7 Fraternal Twins 3 3 3 3 a. After having a sonogram, a pregnant woman learns that she will have twins. What is the probability that she will have identical twins? The probability is seven thirteenths . (Type an integer or a simplified fraction.) b. After studying the sonogram more closely, the physician tells the pregnant woman that she will give birth to twin boys. What is the probability that she will have identical twins? That is, find the probability of identical twins given that the twins consist of two boys. The probability is seven tenths .
Wrong, correct answers 7/13 7/10 REMEMBER TO SIMPLIFY
Winning the jackpot in a particular lottery requires that you select the correct four numbers between 1 and 54 and, in a separate drawing, you must also select the correct single number between 1 and 49. Find the probability of winning the jackpot. The probability of winning the jackpot is StartFraction 1 Over 15,496,299 EndFraction .
wrong, correct answer: 1/15496299 54C4 x 49C1