Discrete Probability - GMAT Quantitative

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Question

Two fair dice are tossed. What is the probability that the difference of the two numbers will be 1?

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Answer

The difference of the two dice will be 1 in case any of the following outcomes occur:

$\left \{ 1,2\right \}, \left \{ 2,1\right \}, \left \{ 2,3\right \}, \left \{ 3,2\right \}, \left \{ 3,4\right \} ,\left \{ 4,3\right \},\left \{ 4,5\right \}, \left \{ 5,4\right \}, \left \{ 5,6\right \} ,\left \{ 6,5\right \}$

This is ten outcomes out of a possible 36, so this yields a probability of

$\frac{10}{36} = \frac{5}{18}$ .

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Question

Some balls are placed in a large box; the balls include one ball marked "10", two balls marked "9", and so forth up to ten balls marked "1". A ball is drawn at random.

is an integer between 1 and 10 inclusive. True or false: the probability that the ball will have the number marked on it is greater than $\frac{1}{10}$ .

Statement 1: is a prime integer.

Statement 2: $N \le 5$

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Answer

The total number of balls in the box will be

.

Since

$1+ 2 + 3 + ...+ N = \frac{N (N + 1)}{2}$ ,

it follows that the number of balls is

$1+ 2 + 3 + ...+ 10= \frac{10 (10 + 1)}{2} = \frac{10 \cdot 11}{2} = 55$ .

The frequencies out of 55 of each outcome from 1 to 10, in order, are as follows:

$\left \{ 10, 9, 8, 7, 6, 5, 4, 3, 2, 1\right \}$

Their respective probabilities are their frequencies divided by 55:

$\left \{ \frac{10}{55}, \frac{9}{55}, \frac{8}{55}, \frac{7}{55}, \frac{6}{55}, \frac{5}{55}, \frac{4}{55}, \frac{3}{55}, \frac{2}{55}, \frac{1}{55} \right \}$ .

The probability that the ball will be marked "5" is

$\frac{6}{55} \approx 0.109 > 0.1 = \frac{1}{10}$ ;

therefore, the probability that the ball will be marked with any given integer less than or equal to 5 will be greater than $\frac{1}{10}$ .

The probability that the ball will be marked "6" is

$\frac{5}{55} \approx 0.091 < 0.1 = \frac{1}{10}$ ;

therefore, the probability that the ball will be marked with any given integer greater than or equal to 6 will be less than $\frac{1}{10}$ .

Therefore, it suffices to know whether the number on the ball is less than or equal to 5. Statement 2 states that the number on the ball is less than or equal to 5, so it is sufficient to answer the question in the affirmative. Statement 1 is insufficient, since there are primes less than or equal to 5 - 2, 3, and 5 - and one prime greater than 5, which is 7.

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Question

Some balls are placed in a large box; the balls include one ball marked "10", two balls marked "9", and so forth up to ten balls marked "1". A ball is drawn at random.

is an integer between 1 and 10 inclusive. True or false: the probability that the ball will have the number marked on it is greater than $\frac{1}{10}$ .

Statement 1: is a perfect square integer.

Statement 2: $N \ge 5$

Tap to reveal answer

Answer

The total number of balls in the box will be

.

Since

$1+ 2 + 3 + ...+ N = \frac{N (N + 1)}{2}$ ,

it follows that the number of balls is

$1+ 2 + 3 + ...+ 10= \frac{10 (10 + 1)}{2} = \frac{10 \cdot 11}{2} = 55$ .

The frequencies out of 55 of each outcome from 1 to 10, in order, is as follows:

$\left \{ 10, 9, 8, 7, 6, 5, 4, 3, 2, 1\right \}$

Their respective probabilities are their frequencies divided by 55:

$\left \{ \frac{10}{55}, \frac{9}{55}, \frac{8}{55}, \frac{7}{55}, \frac{6}{55}, \frac{5}{55}, \frac{4}{55}, \frac{3}{55}, \frac{2}{55}, \frac{1}{55} \right \}$ .

The probability that the ball will be marked "5" is

$\frac{6}{55} \approx 0.109 > 0.1 = \frac{1}{10}$ ;

therefore, the probability that the ball will be marked with any given integer less than or equal to 5 will be greater than $\frac{1}{10}$ .

The probability that the ball will be marked "6" is

$\frac{5}{55} \approx 0.091 < 0.1 = \frac{1}{10}$ ;

therefore, the probability that the ball will be marked with any given integer greater than or equal to 6 will be less than $\frac{1}{10}$ .

Therefore, it suffices to know whether the number on the ball is less than or equal to 5.

Statement 1 alone is insufficient, since there are two perfect square integers from 1 to 5 (1 and 4) and one perfect square integer from 6 to 10 (9). Statement 2 alone is insufficient, since it is not clear whether the number on the ball is 5 or a number greater than 5. However, from the two statements together, it can be inferred that , and that the probability of drawing a ball with this number is $\frac{2}{55} < \frac{1}{10}$ .

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Question

Two fair dice are tossed. What is the probability that the difference of the two numbers will be 1?

Tap to reveal answer

Answer

The difference of the two dice will be 1 in case any of the following outcomes occur:

$\left \{ 1,2\right \}, \left \{ 2,1\right \}, \left \{ 2,3\right \}, \left \{ 3,2\right \}, \left \{ 3,4\right \} ,\left \{ 4,3\right \},\left \{ 4,5\right \}, \left \{ 5,4\right \}, \left \{ 5,6\right \} ,\left \{ 6,5\right \}$

This is ten outcomes out of a possible 36, so this yields a probability of

$\frac{10}{36} = \frac{5}{18}$ .

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Question

Some balls are placed in a large box; the balls include one ball marked "10", two balls marked "9", and so forth up to ten balls marked "1". A ball is drawn at random.

is an integer between 1 and 10 inclusive. True or false: the probability that the ball will have the number marked on it is greater than $\frac{1}{10}$ .

Statement 1: is a prime integer.

Statement 2: $N \le 5$

Tap to reveal answer

Answer

The total number of balls in the box will be

.

Since

$1+ 2 + 3 + ...+ N = \frac{N (N + 1)}{2}$ ,

it follows that the number of balls is

$1+ 2 + 3 + ...+ 10= \frac{10 (10 + 1)}{2} = \frac{10 \cdot 11}{2} = 55$ .

The frequencies out of 55 of each outcome from 1 to 10, in order, are as follows:

$\left \{ 10, 9, 8, 7, 6, 5, 4, 3, 2, 1\right \}$

Their respective probabilities are their frequencies divided by 55:

$\left \{ \frac{10}{55}, \frac{9}{55}, \frac{8}{55}, \frac{7}{55}, \frac{6}{55}, \frac{5}{55}, \frac{4}{55}, \frac{3}{55}, \frac{2}{55}, \frac{1}{55} \right \}$ .

The probability that the ball will be marked "5" is

$\frac{6}{55} \approx 0.109 > 0.1 = \frac{1}{10}$ ;

therefore, the probability that the ball will be marked with any given integer less than or equal to 5 will be greater than $\frac{1}{10}$ .

The probability that the ball will be marked "6" is

$\frac{5}{55} \approx 0.091 < 0.1 = \frac{1}{10}$ ;

therefore, the probability that the ball will be marked with any given integer greater than or equal to 6 will be less than $\frac{1}{10}$ .

Therefore, it suffices to know whether the number on the ball is less than or equal to 5. Statement 2 states that the number on the ball is less than or equal to 5, so it is sufficient to answer the question in the affirmative. Statement 1 is insufficient, since there are primes less than or equal to 5 - 2, 3, and 5 - and one prime greater than 5, which is 7.

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Question

Some balls are placed in a large box; the balls include one ball marked "10", two balls marked "9", and so forth up to ten balls marked "1". A ball is drawn at random.

is an integer between 1 and 10 inclusive. True or false: the probability that the ball will have the number marked on it is greater than $\frac{1}{10}$ .

Statement 1: is a perfect square integer.

Statement 2: $N \ge 5$

Tap to reveal answer

Answer

The total number of balls in the box will be

.

Since

$1+ 2 + 3 + ...+ N = \frac{N (N + 1)}{2}$ ,

it follows that the number of balls is

$1+ 2 + 3 + ...+ 10= \frac{10 (10 + 1)}{2} = \frac{10 \cdot 11}{2} = 55$ .

The frequencies out of 55 of each outcome from 1 to 10, in order, is as follows:

$\left \{ 10, 9, 8, 7, 6, 5, 4, 3, 2, 1\right \}$

Their respective probabilities are their frequencies divided by 55:

$\left \{ \frac{10}{55}, \frac{9}{55}, \frac{8}{55}, \frac{7}{55}, \frac{6}{55}, \frac{5}{55}, \frac{4}{55}, \frac{3}{55}, \frac{2}{55}, \frac{1}{55} \right \}$ .

The probability that the ball will be marked "5" is

$\frac{6}{55} \approx 0.109 > 0.1 = \frac{1}{10}$ ;

therefore, the probability that the ball will be marked with any given integer less than or equal to 5 will be greater than $\frac{1}{10}$ .

The probability that the ball will be marked "6" is

$\frac{5}{55} \approx 0.091 < 0.1 = \frac{1}{10}$ ;

therefore, the probability that the ball will be marked with any given integer greater than or equal to 6 will be less than $\frac{1}{10}$ .

Therefore, it suffices to know whether the number on the ball is less than or equal to 5.

Statement 1 alone is insufficient, since there are two perfect square integers from 1 to 5 (1 and 4) and one perfect square integer from 6 to 10 (9). Statement 2 alone is insufficient, since it is not clear whether the number on the ball is 5 or a number greater than 5. However, from the two statements together, it can be inferred that , and that the probability of drawing a ball with this number is $\frac{2}{55} < \frac{1}{10}$ .

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Question

Two fair dice are tossed. What is the probability that the difference of the two numbers will be 1?

Tap to reveal answer

Answer

The difference of the two dice will be 1 in case any of the following outcomes occur:

$\left \{ 1,2\right \}, \left \{ 2,1\right \}, \left \{ 2,3\right \}, \left \{ 3,2\right \}, \left \{ 3,4\right \} ,\left \{ 4,3\right \},\left \{ 4,5\right \}, \left \{ 5,4\right \}, \left \{ 5,6\right \} ,\left \{ 6,5\right \}$

This is ten outcomes out of a possible 36, so this yields a probability of

$\frac{10}{36} = \frac{5}{18}$ .

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Question

Some balls are placed in a large box; the balls include one ball marked "10", two balls marked "9", and so forth up to ten balls marked "1". A ball is drawn at random.

is an integer between 1 and 10 inclusive. True or false: the probability that the ball will have the number marked on it is greater than $\frac{1}{10}$ .

Statement 1: is a prime integer.

Statement 2: $N \le 5$

Tap to reveal answer

Answer

The total number of balls in the box will be

.

Since

$1+ 2 + 3 + ...+ N = \frac{N (N + 1)}{2}$ ,

it follows that the number of balls is

$1+ 2 + 3 + ...+ 10= \frac{10 (10 + 1)}{2} = \frac{10 \cdot 11}{2} = 55$ .

The frequencies out of 55 of each outcome from 1 to 10, in order, are as follows:

$\left \{ 10, 9, 8, 7, 6, 5, 4, 3, 2, 1\right \}$

Their respective probabilities are their frequencies divided by 55:

$\left \{ \frac{10}{55}, \frac{9}{55}, \frac{8}{55}, \frac{7}{55}, \frac{6}{55}, \frac{5}{55}, \frac{4}{55}, \frac{3}{55}, \frac{2}{55}, \frac{1}{55} \right \}$ .

The probability that the ball will be marked "5" is

$\frac{6}{55} \approx 0.109 > 0.1 = \frac{1}{10}$ ;

therefore, the probability that the ball will be marked with any given integer less than or equal to 5 will be greater than $\frac{1}{10}$ .

The probability that the ball will be marked "6" is

$\frac{5}{55} \approx 0.091 < 0.1 = \frac{1}{10}$ ;

therefore, the probability that the ball will be marked with any given integer greater than or equal to 6 will be less than $\frac{1}{10}$ .

Therefore, it suffices to know whether the number on the ball is less than or equal to 5. Statement 2 states that the number on the ball is less than or equal to 5, so it is sufficient to answer the question in the affirmative. Statement 1 is insufficient, since there are primes less than or equal to 5 - 2, 3, and 5 - and one prime greater than 5, which is 7.

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Question

Some balls are placed in a large box; the balls include one ball marked "10", two balls marked "9", and so forth up to ten balls marked "1". A ball is drawn at random.

is an integer between 1 and 10 inclusive. True or false: the probability that the ball will have the number marked on it is greater than $\frac{1}{10}$ .

Statement 1: is a perfect square integer.

Statement 2: $N \ge 5$

Tap to reveal answer

Answer

The total number of balls in the box will be

.

Since

$1+ 2 + 3 + ...+ N = \frac{N (N + 1)}{2}$ ,

it follows that the number of balls is

$1+ 2 + 3 + ...+ 10= \frac{10 (10 + 1)}{2} = \frac{10 \cdot 11}{2} = 55$ .

The frequencies out of 55 of each outcome from 1 to 10, in order, is as follows:

$\left \{ 10, 9, 8, 7, 6, 5, 4, 3, 2, 1\right \}$

Their respective probabilities are their frequencies divided by 55:

$\left \{ \frac{10}{55}, \frac{9}{55}, \frac{8}{55}, \frac{7}{55}, \frac{6}{55}, \frac{5}{55}, \frac{4}{55}, \frac{3}{55}, \frac{2}{55}, \frac{1}{55} \right \}$ .

The probability that the ball will be marked "5" is

$\frac{6}{55} \approx 0.109 > 0.1 = \frac{1}{10}$ ;

therefore, the probability that the ball will be marked with any given integer less than or equal to 5 will be greater than $\frac{1}{10}$ .

The probability that the ball will be marked "6" is

$\frac{5}{55} \approx 0.091 < 0.1 = \frac{1}{10}$ ;

therefore, the probability that the ball will be marked with any given integer greater than or equal to 6 will be less than $\frac{1}{10}$ .

Therefore, it suffices to know whether the number on the ball is less than or equal to 5.

Statement 1 alone is insufficient, since there are two perfect square integers from 1 to 5 (1 and 4) and one perfect square integer from 6 to 10 (9). Statement 2 alone is insufficient, since it is not clear whether the number on the ball is 5 or a number greater than 5. However, from the two statements together, it can be inferred that , and that the probability of drawing a ball with this number is $\frac{2}{55} < \frac{1}{10}$ .

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Question

Two fair dice are tossed. What is the probability that the difference of the two numbers will be 1?

Tap to reveal answer

Answer

The difference of the two dice will be 1 in case any of the following outcomes occur:

$\left \{ 1,2\right \}, \left \{ 2,1\right \}, \left \{ 2,3\right \}, \left \{ 3,2\right \}, \left \{ 3,4\right \} ,\left \{ 4,3\right \},\left \{ 4,5\right \}, \left \{ 5,4\right \}, \left \{ 5,6\right \} ,\left \{ 6,5\right \}$

This is ten outcomes out of a possible 36, so this yields a probability of

$\frac{10}{36} = \frac{5}{18}$ .

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Question

Some balls are placed in a large box; the balls include one ball marked "10", two balls marked "9", and so forth up to ten balls marked "1". A ball is drawn at random.

is an integer between 1 and 10 inclusive. True or false: the probability that the ball will have the number marked on it is greater than $\frac{1}{10}$ .

Statement 1: is a prime integer.

Statement 2: $N \le 5$

Tap to reveal answer

Answer

The total number of balls in the box will be

.

Since

$1+ 2 + 3 + ...+ N = \frac{N (N + 1)}{2}$ ,

it follows that the number of balls is

$1+ 2 + 3 + ...+ 10= \frac{10 (10 + 1)}{2} = \frac{10 \cdot 11}{2} = 55$ .

The frequencies out of 55 of each outcome from 1 to 10, in order, are as follows:

$\left \{ 10, 9, 8, 7, 6, 5, 4, 3, 2, 1\right \}$

Their respective probabilities are their frequencies divided by 55:

$\left \{ \frac{10}{55}, \frac{9}{55}, \frac{8}{55}, \frac{7}{55}, \frac{6}{55}, \frac{5}{55}, \frac{4}{55}, \frac{3}{55}, \frac{2}{55}, \frac{1}{55} \right \}$ .

The probability that the ball will be marked "5" is

$\frac{6}{55} \approx 0.109 > 0.1 = \frac{1}{10}$ ;

therefore, the probability that the ball will be marked with any given integer less than or equal to 5 will be greater than $\frac{1}{10}$ .

The probability that the ball will be marked "6" is

$\frac{5}{55} \approx 0.091 < 0.1 = \frac{1}{10}$ ;

therefore, the probability that the ball will be marked with any given integer greater than or equal to 6 will be less than $\frac{1}{10}$ .

Therefore, it suffices to know whether the number on the ball is less than or equal to 5. Statement 2 states that the number on the ball is less than or equal to 5, so it is sufficient to answer the question in the affirmative. Statement 1 is insufficient, since there are primes less than or equal to 5 - 2, 3, and 5 - and one prime greater than 5, which is 7.

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Question

Some balls are placed in a large box; the balls include one ball marked "10", two balls marked "9", and so forth up to ten balls marked "1". A ball is drawn at random.

is an integer between 1 and 10 inclusive. True or false: the probability that the ball will have the number marked on it is greater than $\frac{1}{10}$ .

Statement 1: is a perfect square integer.

Statement 2: $N \ge 5$

Tap to reveal answer

Answer

The total number of balls in the box will be

.

Since

$1+ 2 + 3 + ...+ N = \frac{N (N + 1)}{2}$ ,

it follows that the number of balls is

$1+ 2 + 3 + ...+ 10= \frac{10 (10 + 1)}{2} = \frac{10 \cdot 11}{2} = 55$ .

The frequencies out of 55 of each outcome from 1 to 10, in order, is as follows:

$\left \{ 10, 9, 8, 7, 6, 5, 4, 3, 2, 1\right \}$

Their respective probabilities are their frequencies divided by 55:

$\left \{ \frac{10}{55}, \frac{9}{55}, \frac{8}{55}, \frac{7}{55}, \frac{6}{55}, \frac{5}{55}, \frac{4}{55}, \frac{3}{55}, \frac{2}{55}, \frac{1}{55} \right \}$ .

The probability that the ball will be marked "5" is

$\frac{6}{55} \approx 0.109 > 0.1 = \frac{1}{10}$ ;

therefore, the probability that the ball will be marked with any given integer less than or equal to 5 will be greater than $\frac{1}{10}$ .

The probability that the ball will be marked "6" is

$\frac{5}{55} \approx 0.091 < 0.1 = \frac{1}{10}$ ;

therefore, the probability that the ball will be marked with any given integer greater than or equal to 6 will be less than $\frac{1}{10}$ .

Therefore, it suffices to know whether the number on the ball is less than or equal to 5.

Statement 1 alone is insufficient, since there are two perfect square integers from 1 to 5 (1 and 4) and one perfect square integer from 6 to 10 (9). Statement 2 alone is insufficient, since it is not clear whether the number on the ball is 5 or a number greater than 5. However, from the two statements together, it can be inferred that , and that the probability of drawing a ball with this number is $\frac{2}{55} < \frac{1}{10}$ .

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Question

Two fair dice are tossed. What is the probability that the difference of the two numbers will be 1?

Tap to reveal answer

Answer

The difference of the two dice will be 1 in case any of the following outcomes occur:

$\left \{ 1,2\right \}, \left \{ 2,1\right \}, \left \{ 2,3\right \}, \left \{ 3,2\right \}, \left \{ 3,4\right \} ,\left \{ 4,3\right \},\left \{ 4,5\right \}, \left \{ 5,4\right \}, \left \{ 5,6\right \} ,\left \{ 6,5\right \}$

This is ten outcomes out of a possible 36, so this yields a probability of

$\frac{10}{36} = \frac{5}{18}$ .

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Question

Some balls are placed in a large box; the balls include one ball marked "10", two balls marked "9", and so forth up to ten balls marked "1". A ball is drawn at random.

is an integer between 1 and 10 inclusive. True or false: the probability that the ball will have the number marked on it is greater than $\frac{1}{10}$ .

Statement 1: is a prime integer.

Statement 2: $N \le 5$

Tap to reveal answer

Answer

The total number of balls in the box will be

.

Since

$1+ 2 + 3 + ...+ N = \frac{N (N + 1)}{2}$ ,

it follows that the number of balls is

$1+ 2 + 3 + ...+ 10= \frac{10 (10 + 1)}{2} = \frac{10 \cdot 11}{2} = 55$ .

The frequencies out of 55 of each outcome from 1 to 10, in order, are as follows:

$\left \{ 10, 9, 8, 7, 6, 5, 4, 3, 2, 1\right \}$

Their respective probabilities are their frequencies divided by 55:

$\left \{ \frac{10}{55}, \frac{9}{55}, \frac{8}{55}, \frac{7}{55}, \frac{6}{55}, \frac{5}{55}, \frac{4}{55}, \frac{3}{55}, \frac{2}{55}, \frac{1}{55} \right \}$ .

The probability that the ball will be marked "5" is

$\frac{6}{55} \approx 0.109 > 0.1 = \frac{1}{10}$ ;

therefore, the probability that the ball will be marked with any given integer less than or equal to 5 will be greater than $\frac{1}{10}$ .

The probability that the ball will be marked "6" is

$\frac{5}{55} \approx 0.091 < 0.1 = \frac{1}{10}$ ;

therefore, the probability that the ball will be marked with any given integer greater than or equal to 6 will be less than $\frac{1}{10}$ .

Therefore, it suffices to know whether the number on the ball is less than or equal to 5. Statement 2 states that the number on the ball is less than or equal to 5, so it is sufficient to answer the question in the affirmative. Statement 1 is insufficient, since there are primes less than or equal to 5 - 2, 3, and 5 - and one prime greater than 5, which is 7.

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Question

Some balls are placed in a large box; the balls include one ball marked "10", two balls marked "9", and so forth up to ten balls marked "1". A ball is drawn at random.

is an integer between 1 and 10 inclusive. True or false: the probability that the ball will have the number marked on it is greater than $\frac{1}{10}$ .

Statement 1: is a perfect square integer.

Statement 2: $N \ge 5$

Tap to reveal answer

Answer

The total number of balls in the box will be

.

Since

$1+ 2 + 3 + ...+ N = \frac{N (N + 1)}{2}$ ,

it follows that the number of balls is

$1+ 2 + 3 + ...+ 10= \frac{10 (10 + 1)}{2} = \frac{10 \cdot 11}{2} = 55$ .

The frequencies out of 55 of each outcome from 1 to 10, in order, is as follows:

$\left \{ 10, 9, 8, 7, 6, 5, 4, 3, 2, 1\right \}$

Their respective probabilities are their frequencies divided by 55:

$\left \{ \frac{10}{55}, \frac{9}{55}, \frac{8}{55}, \frac{7}{55}, \frac{6}{55}, \frac{5}{55}, \frac{4}{55}, \frac{3}{55}, \frac{2}{55}, \frac{1}{55} \right \}$ .

The probability that the ball will be marked "5" is

$\frac{6}{55} \approx 0.109 > 0.1 = \frac{1}{10}$ ;

therefore, the probability that the ball will be marked with any given integer less than or equal to 5 will be greater than $\frac{1}{10}$ .

The probability that the ball will be marked "6" is

$\frac{5}{55} \approx 0.091 < 0.1 = \frac{1}{10}$ ;

therefore, the probability that the ball will be marked with any given integer greater than or equal to 6 will be less than $\frac{1}{10}$ .

Therefore, it suffices to know whether the number on the ball is less than or equal to 5.

Statement 1 alone is insufficient, since there are two perfect square integers from 1 to 5 (1 and 4) and one perfect square integer from 6 to 10 (9). Statement 2 alone is insufficient, since it is not clear whether the number on the ball is 5 or a number greater than 5. However, from the two statements together, it can be inferred that , and that the probability of drawing a ball with this number is $\frac{2}{55} < \frac{1}{10}$ .

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Question

Two fair dice are tossed. What is the probability that the difference of the two numbers will be 1?

Tap to reveal answer

Answer

The difference of the two dice will be 1 in case any of the following outcomes occur:

$\left \{ 1,2\right \}, \left \{ 2,1\right \}, \left \{ 2,3\right \}, \left \{ 3,2\right \}, \left \{ 3,4\right \} ,\left \{ 4,3\right \},\left \{ 4,5\right \}, \left \{ 5,4\right \}, \left \{ 5,6\right \} ,\left \{ 6,5\right \}$

This is ten outcomes out of a possible 36, so this yields a probability of

$\frac{10}{36} = \frac{5}{18}$ .

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Question

Some balls are placed in a large box; the balls include one ball marked "10", two balls marked "9", and so forth up to ten balls marked "1". A ball is drawn at random.

is an integer between 1 and 10 inclusive. True or false: the probability that the ball will have the number marked on it is greater than $\frac{1}{10}$ .

Statement 1: is a prime integer.

Statement 2: $N \le 5$

Tap to reveal answer

Answer

The total number of balls in the box will be

.

Since

$1+ 2 + 3 + ...+ N = \frac{N (N + 1)}{2}$ ,

it follows that the number of balls is

$1+ 2 + 3 + ...+ 10= \frac{10 (10 + 1)}{2} = \frac{10 \cdot 11}{2} = 55$ .

The frequencies out of 55 of each outcome from 1 to 10, in order, are as follows:

$\left \{ 10, 9, 8, 7, 6, 5, 4, 3, 2, 1\right \}$

Their respective probabilities are their frequencies divided by 55:

$\left \{ \frac{10}{55}, \frac{9}{55}, \frac{8}{55}, \frac{7}{55}, \frac{6}{55}, \frac{5}{55}, \frac{4}{55}, \frac{3}{55}, \frac{2}{55}, \frac{1}{55} \right \}$ .

The probability that the ball will be marked "5" is

$\frac{6}{55} \approx 0.109 > 0.1 = \frac{1}{10}$ ;

therefore, the probability that the ball will be marked with any given integer less than or equal to 5 will be greater than $\frac{1}{10}$ .

The probability that the ball will be marked "6" is

$\frac{5}{55} \approx 0.091 < 0.1 = \frac{1}{10}$ ;

therefore, the probability that the ball will be marked with any given integer greater than or equal to 6 will be less than $\frac{1}{10}$ .

Therefore, it suffices to know whether the number on the ball is less than or equal to 5. Statement 2 states that the number on the ball is less than or equal to 5, so it is sufficient to answer the question in the affirmative. Statement 1 is insufficient, since there are primes less than or equal to 5 - 2, 3, and 5 - and one prime greater than 5, which is 7.

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Question

Some balls are placed in a large box; the balls include one ball marked "10", two balls marked "9", and so forth up to ten balls marked "1". A ball is drawn at random.

is an integer between 1 and 10 inclusive. True or false: the probability that the ball will have the number marked on it is greater than $\frac{1}{10}$ .

Statement 1: is a perfect square integer.

Statement 2: $N \ge 5$

Tap to reveal answer

Answer

The total number of balls in the box will be

.

Since

$1+ 2 + 3 + ...+ N = \frac{N (N + 1)}{2}$ ,

it follows that the number of balls is

$1+ 2 + 3 + ...+ 10= \frac{10 (10 + 1)}{2} = \frac{10 \cdot 11}{2} = 55$ .

The frequencies out of 55 of each outcome from 1 to 10, in order, is as follows:

$\left \{ 10, 9, 8, 7, 6, 5, 4, 3, 2, 1\right \}$

Their respective probabilities are their frequencies divided by 55:

$\left \{ \frac{10}{55}, \frac{9}{55}, \frac{8}{55}, \frac{7}{55}, \frac{6}{55}, \frac{5}{55}, \frac{4}{55}, \frac{3}{55}, \frac{2}{55}, \frac{1}{55} \right \}$ .

The probability that the ball will be marked "5" is

$\frac{6}{55} \approx 0.109 > 0.1 = \frac{1}{10}$ ;

therefore, the probability that the ball will be marked with any given integer less than or equal to 5 will be greater than $\frac{1}{10}$ .

The probability that the ball will be marked "6" is

$\frac{5}{55} \approx 0.091 < 0.1 = \frac{1}{10}$ ;

therefore, the probability that the ball will be marked with any given integer greater than or equal to 6 will be less than $\frac{1}{10}$ .

Therefore, it suffices to know whether the number on the ball is less than or equal to 5.

Statement 1 alone is insufficient, since there are two perfect square integers from 1 to 5 (1 and 4) and one perfect square integer from 6 to 10 (9). Statement 2 alone is insufficient, since it is not clear whether the number on the ball is 5 or a number greater than 5. However, from the two statements together, it can be inferred that , and that the probability of drawing a ball with this number is $\frac{2}{55} < \frac{1}{10}$ .

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Question

Two fair dice are tossed. What is the probability that the difference of the two numbers will be 1?

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Answer

The difference of the two dice will be 1 in case any of the following outcomes occur:

$\left \{ 1,2\right \}, \left \{ 2,1\right \}, \left \{ 2,3\right \}, \left \{ 3,2\right \}, \left \{ 3,4\right \} ,\left \{ 4,3\right \},\left \{ 4,5\right \}, \left \{ 5,4\right \}, \left \{ 5,6\right \} ,\left \{ 6,5\right \}$

This is ten outcomes out of a possible 36, so this yields a probability of

$\frac{10}{36} = \frac{5}{18}$ .

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Question

Some balls are placed in a large box; the balls include one ball marked "10", two balls marked "9", and so forth up to ten balls marked "1". A ball is drawn at random.

is an integer between 1 and 10 inclusive. True or false: the probability that the ball will have the number marked on it is greater than $\frac{1}{10}$ .

Statement 1: is a prime integer.

Statement 2: $N \le 5$

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Answer

The total number of balls in the box will be

.

Since

$1+ 2 + 3 + ...+ N = \frac{N (N + 1)}{2}$ ,

it follows that the number of balls is

$1+ 2 + 3 + ...+ 10= \frac{10 (10 + 1)}{2} = \frac{10 \cdot 11}{2} = 55$ .

The frequencies out of 55 of each outcome from 1 to 10, in order, are as follows:

$\left \{ 10, 9, 8, 7, 6, 5, 4, 3, 2, 1\right \}$

Their respective probabilities are their frequencies divided by 55:

$\left \{ \frac{10}{55}, \frac{9}{55}, \frac{8}{55}, \frac{7}{55}, \frac{6}{55}, \frac{5}{55}, \frac{4}{55}, \frac{3}{55}, \frac{2}{55}, \frac{1}{55} \right \}$ .

The probability that the ball will be marked "5" is

$\frac{6}{55} \approx 0.109 > 0.1 = \frac{1}{10}$ ;

therefore, the probability that the ball will be marked with any given integer less than or equal to 5 will be greater than $\frac{1}{10}$ .

The probability that the ball will be marked "6" is

$\frac{5}{55} \approx 0.091 < 0.1 = \frac{1}{10}$ ;

therefore, the probability that the ball will be marked with any given integer greater than or equal to 6 will be less than $\frac{1}{10}$ .

Therefore, it suffices to know whether the number on the ball is less than or equal to 5. Statement 2 states that the number on the ball is less than or equal to 5, so it is sufficient to answer the question in the affirmative. Statement 1 is insufficient, since there are primes less than or equal to 5 - 2, 3, and 5 - and one prime greater than 5, which is 7.

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Discrete Probability - GMAT Quantitative

Two fair dice are tossed. What is the probability that the difference of the two numbers will be 1?

The difference of the two dice will be 1 in case any of the following outcomes occur: This is ten outcomes out of a possible 36, so this yields a probability of .

Two fair dice are tossed. What is the probability that the difference of the two numbers will be 1?

The difference of the two dice will be 1 in case any of the following outcomes occur: This is ten outcomes out of a possible 36, so this yields a probability of .

Two fair dice are tossed. What is the probability that the difference of the two numbers will be 1?

The difference of the two dice will be 1 in case any of the following outcomes occur: This is ten outcomes out of a possible 36, so this yields a probability of .

Two fair dice are tossed. What is the probability that the difference of the two numbers will be 1?

The difference of the two dice will be 1 in case any of the following outcomes occur: This is ten outcomes out of a possible 36, so this yields a probability of .

Two fair dice are tossed. What is the probability that the difference of the two numbers will be 1?

The difference of the two dice will be 1 in case any of the following outcomes occur: This is ten outcomes out of a possible 36, so this yields a probability of .

Two fair dice are tossed. What is the probability that the difference of the two numbers will be 1?

The difference of the two dice will be 1 in case any of the following outcomes occur: This is ten outcomes out of a possible 36, so this yields a probability of .

Two fair dice are tossed. What is the probability that the difference of the two numbers will be 1?

The difference of the two dice will be 1 in case any of the following outcomes occur: This is ten outcomes out of a possible 36, so this yields a probability of .