Basic Single-Event Probability Practice Questions — GMAT

Question 1Question →

A wooden box contains 15 identical tokens, each marked with a distinct integer from $1$ to $15$ , inclusive. If one token is drawn at random from the box, what is the probability that the integer on the drawn token is a prime number?

\frac{1}{3}

\frac{2}{5}

\frac{7}{15}

\frac{8}{15}

\frac{2}{3}

Show answer & explanation

Answer: $\frac{2}{5}$

Answer

The probability that the integer on the drawn token is a prime number is

\frac{2}{5}

.

The total number of possible outcomes when choosing one token from

15

tokens is

15

. The prime numbers between

1

and

15

, inclusive, are

2, 3, 5, 7, 11,

and

13

. There are

6

favorable outcomes. The probability is therefore

\frac{6}{15}

, which simplifies to

\frac{2}{5}

.

Step-by-Step Solution

1

Determine the total number of possible outcomes (the denominator).

The total number of tokens is

15

, so

N = 15

.

The sample space consists of all integers from

1

to

15

, inclusive.

2

Identify and count all prime numbers in the set from

1

to

15

, inclusive.

The prime numbers in this set are

2, 3, 5, 7, 11,

and

13

. Thus, there are

6

prime numbers.

A prime number is an integer strictly greater than

1

that has exactly two distinct positive divisors:

1

and itself. Note that

1

is not prime.

3

Calculate the basic probability

P(\text{Prime}) = \frac{\text{Number of Favorable Outcomes}}{\text{Total Number of Possible Outcomes}}

.

P(\text{Prime}) = \frac{6}{15} = \frac{2}{5}

.

Dividing the favorable count

6

by total count

15

and simplifying by dividing numerator and denominator by

3

yields

\frac{2}{5}

.

Key Concept

Basic Single-Event Probability and Definition of Prime Numbers

Estimated Time:45s

Question 2Question →

An integer $n$ is selected at random from the set $\{1, 2, 3, \dots, 500\}$ . What is the probability that $n$ is a perfect square that leaves a remainder of $1$ when divided by $3$ ?

Show answer & explanation

Answer: 0.03

Answer

0.03

The sample space consists of 500 equally likely outcomes. Perfect squares in this range are generated by integers

k

from 1 to 22. Evaluating

k^2 \pmod 3

shows that

k^2

has a remainder of 1 when divided by 3 for all integers

k

not divisible by 3. Excluding the 7 multiples of 3 from the 22 possible values of

k

leaves 15 favorable integers. Dividing the 15 favorable outcomes by the 500 total outcomes gives a probability of 0.03.

Step-by-Step Solution

1

Determine total possible outcomes.

The total number of integers in the set

\{1, 2, 3, \dots, 500\}

is

N = 500

.

Each integer in the set is equally likely to be selected.

2

Find the range of square roots for perfect squares in the set.

Since

22^2 = 484 \le 500

and

23^2 = 529 > 500

,

n = k^2

for

k \in \{1, 2, \dots, 22\}

.

There are 22 perfect squares in the given sample space.

3

Apply remainder constraints to

k^2

.

A square

k^2

leaves a remainder of

1

when divided by

3

if and only if

k

is not divisible by

3

.

If

k

is a multiple of 3,

k^2

is divisible by 3 (remainder 0). If

k

is not a multiple of 3,

k \equiv 1

or

2 \pmod 3

, so

k^2 \equiv 1 \pmod 3

.

4

Count the number of favorable outcomes.

The multiples of 3 up to 22 are

3, 6, 9, 12, 15, 18, 21

(7 numbers). Thus, there are

22 - 7 = 15

values of

k

that satisfy the condition.

Favorable outcomes correspond exactly to integers

k \in \{1, 2, \dots, 22\}

that are not multiples of 3.

5

Compute the single-event probability.

Probability

= \frac{15}{500} = 0.03

.

The probability of a single event is the number of favorable outcomes divided by the total number of outcomes.

Key Concept

Basic Single-Event Probability with Number Properties

Question 3Question →

A computer program randomly selects a single positive integer $x$ from the set $\{1, 2, 3, \dots, 300\}$ , where each integer has an equal probability of being chosen. What is the probability that $x$ is divisible by either 6 or 9, but not by 4?

Show answer & explanation

Answer: 0.14

Answer

The probability that

x

is divisible by 6 or 9, but not by 4, is

0.14

(or

\frac{7}{50}

).

To calculate the probability, determine the ratio of favorable outcomes to total outcomes (300). First, count integers divisible by 6 (50) or 9 (33). Subtracting the 16 multiples of 18 counted in both yields 67 integers. Next, identify how many of these 67 integers are divisible by 4: there are 25 multiples of 12 and 8 multiples of 36 (all 8 of which are already included in the multiples of 12), resulting in 25 integers divisible by 4. Subtracting 25 from 67 leaves 42 favorable integers. The probability is

42 / 300 = 0.14

.

Step-by-Step Solution

1

Determine the total sample space size.

The total number of equally likely outcomes is

|S| = 300

.

Single-event probability requires dividing the count of favorable outcomes by the total number of outcomes in the set.

2

Calculate the count of integers divisible by 6 or 9.

Count =

50 + 33 - 16 = 67

.

Apply the Principle of Inclusion-Exclusion: sum the multiples of 6 and 9, then subtract the multiples of

\text{lcm}(6, 9) = 18

to eliminate double counting.

3

Exclude integers from this set that are divisible by 4.

Count of multiples of 4 to exclude =

25 + 8 - 8 = 25

.

Integers divisible by both 6 and 4 are multiples of

\text{lcm}(6, 4) = 12

. Integers divisible by both 9 and 4 are multiples of

\text{lcm}(9, 4) = 36

. Using inclusion-exclusion,

25 + 8 - 8 = 25

integers must be excluded.

4

Find the number of favorable outcomes and compute the probability.

Favorable outcomes

|E| = 67 - 25 = 42

. Probability

P(E) = \frac{42}{300} = 0.14

.

Divide the favorable set size by the total sample space size.

Key Concept

Basic Single-Event Probability with Set Inclusion-Exclusion and Number Properties

Question 4Question →

A container holds 60 cards, each marked with a distinct integer from 1 to 60, inclusive. If one card is selected at random, the probability that the integer on the selected card is divisible by 5, but divisible by neither 4 nor 6, can be expressed as a simplified fraction $\frac{a}{b}$ , where $a$ and $b$ are positive integers with no common factors greater than 1. What is the value of $a + b$ ?

Show answer & explanation

Answer: 17

Answer

The value of

a + b

is 17.

There are 60 equally likely outcomes. Out of 12 multiples of 5, exactly 4 integers (20, 30, 40, and 60) are divisible by 4 or 6. Thus, 12 - 4 = 8 integers satisfy the condition of being divisible by 5 but neither 4 nor 6. The single-event probability is 8/60 = 2/15. In simplest form, a = 2 and b = 15, yielding a + b = 17.

Step-by-Step Solution

1

Determine total number of possible outcomes.

Total outcomes

N = 60

.

Each of the 60 cards is equally likely to be selected.

2

Identify all multiples of 5 in the set

\{1, 2, \dots, 60\}

.

Multiples of 5:

\{5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60\}

, containing

\frac{60}{5} = 12

integers.

These represent all potential candidates before applying exclusion criteria.

3

Identify multiples of 5 that are also divisible by 4 or by 6.

Multiples of 5 divisible by 4:

\text{LCM}(5,4) = 20 \rightarrow \{20, 40, 60\}

(3 integers). Multiples of 5 divisible by 6:

\text{LCM}(5,6) = 30 \rightarrow \{30, 60\}

(2 integers). Overlap (divisible by both 4 and 6):

\text{LCM}(5,4,6) = 60 \rightarrow \{60\}

(1 integer).

An integer is divisible by both 5 and 4 if it is a multiple of 20, and by both 5 and 6 if it is a multiple of 30.

4

Apply the inclusion-exclusion principle to find the count of invalid multiples of 5.

Invalid multiples =

3 + 2 - 1 = 4

integers (

\{20, 30, 40, 60\}

). Favorable outcomes =

12 - 4 = 8

integers (

\{5, 10, 15, 25, 35, 45, 50, 55\}

).

The integer 60 is divisible by both 4 and 6 and must not be double-subtracted.

5

Calculate the probability, express it as a simplified fraction

\frac{a}{b}

, and compute

a + b

.

Probability

P = \frac{8}{60} = \frac{2}{15}

. Thus,

a = 2

,

b = 15

, and

a + b = 2 + 15 = 17

.

Dividing the numerator and denominator by 4 simplifies

\frac{8}{60}

to

\frac{2}{15}

, where

\text{gcd}(2, 15) = 1

.

Key Concept

Basic Single-Event Probability with Multiple Divisibility Constraints

Question 5Question →

A box contains 20 cards numbered sequentially from 1 through 20, inclusive. If one card is drawn at random from the box, what is the probability that the number on the drawn card is a prime number? Express your answer as a decimal.

Show answer & explanation

Answer: 0.4

Answer

0.4

The total number of possible outcomes when selecting one card from 20 is 20. The prime numbers between 1 and 20 inclusive are 2, 3, 5, 7, 11, 13, 17, and 19, giving 8 favorable outcomes (remembering that 1 is not prime). The single-event probability is calculated by dividing the number of favorable outcomes by the total number of outcomes, yielding

8 / 20 = 0.4

.

Step-by-Step Solution

1

Determine the total number of possible outcomes in the sample space.

The sample space consists of 20 equally likely outcomes (integers 1 through 20).

Calculating single-event probability requires establishing the size of the total outcome space

N

.

2

Count the number of prime numbers in the set

\{1, 2, \dots, 20\}

.

There are 8 prime numbers:

\{2, 3, 5, 7, 11, 13, 17, 19\}

.

By definition, a prime number is an integer greater than 1 with exactly two positive divisors: 1 and itself. Thus, 1 is excluded.

3

Compute the probability using

P(E) = \frac{\text{Favorable Outcomes}}{\text{Total Outcomes}}

.

P = \frac{8}{20} = 0.4

.

Directly apply the basic single-event probability formula.

Key Concept

Basic Single-Event Probability and Prime Number Identification

Estimated Time:45s

Question 6Question →

In a game, a fair wheel lands on a single integer from $1$ to $100$ , inclusive, with each integer being equally likely to be selected. What is the probability that the integer selected is either a prime number or leaves a remainder of $1$ when divided by $5$ , but not both?

\frac{7}{20}

\frac{17}{50}

\frac{2}{5}

\frac{9}{20}

\frac{1}{20}

Show answer & explanation

Answer: $\frac{7}{20}$

Answer

The correct answer is 7/20.

The total number of possible outcomes is 100. There are 25 prime numbers in the range 1 to 100, and 20 numbers that leave a remainder of 1 when divided by 5. Exactly 5 numbers (11, 31, 41, 61, 71) meet both conditions. To find the numbers that meet either condition but NOT both, subtract the 5 overlapping numbers from each category: 20 numbers are prime only, and 15 numbers leave a remainder of 1 only. The number of favorable outcomes is 20 + 15 = 35. Therefore, the single-event probability is 35/100 = 7/20.

Step-by-Step Solution

1

Determine the total sample space size.

The total number of possible outcomes is

N = 100

.

Each integer from 1 to 100 inclusive is equally likely.

2

Identify and count all prime numbers in the set

\{1, 2, \dots, 100\}

.

There are 25 prime numbers:

\{2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97\}

.

Note that 1 is not a prime number.

3

Identify and count all integers in

\{1, 2, \dots, 100\}

that leave a remainder of 1 when divided by 5.

There are 20 such integers:

\{1, 6, 11, 16, 21, 26, 31, 36, 41, 46, 51, 56, 61, 66, 71, 76, 81, 86, 91, 96\}

.

These integers take the form

5k + 1

for integers

k \ge 0

where

1 \le 5k + 1 \le 100

.

4

Find the intersection of both sets (numbers that are prime AND leave a remainder of 1 when divided by 5).

The 5 numbers satisfying both conditions are

\{11, 31, 41, 61, 71\}

.

Any integer leaving a remainder of 1 when divided by 5 must end in 1 or 6. No prime ends in 6. The primes ending in 1 in this range are 11, 31, 41, 61, and 71.

5

Calculate the number of favorable outcomes satisfying 'either condition, but not both' (symmetric difference).

Favorable outcomes =

(25 - 5) + (20 - 5) = 20 + 15 = 35

.

We must count numbers that are only prime (20) plus numbers that are only remainder 1 (15), excluding the 5 overlapping numbers.

6

Calculate the single-event probability.

P = \frac{35}{100} = \frac{7}{20}

.

Divide favorable outcomes by total sample space size.

Key Concept

Basic single-event probability with set operations and number properties (prime numbers, remainders).

Estimated Time:2m 0s

Question 7Question →

A box contains 90 tokens, numbered sequentially from $1$ to $90$ , inclusive. If one token is drawn at random from the box, what is the probability that the number on the token is a multiple of $4$ or $6$ , but not a multiple of $8$ ?

\frac{13}{45}

\frac{1}{3}

\frac{19}{90}

\frac{2}{9}

\frac{7}{30}

Show answer & explanation

Answer: $\frac{19}{90}$

Answer

The probability that the selected number is a multiple of 4 or 6, but not a multiple of 8, is

\frac{19}{90}

.

To find the probability, determine the number of favorable outcomes out of 90 total outcomes. The number of multiples of 4 is 22, and the number of multiples of 6 is 15. The numbers that are multiples of both 4 and 6 are multiples of 12, of which there are 7. By inclusion-exclusion, the number of integers that are multiples of 4 or 6 is

22 + 15 - 7 = 30

. Since all multiples of 8 are automatically multiples of 4, all 11 multiples of 8 in the range are included in these 30 numbers. Excluding the multiples of 8 leaves

30 - 11 = 19

favorable integers. Thus, the probability is

\frac{19}{90}

.

Step-by-Step Solution

1

Find the count of multiples of 4 and multiples of 6 in the range from 1 to 90.

Multiples of 4:

\lfloor \frac{90}{4} \rfloor = 22

. Multiples of 6:

\lfloor \frac{90}{6} \rfloor = 15

.

Identify the size of each individual set of multiples.

2

Calculate the number of elements in the union of multiples of 4 or 6 using the Inclusion-Exclusion Principle.

Multiples of both 4 and 6 are multiples of

\text{lcm}(4,6) = 12

. Count of multiples of 12:

\lfloor \frac{90}{12} \rfloor = 7

. Union size:

22 + 15 - 7 = 30

.

Avoid double-counting numbers that are divisible by both 4 and 6.

3

Identify and subtract the multiples of 8 within this union.

Multiples of 8 in the range:

\lfloor \frac{90}{8} \rfloor = 11

. Since every multiple of 8 is also a multiple of 4, all 11 multiples of 8 are contained within the union. Favorable outcomes:

30 - 11 = 19

.

Filter out numbers that satisfy the exclusion condition (multiples of 8).

4

Compute the single-event probability.

P = \frac{19}{90}

.

Divide the number of favorable outcomes (19) by the total sample space (90).

Key Concept

Basic Single-Event Probability with Principle of Inclusion-Exclusion and Set Restriction

Estimated Time:2m 0s

Question 8Question →

An integer $n$ is chosen at random from the set of all positive integers less than or equal to $120$ . What is the probability that $n$ is a multiple of either $3$ or $5$ , but not a multiple of $15$ ?

Show answer & explanation

Answer: 0.4

Answer

The probability that the selected integer is a multiple of either 3 or 5, but not 15, is 0.4 (or 2/5).

The total number of integers from 1 to 120 is 120. Multiples of 3 up to 120 total 40, multiples of 5 total 24, and multiples of 15 total 8. Integers that are multiples of 3 but not 15 number 40 - 8 = 32. Integers that are multiples of 5 but not 15 number 24 - 8 = 16. The total number of favorable outcomes is 32 + 16 = 48. Thus, the probability is 48/120 = 2/5 = 0.4.

Step-by-Step Solution

1

Count total possible outcomes in the sample space.

The total number of integers from 1 to 120 is 120.

Each integer in the set {1, 2, ..., 120} is equally likely to be selected.

2

Count the number of multiples of 3, 5, and 15 within the range.

Multiples of 3: 40; Multiples of 5: 24; Multiples of 15: 8.

Since 120 is divisible by 3, 5, and 15, the count of multiples of k up to 120 is 120/k.

3

Calculate the number of integers that are multiples of 3 or 5, but not 15.

Number of favorable outcomes = (Multiples of 3 only) + (Multiples of 5 only) = (40 - 8) + (24 - 8) = 32 + 16 = 48.

Multiples of 15 are common multiples of both 3 and 5 and must be excluded completely according to the condition 'not a multiple of 15'.

4

Compute the single-event probability.

Probability = 48 / 120 = 2 / 5 = 0.4.

Probability of a single event is defined as the ratio of favorable outcomes to total possible outcomes.

Key Concept

Basic Single-Event Probability with Set Restrictions

Question 9Question →

The table below categorizes 100 analysts at a consulting firm by their department and experience level:

Department	Junior (1–3 yrs)	Senior (4–7 yrs)	Lead (8+ yrs)	Total
Technology	14	16	10	40
Analytics	12	18	5	35
Operations	9	11	5	25
Total	35	45	20	100

If one analyst is selected at random from this group, what is the probability that the selected analyst works in the Analytics department or has Lead experience, but NOT both?

Show answer & explanation

Answer: 0.45

Answer

0.45

To find the probability of selecting an analyst who is either in the Analytics department or at the Lead level, but not both, we count the analysts in Analytics who are not Leads (

12 + 18 = 30

) and the analysts at the Lead level who are not in Analytics (

10 + 5 = 15

). The total number of favorable outcomes is

30 + 15 = 45

. Dividing by the total pool of

100

analysts yields a probability of

\frac{45}{100} = 0.45

.

Step-by-Step Solution

1

Determine the total size of the sample space.

Total analysts

N = 100

.

Basic probability requires dividing favorable outcomes by total possible outcomes.

2

Calculate the number of analysts satisfying 'Analytics, but NOT Lead'.

Junior Analytics (

12

) + Senior Analytics (

18

) =

30

.

Excludes the

5

Lead analysts in the Analytics department.

3

Calculate the number of analysts satisfying 'Lead, but NOT Analytics'.

Technology Lead (

10

) + Operations Lead (

5

) =

15

.

Excludes the

5

Lead analysts in the Analytics department.

4

Sum the non-overlapping favorable counts and compute probability.

Favorable outcomes

= 30 + 15 = 45

; Probability

= \frac{45}{100} = 0.45

.

Probability of a single event is defined as

\frac{\text{Favorable Outcomes}}{\text{Total Outcomes}}

.

Key Concept

Basic Single-Event Probability from Two-Way Tabular Data with Mutually Exclusive Set Conditions

Question 10Question →

A quality control analyst randomly selects one component from a shipment containing components labeled with distinct integer batch numbers from $21$ to $100$ , inclusive. What is the probability that the batch number of the selected component is a prime number?

\frac{1}{5}

\frac{17}{80}

\frac{17}{79}

\frac{9}{40}

\frac{17}{63}

Show answer & explanation

Answer: $\frac{17}{80}$

Answer

\frac{17}{80}

The total number of batch numbers from 21 through 100 inclusive is given by

100 - 21 + 1 = 80

. The prime numbers in this range are 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, and 97, which count to 17 prime numbers. Dividing the 17 favorable outcomes by the 80 total outcomes gives a probability of

\frac{17}{80}

.

Step-by-Step Solution

1

Determine the total number of components (the sample space size).

Total components

N = 100 - 21 + 1 = 80

.

For an inclusive range of integers from

a

to

b

, the number of elements is

b - a + 1

.

2

Identify all prime numbers between

21

and

100

, inclusive.

The prime numbers in this range are:

23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97

. There are

17

prime numbers in total.

Each listed number has exactly two distinct positive divisors:

1

and itself.

3

Calculate the probability of selecting a prime batch number.

P(\text{Prime}) = \frac{\text{Number of Prime Batch Numbers}}{\text{Total Number of Batch Numbers}} = \frac{17}{80}

.

Probability of a single event is the ratio of favorable outcomes to total possible outcomes.

Key Concept

Basic Single-Event Probability

Question 11Question →

An integer $n$ is randomly selected from the set of all integers from $10$ to $59$ , inclusive. What is the probability that $n$ is a prime number whose units digit is $3$ ?

\frac{1}{10}

\frac{4}{49}

\frac{2}{25}

\frac{2}{23}

\frac{3}{50}

Show answer & explanation

Answer: $\frac{2}{25}$

Answer

The probability that the selected integer is a prime number whose units digit is

3

is

\frac{2}{25}

.

The correct answer identifies that there are 50 total integers in the inclusive range from 10 to 59. Among the numbers ending in 3 (

13, 23, 33, 43, 53

), exactly four are prime (

13, 23, 43, 53

), while 33 is composite. Dividing 4 favorable outcomes by 50 total outcomes yields

\frac{4}{50}

, which simplifies to

\frac{2}{25}

.

Step-by-Step Solution

1

Determine the total number of outcomes in the set.

The set contains

59 - 10 + 1 = 50

integers.

For an inclusive set of integers from

a

to

b

, the total number of integers is

b - a + 1

.

2

Identify all integers in the set with a units digit of

3

.

The candidate integers ending in 3 are

13, 23, 33, 43, 53

.

These are all two-digit numbers in the range

[10, 59]

ending with

3

.

3

Determine which candidate integers are prime numbers.

The prime numbers are

13, 23, 43,

and

53

(a total of 4 favorable outcomes). Note that

33 = 3 \times 11

, so it is composite.

A prime number is an integer greater than 1 that has no positive divisors other than 1 and itself.

4

Calculate the single-event probability and simplify the fraction.

\text{Probability} = \frac{4}{50} = \frac{2}{25}

.

Probability is defined as the number of favorable outcomes divided by the total number of possible outcomes.

Key Concept

Basic Single-Event Probability

Estimated Time:1m 30s

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