Arithmetic - GMAT Quantitative
Card 0 of 2420
A Veggie Platter at the restaurant where Georgia likes to eat comprises four different vegetables. How many possible Veggie Platters can Georgia choose from?
Statement 1: Half the choices are listed as being "hearth healthy".
Statement 2: Seven of the choices are not listed as being "hearth healthy".
A Veggie Platter at the restaurant where Georgia likes to eat comprises four different vegetables. How many possible Veggie Platters can Georgia choose from?
Statement 1: Half the choices are listed as being "hearth healthy".
Statement 2: Seven of the choices are not listed as being "hearth healthy".
The only thing needed in order to answer the question is the number of vegetables from which Georgia has to choose. Neither statement alone gives this information.
However, it is possible to deduce this information from both statements together. From Statement 1, half the choices are "hearth healthy"; therefore, half are not. From Statement 2, this amounts to seven choices, so the total number of vegetables on the menu is fourteen, and there are a total of 
possible Veggie Platters.
The only thing needed in order to answer the question is the number of vegetables from which Georgia has to choose. Neither statement alone gives this information.
However, it is possible to deduce this information from both statements together. From Statement 1, half the choices are "hearth healthy"; therefore, half are not. From Statement 2, this amounts to seven choices, so the total number of vegetables on the menu is fourteen, and there are a total of
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A Lunch Deal Meal at the Korean restaurant where Mickey likes to eat comprises one appetizer, one entree, and one beverage. How many possible Lunch Deal Meals can Mickey choose from?
Statement 1: The number of appetizers, the number of entrees, and the number of beverages are the same.
Statement 2: If Mickey orders one of the shrimp dishes, there is a $1 upcharge; if he orders one of the other six entrees, there is no upcharge.
A Lunch Deal Meal at the Korean restaurant where Mickey likes to eat comprises one appetizer, one entree, and one beverage. How many possible Lunch Deal Meals can Mickey choose from?
Statement 1: The number of appetizers, the number of entrees, and the number of beverages are the same.
Statement 2: If Mickey orders one of the shrimp dishes, there is a $1 upcharge; if he orders one of the other six entrees, there is no upcharge.
By the multiplication principle, the number of possible Lunch Deal Meals is the product of the number of appetizers, the number of entrees, and the number of beverages.
Assume both statements are true. From Statement 1 it can be determined that there are an equal number of choices from each column, but the actual number of choices still needs to be determined. From Statement 2, it can be determined only that there are at least eight entrees - it is given that there are six dishes without shirimp, but the number of dishes with shrimp is not given (only that there must be more than two, since a plural is given). The statements together give insufficient information.
By the multiplication principle, the number of possible Lunch Deal Meals is the product of the number of appetizers, the number of entrees, and the number of beverages.
Assume both statements are true. From Statement 1 it can be determined that there are an equal number of choices from each column, but the actual number of choices still needs to be determined. From Statement 2, it can be determined only that there are at least eight entrees - it is given that there are six dishes without shirimp, but the number of dishes with shrimp is not given (only that there must be more than two, since a plural is given). The statements together give insufficient information.
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A Lunch Deal Meal at the Chinese restaurant where Mickey likes to eat comprises one appetizer, one entree, and one beverage. How many possible Lunch Meal Deals can Mickey choose from?
Statement 1: The number of appetizers and the number of beverages from which Mickey may choose are both equal to the number of entrees that do not have shrimp.
Statement 2: Mickey may choose any of ten entrees, but if he chooses any of the three entrees with shrimp, he must pay an upcharge of $1.
A Lunch Deal Meal at the Chinese restaurant where Mickey likes to eat comprises one appetizer, one entree, and one beverage. How many possible Lunch Meal Deals can Mickey choose from?
Statement 1: The number of appetizers and the number of beverages from which Mickey may choose are both equal to the number of entrees that do not have shrimp.
Statement 2: Mickey may choose any of ten entrees, but if he chooses any of the three entrees with shrimp, he must pay an upcharge of $1.
By the multiplication principle, the number of possible Lunch Deal Meals is the product of the number of appetizers, the number of entrees, and the number of beverages. Neither statement alone gives all three numbers.
Assume both statements are true. From Statement 2, there are ten entrees on the menu, seven of which do not have shrimp (note that the upcharge is irrelevant). From Statement 1, the number of appetizers and the number of beverages are both equal to the number of entrees that do not have shrimp, which is seven. Therefore, the number of possible Lunch Deal Meals is 
.
By the multiplication principle, the number of possible Lunch Deal Meals is the product of the number of appetizers, the number of entrees, and the number of beverages. Neither statement alone gives all three numbers.
Assume both statements are true. From Statement 2, there are ten entrees on the menu, seven of which do not have shrimp (note that the upcharge is irrelevant). From Statement 1, the number of appetizers and the number of beverages are both equal to the number of entrees that do not have shrimp, which is seven. Therefore, the number of possible Lunch Deal Meals is
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A Lunch Deal Meal at the Chinese restaurant where Phyllis likes to eat comprises one appetizer, one entree, and one beverage. How many possible Lunch Deal Meals can Phyllis choose from?
Statement 1: If Phyllis orders either shrimp-based appetizer or one of the three shrimp-based entrees, she will pay extra.
Statement 2: If Phyllis doesn't want shrimp, she has five appetizers and seven entrees to choose from.
A Lunch Deal Meal at the Chinese restaurant where Phyllis likes to eat comprises one appetizer, one entree, and one beverage. How many possible Lunch Deal Meals can Phyllis choose from?
Statement 1: If Phyllis orders either shrimp-based appetizer or one of the three shrimp-based entrees, she will pay extra.
Statement 2: If Phyllis doesn't want shrimp, she has five appetizers and seven entrees to choose from.
By the multiplication principle, the number of possible Lunch Deal Meals is the product of the number of appetizers, the number of entrees, and the number of beverages. The two statements together give the number of appetizers - two with shrimp and five without for a total of seven - and the number of entrees - three with shrimp and seven without - but not the number of beverages.
By the multiplication principle, the number of possible Lunch Deal Meals is the product of the number of appetizers, the number of entrees, and the number of beverages. The two statements together give the number of appetizers - two with shrimp and five without for a total of seven - and the number of entrees - three with shrimp and seven without - but not the number of beverages.
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Jared, a jewelry salesman, makes a 10% commission on the selling price of all the jewelry he sells. Last month, he earned $4,000 in commission. What is the total selling price of the jewelry he sold this month?
1. His sales this month were 25% more than his sales last month.
2. He earned $5,000 in commission this month.
Jared, a jewelry salesman, makes a 10% commission on the selling price of all the jewelry he sells. Last month, he earned $4,000 in commission. What is the total selling price of the jewelry he sold this month?
1. His sales this month were 25% more than his sales last month.
2. He earned $5,000 in commission this month.
Each statement has enough information to calculate the answer.
Using statement 1, we know that he sold 25% more than last month. Given in the question, last month he earned $4,000. If $4,000 is 10% of the total price of his jewelry sold last month, we do some math (below) to find he sold $40,000 worth of jewelry. Let 
be the total price of jewelry sold last month. Then,
so
Increasing that by 25% we see that
NOTE: when multiplying the percents, make sure you use the decimal values.
Using statement 2, we know that we can just calculate the total selling price of the jewelry he sold this month based on the commission percentage. So, if we let 
equal the total price of jewelry sold this month, we get 
or
Therefore we see we can use either statement individually to find what we are looking for.
Each statement has enough information to calculate the answer.
Using statement 1, we know that he sold 25% more than last month. Given in the question, last month he earned $4,000. If $4,000 is 10% of the total price of his jewelry sold last month, we do some math (below) to find he sold $40,000 worth of jewelry. Let
Increasing that by 25% we see that
NOTE: when multiplying the percents, make sure you use the decimal values.
Using statement 2, we know that we can just calculate the total selling price of the jewelry he sold this month based on the commission percentage. So, if we let
or
Therefore we see we can use either statement individually to find what we are looking for.
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How many subsets does set 
have?
Statement 1: 
has eight elements.
Statement 2: 
is the set of all prime numbers between 1 and 20.
How many subsets does set
Statement 1:
Statement 2:
The number of subsets of any set can be calculated by raising 2 to the power of the number of elements in the set. The first statement gives you that information immediately. The second gives you enough information to find the number of elements, as there are eight primes between 1 and 20: 2, 3, 5, 7, 11, 13, 17, and 19. From either statement alone, you can deduce the answer to be 
.
The number of subsets of any set can be calculated by raising 2 to the power of the number of elements in the set. The first statement gives you that information immediately. The second gives you enough information to find the number of elements, as there are eight primes between 1 and 20: 2, 3, 5, 7, 11, 13, 17, and 19. From either statement alone, you can deduce the answer to be
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Let 
be the set of all of the multiples of 3 between 29 and 50. How many subsets of 
can be formed?
Let
The multiples of 3 between 29 and 50 are 30, 33, 36, 39, 42, 45, and 48 - therefore, 
has seven elements total.
The number of subsets in a set can be calculated by raising 2 to the power of the number of elements. Therefore, the answer to our question is 
.
The multiples of 3 between 29 and 50 are 30, 33, 36, 39, 42, 45, and 48 - therefore,
The number of subsets in a set can be calculated by raising 2 to the power of the number of elements. Therefore, the answer to our question is
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Data sufficiency question- do not actually solve the question
A bag of marbles consist of a mixture of black and red marbles. What is the probability of choosing a red marble followed by a black marble?
1. The probability of choosing a black marble first is 
.
2. There are 10 black marbles in the bag.
Data sufficiency question- do not actually solve the question
A bag of marbles consist of a mixture of black and red marbles. What is the probability of choosing a red marble followed by a black marble?
1. The probability of choosing a black marble first is
2. There are 10 black marbles in the bag.
From statement 1, we know the probabilty of choosing the first marble. However, since the marble is not replaced, it is impossible to calculate the probability of choosing the second marble. By knowing the information in statement 2 combined with statement 1, we can calculate the total number of marbles initially present.
From statement 1, we know the probabilty of choosing the first marble. However, since the marble is not replaced, it is impossible to calculate the probability of choosing the second marble. By knowing the information in statement 2 combined with statement 1, we can calculate the total number of marbles initially present.
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A certain major league baseball player gets on base 25% of the time (once every 4 times at bat).
For any game where he comes to bat 5 times, what is the probability that he will get on base either 3 or 4 times? - Hint – add the probability of 3 to the probability of 4.
A certain major league baseball player gets on base 25% of the time (once every 4 times at bat).
For any game where he comes to bat 5 times, what is the probability that he will get on base either 3 or 4 times? - Hint – add the probability of 3 to the probability of 4.
Binomial Table
Binomial Table
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Assume that we are the immortal gods of statistics and we know the following population statistics:
-
average driving speed for women=50 mph with a standard deviation of 12
-
average driving speed for men=45 mph with a standard deviation of 11
We look down from our statistical Mount Olympus and notice that the Earth mortals have randomly sampled 60 women and 65 men in an attempt to detect a significant difference in the average driving speed.
What is the probability that the Earth mortals will properly reject the assumption (i.e. the null hypothesis) that there is no significant difference between the average driving speeds. The Earth mortals have decided to use a 2-tailed 95% confidence test.
Assume that we are the immortal gods of statistics and we know the following population statistics:
-
average driving speed for women=50 mph with a standard deviation of 12
-
average driving speed for men=45 mph with a standard deviation of 11
We look down from our statistical Mount Olympus and notice that the Earth mortals have randomly sampled 60 women and 65 men in an attempt to detect a significant difference in the average driving speed.
What is the probability that the Earth mortals will properly reject the assumption (i.e. the null hypothesis) that there is no significant difference between the average driving speeds. The Earth mortals have decided to use a 2-tailed 95% confidence test.
standard deviation of the difference between the sample means =
at 95% (2-tailed) = 1.96
the sample difference must be 4 or greater
(note: the probability of the sample difference being -4 or less is so small (4.5 standard deviations) that it will be ignored and we will only consider the probability that the difference is 4 or more.)
the probablity of the sample difference being 4 or greater (knowing that the population difference is 5) =
the table shows that .3156 lies below -.48, so, .6844 lies above -.48
In English - there is a .6844 probability that the 2 sample means will yield a sample difference that is 1.96 or more standard deviations above 0.
standard deviation of the difference between the sample means =
the sample difference must be 4 or greater
(note: the probability of the sample difference being -4 or less is so small (4.5 standard deviations) that it will be ignored and we will only consider the probability that the difference is 4 or more.)
the probablity of the sample difference being 4 or greater (knowing that the population difference is 5) =
the table shows that .3156 lies below -.48, so, .6844 lies above -.48
In English - there is a .6844 probability that the 2 sample means will yield a sample difference that is 1.96 or more standard deviations above 0.
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In a popular state lottery game, a player selects 5 numbers (on 1 ticket) out of a possible 39 numbers. There are 575,757 possible 5 number combinations.
So, the odds are 575,757 to 1 against winning.
What are the odds of getting 4 of the 5 numbers correct on 1 ticket?
In a popular state lottery game, a player selects 5 numbers (on 1 ticket) out of a possible 39 numbers. There are 575,757 possible 5 number combinations.
So, the odds are 575,757 to 1 against winning.
What are the odds of getting 4 of the 5 numbers correct on 1 ticket?
575,757 must be divided by -
575,757 must be divided by -
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One card from another deck is added to a standard deck of fifty-two cards. The cards are shuffled and one card is removed.
A card is then drawn at random. What is the probability that that card is an ace?
Statement 1: The card that was added was a spade.
Statetment 2: The card that was removed was a jack.
One card from another deck is added to a standard deck of fifty-two cards. The cards are shuffled and one card is removed.
A card is then drawn at random. What is the probability that that card is an ace?
Statement 1: The card that was added was a spade.
Statetment 2: The card that was removed was a jack.
You need to know two things to answer this question - the rank of the added card, and the rank of the removed card. The second statement is useful but not sufficient; the first is irrelevant to the question.
You need to know two things to answer this question - the rank of the added card, and the rank of the removed card. The second statement is useful but not sufficient; the first is irrelevant to the question.
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Your friend at work submits a bold hypothesis. He suggests that the number of sales per day follow the pattern - Monday-10%, Tuesday-10%, Wednesday-10%, Thursday 35% and Friday 35%.
You and he then record the number of sales for the following week: Monday-120, Tuesday-85, Wednesday-105, Thursday-325 and Friday-365.
After viewing the observed data, your friend expresses serious concern regarding his hypothesis.
You can help; you can tell him the probability of the observed data occuring if the hypothesis is true. Hint - Excel ChiTest.
Your friend at work submits a bold hypothesis. He suggests that the number of sales per day follow the pattern - Monday-10%, Tuesday-10%, Wednesday-10%, Thursday 35% and Friday 35%.
You and he then record the number of sales for the following week: Monday-120, Tuesday-85, Wednesday-105, Thursday-325 and Friday-365.
After viewing the observed data, your friend expresses serious concern regarding his hypothesis.
You can help; you can tell him the probability of the observed data occuring if the hypothesis is true. Hint - Excel ChiTest.
Use Excel ChiTest to get the .063 probability. If you are old-fashioned, you can also obtain the Chi-Squared number (8.928) by using ChiInv; but, it is not needed
Use Excel ChiTest to get the .063 probability. If you are old-fashioned, you can also obtain the Chi-Squared number (8.928) by using ChiInv; but, it is not needed
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A certain tutor boasts that his 2 week training program will increase a student's score on a 2400 point exam by at least 100 points (4.167%). A 10 student 'before-and-after' study was conducted to validate the claim. The following results were obtained - the 3 columns represent the before, afer and increase numbers for each of the 10 students:
1300 1340 40
1670 1790 120
1500 1710 210
1360 1660 300
1580 1730 150
1160 1320 160
1910 2100 190
1410 1490 80
1710 1880 170
1990 2060 70
Assume the null Hypothesis:
'The average increase is less than 100 points'
What is the highest level of significance (p-value) at which the null hypothesis will be rejected?
A certain tutor boasts that his 2 week training program will increase a student's score on a 2400 point exam by at least 100 points (4.167%). A 10 student 'before-and-after' study was conducted to validate the claim. The following results were obtained - the 3 columns represent the before, afer and increase numbers for each of the 10 students:
1300 1340 40
1670 1790 120
1500 1710 210
1360 1660 300
1580 1730 150
1160 1320 160
1910 2100 190
1410 1490 80
1710 1880 170
1990 2060 70
Assume the null Hypothesis:
'The average increase is less than 100 points'
What is the highest level of significance (p-value) at which the null hypothesis will be rejected?
Using Excel, the average increase (column 3) is 149 and Standard Deviation of the increases is 76.37
Using Excel - a t value of 2.03 for 9 degrees of freedom = .036
Using Excel, the average increase (column 3) is 149 and Standard Deviation of the increases is 76.37
Using Excel - a t value of 2.03 for 9 degrees of freedom = .036
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A test for a new drug was conducted. In the control or placebo group, 7 of 210 participants experienced positive results. The group that took the drug experienced 27 out of 374 positive resluts.
The placebo group had a sucess rate of .0333 and the drug group had a success rate of .0722. The difference is .0389 and the overall percentage (for both groups combined) is .0582
At what level is the difference of .0389 significant? Asked another way - what is the p-value for .0389?
A test for a new drug was conducted. In the control or placebo group, 7 of 210 participants experienced positive results. The group that took the drug experienced 27 out of 374 positive resluts.
The placebo group had a sucess rate of .0333 and the drug group had a success rate of .0722. The difference is .0389 and the overall percentage (for both groups combined) is .0582
At what level is the difference of .0389 significant? Asked another way - what is the p-value for .0389?
The standard error of the difference (.0389) =
test statistic -
from the table (or excel NormsDist) - Z=1.9245 translates to .9729
The standard error of the difference (.0389) =
test statistic -
from the table (or excel NormsDist) - Z=1.9245 translates to .9729
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A type 1 error (False Alarm or 'Convicting the innocent man') occurs when we incorrectly reject a true null hypothesis.
A type 2 error (failure to detect) occurs when we fail to reject a false null hypothesis.
Which one of the following 5 statements is false?
Note - only 1 of the statements is false.
A) For a given sample size (n=100), decreasing the significane level (from .05 to .01) will decrease the chance of a type 1 error.
B) For a given sample size (n=100), increasing the significane level (from .01 to .05) will decrease the chance of a type 2 error.
C) The ability to correctly detect a false null hypothesis is called the 'Power' of a test.
D) Increasing sample size (from 100 to 120) will always decrease the chance of both a type 1 error and a type 2 error.
E) None of the above statements are true.
A type 1 error (False Alarm or 'Convicting the innocent man') occurs when we incorrectly reject a true null hypothesis.
A type 2 error (failure to detect) occurs when we fail to reject a false null hypothesis.
Which one of the following 5 statements is false?
Note - only 1 of the statements is false.
A) For a given sample size (n=100), decreasing the significane level (from .05 to .01) will decrease the chance of a type 1 error.
B) For a given sample size (n=100), increasing the significane level (from .01 to .05) will decrease the chance of a type 2 error.
C) The ability to correctly detect a false null hypothesis is called the 'Power' of a test.
D) Increasing sample size (from 100 to 120) will always decrease the chance of both a type 1 error and a type 2 error.
E) None of the above statements are true.
Statements A, B, C, and D are all true - so -
The only false statement is E (the statement that declares that A and B and C and D are all false)
Statements A, B, C, and D are all true - so -
The only false statement is E (the statement that declares that A and B and C and D are all false)
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Four coins - a fair penny, a loaded penny, a fair nickel, and a loaded nickel - are tossed. What is the probability that all four will come up heads?
Statement 1: Yesterday, out of 100 tosses, the loaded penny came up heads 70 times.
Statement 2: Yesterday, out of 100 tosses, the loaded nickel came up heads 40 times.
Four coins - a fair penny, a loaded penny, a fair nickel, and a loaded nickel - are tossed. What is the probability that all four will come up heads?
Statement 1: Yesterday, out of 100 tosses, the loaded penny came up heads 70 times.
Statement 2: Yesterday, out of 100 tosses, the loaded nickel came up heads 40 times.
While experiments such as repeated tossings can give an idea of the probability that a coin will come up heads or tails, they do not provide a definitive answer, so neither statement is helpful here.
While experiments such as repeated tossings can give an idea of the probability that a coin will come up heads or tails, they do not provide a definitive answer, so neither statement is helpful here.
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Jerry is a Cardinal fan and he and his family live on a street with 9 other families that are all Cardinal fans. One block to the north, there are 11 families and they are all Cub fans. These 21 households all buy their lawn fertilizer from Ben's Lawn and Garden Shop. Jerry suspects that Ben (who is originally from Chicago) is a Cub fan and that he provides better fertilizer to the Cub fans than to the Cardinal fans, while charging the same price for all.
Last Saturday everyone in town mowed their lawn. At 2:00 AM Sunday morning, Jerry snuck around town and weighed all of the grass clippings for the 21 households in question.
The weights (in lbs) of the grass clippings for the 10 Cardinal homes were:
82, 85, 90, 74, 80, 89, 75, 81, 93, 75
The weights (in lbs) of the grass clippings for the 11 Cub homes were:
90, 87, 93, 75, 88, 96, 90, 82, 95, 97, 78
The Cardinal average was 82.4; the Cub average was 88.27.
At what level is the 5.87 lb difference significant? - Asked another way - what is the p value for the 5.87 lb difference.
Jerry is a Cardinal fan and he and his family live on a street with 9 other families that are all Cardinal fans. One block to the north, there are 11 families and they are all Cub fans. These 21 households all buy their lawn fertilizer from Ben's Lawn and Garden Shop. Jerry suspects that Ben (who is originally from Chicago) is a Cub fan and that he provides better fertilizer to the Cub fans than to the Cardinal fans, while charging the same price for all.
Last Saturday everyone in town mowed their lawn. At 2:00 AM Sunday morning, Jerry snuck around town and weighed all of the grass clippings for the 21 households in question.
The weights (in lbs) of the grass clippings for the 10 Cardinal homes were:
82, 85, 90, 74, 80, 89, 75, 81, 93, 75
The weights (in lbs) of the grass clippings for the 11 Cub homes were:
90, 87, 93, 75, 88, 96, 90, 82, 95, 97, 78
The Cardinal average was 82.4; the Cub average was 88.27.
At what level is the 5.87 lb difference significant? - Asked another way - what is the p value for the 5.87 lb difference.
Cub variance = 53.218; Cardinal variance = 45.378.
The standard deviation of the difference (5.87) is:
TDIST of 1.918 with 19 Degrees of Freedom = .035
So, we would reject the null hypothesis (the hypothesis that claims that the means are equal) at 95% confidence (p=.05) and not reject at 99% (p=.01)
Cub variance = 53.218; Cardinal variance = 45.378.
The standard deviation of the difference (5.87) is:
TDIST of 1.918 with 19 Degrees of Freedom = .035
So, we would reject the null hypothesis (the hypothesis that claims that the means are equal) at 95% confidence (p=.05) and not reject at 99% (p=.01)
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A marble is selected at random from a box of red, yellow, and blue marbles. What is the probability that the marble is yellow?
-
There are ten blue marbles in the box.
-
There are eight red marbles in the box.
A marble is selected at random from a box of red, yellow, and blue marbles. What is the probability that the marble is yellow?
-
There are ten blue marbles in the box.
-
There are eight red marbles in the box.
To determine the probability that the marble is yellow we need to know two things: the number of yellow marbles, and the number of marbles total. The first quantity divided by the last quantity is our probablility.
But the two given statements together only tell us that eighteen marbles are not yellow. This is not enough information. For example, if there are two yellow marbles, the probability of drawing a yellow marble is 
. But if there are twenty-two yellow marbles, the probability of drawing a yellow marble is 
Therefore, the answer is that both statements together are insufficient to answer the question.
To determine the probability that the marble is yellow we need to know two things: the number of yellow marbles, and the number of marbles total. The first quantity divided by the last quantity is our probablility.
But the two given statements together only tell us that eighteen marbles are not yellow. This is not enough information. For example, if there are two yellow marbles, the probability of drawing a yellow marble is
Therefore, the answer is that both statements together are insufficient to answer the question.
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Several decks of playing cards are shuffled together. One card is drawn, shown, and put aside. Another card is dealt. What is the probability that the dealt card is red, assuming the first card is known?
-
The card removed before the deal was red.
-
The cards were shuffled again between the draw and the deal.
Several decks of playing cards are shuffled together. One card is drawn, shown, and put aside. Another card is dealt. What is the probability that the dealt card is red, assuming the first card is known?
-
The card removed before the deal was red.
-
The cards were shuffled again between the draw and the deal.
To answer this question you need to know two things: the number of red cards left and the number of total cards left. The second statement is irrelevant, as a reshuffle does not change the composition of the deck. The first statement tells you that there is one fewer red card than black cards, but it does not tell you how many of each there are, as you do not know how many decks of cards there were.
And that information, which is not given, affects the answer. For example, if there were four decks, there were 103 red cards out of 207; if there were six decks, there were 155 red cards out of 311. The probabilities would be, respectively,
and
,
a small difference, but nonetheless, a difference.
The correct answer is that both statements together are insufficient to answer the question.
To answer this question you need to know two things: the number of red cards left and the number of total cards left. The second statement is irrelevant, as a reshuffle does not change the composition of the deck. The first statement tells you that there is one fewer red card than black cards, but it does not tell you how many of each there are, as you do not know how many decks of cards there were.
And that information, which is not given, affects the answer. For example, if there were four decks, there were 103 red cards out of 207; if there were six decks, there were 155 red cards out of 311. The probabilities would be, respectively,
and
a small difference, but nonetheless, a difference.
The correct answer is that both statements together are insufficient to answer the question.
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