DSQ: Calculating probability - GMAT Quantitative

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Question

A fair coin is tossed a given number of times. What is the probability that three or more of the tosses will come up heads?

The number of times the coin is tossed is four.

The previous day, the coin was tossed ten times, and heads came up eight times.

Answer

This is a binomial probability experiment.

The probability that a fair coin will come up heads in any single toss is $\frac{1}{2}$ .

If the coin is tossed four times, then the probability that heads will appear three times is:

$P(x=3) = C(4,3) \left (\frac{1}{2} \right )}^{3}\left (\frac{1}{2} \right )}^{4-3}$

$= \frac{4!}{3! (4-3)!}\left (\frac{1}{2} \right )}^{3}\left (\frac{1}{2} \right )}^{1}$

$= \frac{4!}{3! 1!}\left (\frac{1}{2} \right )}^{4}$

$= 4\cdot \left (\frac{1}{16} \right )}=\frac{1}{4}$

The probability that heads will appear four times is:

$P(x=4) = C(4,4) \left (\frac{1}{2} \right )}^{4}\left (\frac{1}{2} \right )}^{4-4}$

$= \frac{4!}{4! (4-4)!}\left (\frac{1}{2} \right )}^{4}\left (\frac{1}{2} \right )}^{0}$

$= \frac{4!}{4! 0!}\left (\frac{1}{2} \right )}^{4}$

$= 1\cdot \left (\frac{1}{16} \right )}=\frac{1}{16}$

So the probability that heads will appear at least three times is

$P(x\geq 3) = \frac{1}{4} + \frac{1}{16}= \frac{5}{16}$

The second statement is irrelevant; the result of a previous trial has no influence on that of the current trial. It has already been established in the problem that the coin is fair.

The answer is that Statement 1 alone is sufficient to answer the question, but not Statement 2.

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Question

A monster is holding a brown paper bag filled with cookies. Inside the bag there are two types of cookies, chocolate chip and oatmeal raisin. If the monster pulls one cookie out at random, what is the probability that he pulls out a chocolate chip cookie.

1. There are a total of 24 cookies in the bag.

2. There are 8 more chocolate chip cookies in the bag than oatmeal raisin cookies.

Answer

We need both statements to get the answer.

Statement 1 alone is not enough. Having 24 total cookies does not tell us the proportions of each type of cookie. We could have 1 chocolate chip cookie and 23 oatmeal raisin, or vice versa.

Statement 2 alone is not sufficient either. Knowing that there are 8 more chocolate chip cookies than oatmeal raisin without knowing the total amount in the bag does not help. It could be 1 oatmeal raisin cookie and 9 chocolate chip, or it could be 100 oatmeal raisin cookies and 108 chocolate chip cookies (although that is a pretty big bag!)

Only by having both statements together can we find the right proportion. If we let be the number of chocolate chip cookies, and be the number of oatmeal raisin, we can set two equations and find the right ratio. We know that

from statement 1, and

from statement 2.

Substituting into the first equation we get

Therefore, we get 8 oatmeal raisin cookies, and 16 chocolate chip cookies. The answer is

$\frac{16}{24}=\frac{2}{3}$

or 66.67% chance that the monster pulls out a chocolate chip cookie.

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Question

Choose the answer that best describes sufficient data to solve the problem.

A student body is polled regarding their favorite campus food. Of those who answered "Pizza" half were male and half were female. Pizza received 24 votes from men.

I. Pizza received the votes of 1 out of every 5 men and 3 out of every 10 women.

II. The student population was found to be 60% male and 40% female

How many students altogether were polled?

Answer

Pizza received 24 votes from men. Since half of those who responded pizza were women, there must be 24 votes from women.

If 1 out of 5 men voted for pizza, this means we have $\ 24\cdot 5 = 120$ men. If 3 out of 10 women voted for pizza, we have $\frac{10}{3}\cdot 24 = 80$ women for a total of 200 people polled.

If our population is instead given as 60% male and 40% female, we still have 24 men and 24 women that voted for pizza, but while we can determine the ratio of people who voted, we cannot get a precise number. For example, while 80 women and 120 men works, we could also have 160 women and 240 men as our total poll (with 1 out of 10 men and 3 out of 20 women voting pizza).

Thus the first condition is sufficient while the second is not.

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Question

Find the probability.

Statement 1): Given that the last flips of a coin were heads, the next coin flip is heads.

Statement 2): Given that the last days were sunny skies, the next day is sunny.

Answer

This question deals with conditional probability.

Statement 1) consists of independent events. The past trials will not affect the future trials of the coin. The odds of flipping a coin and land on heads will be $\frac{1}{2}$ .

Statement 2) also consists of independent events. However, given that the previous 10 days were sunny does not necessarily indicate that the next day will guarantee that it will be a sunny day. The probability is unknown because the statement does not give any information about the percentage of a sunny sky on any particular day.

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Question

Jenkins has a box of 36 guitar picks. Each pick is one of three colors: blue, black, or orange.

I) There are 14 more orange picks than blue picks.

II) There are 12 black picks.

What are the odds of randomly selecting a black pick, followed by a blue pick, followed by an orange pick?

Answer

To find the probability of several events, we need to find the probability of each event separately. First, we need to know how many there are of each color pick.

Statement I relates the number of orange and blue picks.

Statement II tells us the number of black picks.

$P_{orange}+P_{black}+P_{blue}=36$

From Statement I:

$P_{orange}=P_{blue}+14$

From Statement II:

$P_{black}=12$

Using Statement I, Statement II, and the given info, we can substitute $P_{blue}+14$ from Statement I into Statement II for $P_{orange}$ and get the following:

$P_{blue}+14+P_{blue}+12=36$

We can then solve for $P_{blue}$ :

$2P_{blue}+26=36$

$2P_{blue}=10$

$P_{blue}=5$

Since we are told in Statement I that there are 14 more blue picks than orange picks, we can then find the number of orange picks:

$P_{orange}=P_{blue}+14$

$P_{orange}=5+14$

$P_{orange}=19$

So, to find the odds of multiple events, multiply the probability of each event together:

$\frac{12}{36}\cdot \frac{5}{35}\cdot \frac{19}{34}=\frac{1140}{42840}=\frac{19}{714}$

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Question

A fair coin is tossed a given number of times. What is the probability that three or more of the tosses will come up heads?

The number of times the coin is tossed is four.

The previous day, the coin was tossed ten times, and heads came up eight times.

Answer

This is a binomial probability experiment.

The probability that a fair coin will come up heads in any single toss is $\frac{1}{2}$ .

If the coin is tossed four times, then the probability that heads will appear three times is:

$P(x=3) = C(4,3) \left (\frac{1}{2} \right )}^{3}\left (\frac{1}{2} \right )}^{4-3}$

$= \frac{4!}{3! (4-3)!}\left (\frac{1}{2} \right )}^{3}\left (\frac{1}{2} \right )}^{1}$

$= \frac{4!}{3! 1!}\left (\frac{1}{2} \right )}^{4}$

$= 4\cdot \left (\frac{1}{16} \right )}=\frac{1}{4}$

The probability that heads will appear four times is:

$P(x=4) = C(4,4) \left (\frac{1}{2} \right )}^{4}\left (\frac{1}{2} \right )}^{4-4}$

$= \frac{4!}{4! (4-4)!}\left (\frac{1}{2} \right )}^{4}\left (\frac{1}{2} \right )}^{0}$

$= \frac{4!}{4! 0!}\left (\frac{1}{2} \right )}^{4}$

$= 1\cdot \left (\frac{1}{16} \right )}=\frac{1}{16}$

So the probability that heads will appear at least three times is

$P(x\geq 3) = \frac{1}{4} + \frac{1}{16}= \frac{5}{16}$

The second statement is irrelevant; the result of a previous trial has no influence on that of the current trial. It has already been established in the problem that the coin is fair.

The answer is that Statement 1 alone is sufficient to answer the question, but not Statement 2.

Compare your answer with the correct one above

Question

A monster is holding a brown paper bag filled with cookies. Inside the bag there are two types of cookies, chocolate chip and oatmeal raisin. If the monster pulls one cookie out at random, what is the probability that he pulls out a chocolate chip cookie.

1. There are a total of 24 cookies in the bag.

2. There are 8 more chocolate chip cookies in the bag than oatmeal raisin cookies.

Answer

We need both statements to get the answer.

Statement 1 alone is not enough. Having 24 total cookies does not tell us the proportions of each type of cookie. We could have 1 chocolate chip cookie and 23 oatmeal raisin, or vice versa.

Statement 2 alone is not sufficient either. Knowing that there are 8 more chocolate chip cookies than oatmeal raisin without knowing the total amount in the bag does not help. It could be 1 oatmeal raisin cookie and 9 chocolate chip, or it could be 100 oatmeal raisin cookies and 108 chocolate chip cookies (although that is a pretty big bag!)

Only by having both statements together can we find the right proportion. If we let be the number of chocolate chip cookies, and be the number of oatmeal raisin, we can set two equations and find the right ratio. We know that

from statement 1, and

from statement 2.

Substituting into the first equation we get

Therefore, we get 8 oatmeal raisin cookies, and 16 chocolate chip cookies. The answer is

$\frac{16}{24}=\frac{2}{3}$

or 66.67% chance that the monster pulls out a chocolate chip cookie.

Compare your answer with the correct one above

Question

Choose the answer that best describes sufficient data to solve the problem.

A student body is polled regarding their favorite campus food. Of those who answered "Pizza" half were male and half were female. Pizza received 24 votes from men.

I. Pizza received the votes of 1 out of every 5 men and 3 out of every 10 women.

II. The student population was found to be 60% male and 40% female

How many students altogether were polled?

Answer

Pizza received 24 votes from men. Since half of those who responded pizza were women, there must be 24 votes from women.

If 1 out of 5 men voted for pizza, this means we have $\ 24\cdot 5 = 120$ men. If 3 out of 10 women voted for pizza, we have $\frac{10}{3}\cdot 24 = 80$ women for a total of 200 people polled.

If our population is instead given as 60% male and 40% female, we still have 24 men and 24 women that voted for pizza, but while we can determine the ratio of people who voted, we cannot get a precise number. For example, while 80 women and 120 men works, we could also have 160 women and 240 men as our total poll (with 1 out of 10 men and 3 out of 20 women voting pizza).

Thus the first condition is sufficient while the second is not.

Compare your answer with the correct one above

Question

Find the probability.

Statement 1): Given that the last flips of a coin were heads, the next coin flip is heads.

Statement 2): Given that the last days were sunny skies, the next day is sunny.

Answer

This question deals with conditional probability.

Statement 1) consists of independent events. The past trials will not affect the future trials of the coin. The odds of flipping a coin and land on heads will be $\frac{1}{2}$ .

Statement 2) also consists of independent events. However, given that the previous 10 days were sunny does not necessarily indicate that the next day will guarantee that it will be a sunny day. The probability is unknown because the statement does not give any information about the percentage of a sunny sky on any particular day.

Compare your answer with the correct one above

Question

Jenkins has a box of 36 guitar picks. Each pick is one of three colors: blue, black, or orange.

I) There are 14 more orange picks than blue picks.

II) There are 12 black picks.

What are the odds of randomly selecting a black pick, followed by a blue pick, followed by an orange pick?

Answer

To find the probability of several events, we need to find the probability of each event separately. First, we need to know how many there are of each color pick.

Statement I relates the number of orange and blue picks.

Statement II tells us the number of black picks.

$P_{orange}+P_{black}+P_{blue}=36$

From Statement I:

$P_{orange}=P_{blue}+14$

From Statement II:

$P_{black}=12$

Using Statement I, Statement II, and the given info, we can substitute $P_{blue}+14$ from Statement I into Statement II for $P_{orange}$ and get the following:

$P_{blue}+14+P_{blue}+12=36$

We can then solve for $P_{blue}$ :

$2P_{blue}+26=36$

$2P_{blue}=10$

$P_{blue}=5$

Since we are told in Statement I that there are 14 more blue picks than orange picks, we can then find the number of orange picks:

$P_{orange}=P_{blue}+14$

$P_{orange}=5+14$

$P_{orange}=19$

So, to find the odds of multiple events, multiply the probability of each event together:

$\frac{12}{36}\cdot \frac{5}{35}\cdot \frac{19}{34}=\frac{1140}{42840}=\frac{19}{714}$

Compare your answer with the correct one above

Question

A fair coin is tossed a given number of times. What is the probability that three or more of the tosses will come up heads?

The number of times the coin is tossed is four.

The previous day, the coin was tossed ten times, and heads came up eight times.

Answer

This is a binomial probability experiment.

The probability that a fair coin will come up heads in any single toss is $\frac{1}{2}$ .

If the coin is tossed four times, then the probability that heads will appear three times is:

$P(x=3) = C(4,3) \left (\frac{1}{2} \right )}^{3}\left (\frac{1}{2} \right )}^{4-3}$

$= \frac{4!}{3! (4-3)!}\left (\frac{1}{2} \right )}^{3}\left (\frac{1}{2} \right )}^{1}$

$= \frac{4!}{3! 1!}\left (\frac{1}{2} \right )}^{4}$

$= 4\cdot \left (\frac{1}{16} \right )}=\frac{1}{4}$

The probability that heads will appear four times is:

$P(x=4) = C(4,4) \left (\frac{1}{2} \right )}^{4}\left (\frac{1}{2} \right )}^{4-4}$

$= \frac{4!}{4! (4-4)!}\left (\frac{1}{2} \right )}^{4}\left (\frac{1}{2} \right )}^{0}$

$= \frac{4!}{4! 0!}\left (\frac{1}{2} \right )}^{4}$

$= 1\cdot \left (\frac{1}{16} \right )}=\frac{1}{16}$

So the probability that heads will appear at least three times is

$P(x\geq 3) = \frac{1}{4} + \frac{1}{16}= \frac{5}{16}$

The second statement is irrelevant; the result of a previous trial has no influence on that of the current trial. It has already been established in the problem that the coin is fair.

The answer is that Statement 1 alone is sufficient to answer the question, but not Statement 2.

Compare your answer with the correct one above

Question

A monster is holding a brown paper bag filled with cookies. Inside the bag there are two types of cookies, chocolate chip and oatmeal raisin. If the monster pulls one cookie out at random, what is the probability that he pulls out a chocolate chip cookie.

1. There are a total of 24 cookies in the bag.

2. There are 8 more chocolate chip cookies in the bag than oatmeal raisin cookies.

Answer

We need both statements to get the answer.

Statement 1 alone is not enough. Having 24 total cookies does not tell us the proportions of each type of cookie. We could have 1 chocolate chip cookie and 23 oatmeal raisin, or vice versa.

Statement 2 alone is not sufficient either. Knowing that there are 8 more chocolate chip cookies than oatmeal raisin without knowing the total amount in the bag does not help. It could be 1 oatmeal raisin cookie and 9 chocolate chip, or it could be 100 oatmeal raisin cookies and 108 chocolate chip cookies (although that is a pretty big bag!)

Only by having both statements together can we find the right proportion. If we let be the number of chocolate chip cookies, and be the number of oatmeal raisin, we can set two equations and find the right ratio. We know that

from statement 1, and

from statement 2.

Substituting into the first equation we get

Therefore, we get 8 oatmeal raisin cookies, and 16 chocolate chip cookies. The answer is

$\frac{16}{24}=\frac{2}{3}$

or 66.67% chance that the monster pulls out a chocolate chip cookie.

Compare your answer with the correct one above

Question

Choose the answer that best describes sufficient data to solve the problem.

A student body is polled regarding their favorite campus food. Of those who answered "Pizza" half were male and half were female. Pizza received 24 votes from men.

I. Pizza received the votes of 1 out of every 5 men and 3 out of every 10 women.

II. The student population was found to be 60% male and 40% female

How many students altogether were polled?

Answer

Pizza received 24 votes from men. Since half of those who responded pizza were women, there must be 24 votes from women.

If 1 out of 5 men voted for pizza, this means we have $\ 24\cdot 5 = 120$ men. If 3 out of 10 women voted for pizza, we have $\frac{10}{3}\cdot 24 = 80$ women for a total of 200 people polled.

If our population is instead given as 60% male and 40% female, we still have 24 men and 24 women that voted for pizza, but while we can determine the ratio of people who voted, we cannot get a precise number. For example, while 80 women and 120 men works, we could also have 160 women and 240 men as our total poll (with 1 out of 10 men and 3 out of 20 women voting pizza).

Thus the first condition is sufficient while the second is not.

Compare your answer with the correct one above

Question

Find the probability.

Statement 1): Given that the last flips of a coin were heads, the next coin flip is heads.

Statement 2): Given that the last days were sunny skies, the next day is sunny.

Answer

This question deals with conditional probability.

Statement 1) consists of independent events. The past trials will not affect the future trials of the coin. The odds of flipping a coin and land on heads will be $\frac{1}{2}$ .

Statement 2) also consists of independent events. However, given that the previous 10 days were sunny does not necessarily indicate that the next day will guarantee that it will be a sunny day. The probability is unknown because the statement does not give any information about the percentage of a sunny sky on any particular day.

Compare your answer with the correct one above

Question

Jenkins has a box of 36 guitar picks. Each pick is one of three colors: blue, black, or orange.

I) There are 14 more orange picks than blue picks.

II) There are 12 black picks.

What are the odds of randomly selecting a black pick, followed by a blue pick, followed by an orange pick?

Answer

To find the probability of several events, we need to find the probability of each event separately. First, we need to know how many there are of each color pick.

Statement I relates the number of orange and blue picks.

Statement II tells us the number of black picks.

$P_{orange}+P_{black}+P_{blue}=36$

From Statement I:

$P_{orange}=P_{blue}+14$

From Statement II:

$P_{black}=12$

Using Statement I, Statement II, and the given info, we can substitute $P_{blue}+14$ from Statement I into Statement II for $P_{orange}$ and get the following:

$P_{blue}+14+P_{blue}+12=36$

We can then solve for $P_{blue}$ :

$2P_{blue}+26=36$

$2P_{blue}=10$

$P_{blue}=5$

Since we are told in Statement I that there are 14 more blue picks than orange picks, we can then find the number of orange picks:

$P_{orange}=P_{blue}+14$

$P_{orange}=5+14$

$P_{orange}=19$

So, to find the odds of multiple events, multiply the probability of each event together:

$\frac{12}{36}\cdot \frac{5}{35}\cdot \frac{19}{34}=\frac{1140}{42840}=\frac{19}{714}$

Compare your answer with the correct one above

Question

A fair coin is tossed a given number of times. What is the probability that three or more of the tosses will come up heads?

The number of times the coin is tossed is four.

The previous day, the coin was tossed ten times, and heads came up eight times.

Answer

This is a binomial probability experiment.

The probability that a fair coin will come up heads in any single toss is $\frac{1}{2}$ .

If the coin is tossed four times, then the probability that heads will appear three times is:

$P(x=3) = C(4,3) \left (\frac{1}{2} \right )}^{3}\left (\frac{1}{2} \right )}^{4-3}$

$= \frac{4!}{3! (4-3)!}\left (\frac{1}{2} \right )}^{3}\left (\frac{1}{2} \right )}^{1}$

$= \frac{4!}{3! 1!}\left (\frac{1}{2} \right )}^{4}$

$= 4\cdot \left (\frac{1}{16} \right )}=\frac{1}{4}$

The probability that heads will appear four times is:

$P(x=4) = C(4,4) \left (\frac{1}{2} \right )}^{4}\left (\frac{1}{2} \right )}^{4-4}$

$= \frac{4!}{4! (4-4)!}\left (\frac{1}{2} \right )}^{4}\left (\frac{1}{2} \right )}^{0}$

$= \frac{4!}{4! 0!}\left (\frac{1}{2} \right )}^{4}$

$= 1\cdot \left (\frac{1}{16} \right )}=\frac{1}{16}$

So the probability that heads will appear at least three times is

$P(x\geq 3) = \frac{1}{4} + \frac{1}{16}= \frac{5}{16}$

The second statement is irrelevant; the result of a previous trial has no influence on that of the current trial. It has already been established in the problem that the coin is fair.

The answer is that Statement 1 alone is sufficient to answer the question, but not Statement 2.

Compare your answer with the correct one above

Question

A monster is holding a brown paper bag filled with cookies. Inside the bag there are two types of cookies, chocolate chip and oatmeal raisin. If the monster pulls one cookie out at random, what is the probability that he pulls out a chocolate chip cookie.

1. There are a total of 24 cookies in the bag.

2. There are 8 more chocolate chip cookies in the bag than oatmeal raisin cookies.

Answer

We need both statements to get the answer.

Statement 1 alone is not enough. Having 24 total cookies does not tell us the proportions of each type of cookie. We could have 1 chocolate chip cookie and 23 oatmeal raisin, or vice versa.

Statement 2 alone is not sufficient either. Knowing that there are 8 more chocolate chip cookies than oatmeal raisin without knowing the total amount in the bag does not help. It could be 1 oatmeal raisin cookie and 9 chocolate chip, or it could be 100 oatmeal raisin cookies and 108 chocolate chip cookies (although that is a pretty big bag!)

Only by having both statements together can we find the right proportion. If we let be the number of chocolate chip cookies, and be the number of oatmeal raisin, we can set two equations and find the right ratio. We know that

from statement 1, and

from statement 2.

Substituting into the first equation we get

Therefore, we get 8 oatmeal raisin cookies, and 16 chocolate chip cookies. The answer is

$\frac{16}{24}=\frac{2}{3}$

or 66.67% chance that the monster pulls out a chocolate chip cookie.

Compare your answer with the correct one above

Question

Choose the answer that best describes sufficient data to solve the problem.

A student body is polled regarding their favorite campus food. Of those who answered "Pizza" half were male and half were female. Pizza received 24 votes from men.

I. Pizza received the votes of 1 out of every 5 men and 3 out of every 10 women.

II. The student population was found to be 60% male and 40% female

How many students altogether were polled?

Answer

Pizza received 24 votes from men. Since half of those who responded pizza were women, there must be 24 votes from women.

If 1 out of 5 men voted for pizza, this means we have $\ 24\cdot 5 = 120$ men. If 3 out of 10 women voted for pizza, we have $\frac{10}{3}\cdot 24 = 80$ women for a total of 200 people polled.

If our population is instead given as 60% male and 40% female, we still have 24 men and 24 women that voted for pizza, but while we can determine the ratio of people who voted, we cannot get a precise number. For example, while 80 women and 120 men works, we could also have 160 women and 240 men as our total poll (with 1 out of 10 men and 3 out of 20 women voting pizza).

Thus the first condition is sufficient while the second is not.

Compare your answer with the correct one above

Question

Find the probability.

Statement 1): Given that the last flips of a coin were heads, the next coin flip is heads.

Statement 2): Given that the last days were sunny skies, the next day is sunny.

Answer

This question deals with conditional probability.

Statement 1) consists of independent events. The past trials will not affect the future trials of the coin. The odds of flipping a coin and land on heads will be $\frac{1}{2}$ .

Statement 2) also consists of independent events. However, given that the previous 10 days were sunny does not necessarily indicate that the next day will guarantee that it will be a sunny day. The probability is unknown because the statement does not give any information about the percentage of a sunny sky on any particular day.

Compare your answer with the correct one above

Question

Jenkins has a box of 36 guitar picks. Each pick is one of three colors: blue, black, or orange.

I) There are 14 more orange picks than blue picks.

II) There are 12 black picks.

What are the odds of randomly selecting a black pick, followed by a blue pick, followed by an orange pick?

Answer

To find the probability of several events, we need to find the probability of each event separately. First, we need to know how many there are of each color pick.

Statement I relates the number of orange and blue picks.

Statement II tells us the number of black picks.

$P_{orange}+P_{black}+P_{blue}=36$

From Statement I:

$P_{orange}=P_{blue}+14$

From Statement II:

$P_{black}=12$

Using Statement I, Statement II, and the given info, we can substitute $P_{blue}+14$ from Statement I into Statement II for $P_{orange}$ and get the following:

$P_{blue}+14+P_{blue}+12=36$

We can then solve for $P_{blue}$ :

$2P_{blue}+26=36$

$2P_{blue}=10$

$P_{blue}=5$

Since we are told in Statement I that there are 14 more blue picks than orange picks, we can then find the number of orange picks:

$P_{orange}=P_{blue}+14$

$P_{orange}=5+14$

$P_{orange}=19$

So, to find the odds of multiple events, multiply the probability of each event together:

$\frac{12}{36}\cdot \frac{5}{35}\cdot \frac{19}{34}=\frac{1140}{42840}=\frac{19}{714}$

Compare your answer with the correct one above

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