Complementary Probability and At-Least-One Scenarios Practice Questions — GMAT

Question 1Question →

A bag contains 5 red marbles and 5 blue marbles. If 2 marbles are randomly selected from the bag one after another without replacement, what is the probability that at least one of the selected marbles is red?

\frac{7}{9}

\frac{2}{9}

\frac{5}{9}

\frac{3}{4}

\frac{1}{4}

Show answer & explanation

Answer: $\frac{7}{9}$

Answer

\frac{7}{9}

To find the probability of drawing at least one red marble, calculate the probability of the complementary event (drawing two blue marbles) and subtract it from 1. The probability of selecting a blue marble on the first draw is

\frac{5}{10}

, and on the second draw without replacement it is

\frac{4}{9}

. The probability of both marbles being blue is

\frac{5}{10} \times \frac{4}{9} = \frac{2}{9}

. Subtracting this from 1 gives

1 - \frac{2}{9} = \frac{7}{9}

.

Step-by-Step Solution

1

Identify the complementary outcome

The event 'at least one marble is red' is complementary to the event 'no marbles are red' (i.e., 'both marbles are blue').

Using the relationship

P(\text{at least one red}) = 1 - P(\text{both blue})

is simpler than summing individual favorable cases.

2

Calculate the probability of drawing two blue marbles without replacement

The probability that the first marble is blue is

\frac{5}{10} = \frac{1}{2}

. The probability that the second marble is blue is

\frac{4}{9}

. Therefore,

P(\text{both blue}) = \frac{5}{10} \times \frac{4}{9} = \frac{20}{90} = \frac{2}{9}

.

Since draws are without replacement, the sample space and number of blue marbles decrease by 1 after the first draw.

3

Subtract the complementary probability from 1

P(\text{at least one red}) = 1 - \frac{2}{9} = \frac{7}{9}

.

The sum of complementary probabilities is always equal to 1.

Key Concept

Complementary Probability:

P(\text{at least one}) = 1 - P(\text{none})

Question 2Question →

An automated security gate opens successfully on any given approach with a probability of $0.8$ , independently of other attempts. If a driver approaches the gate $3$ times, the probability that the gate opens successfully at least once is $0.992$ .

True

False

Show answer & explanation

Answer: True

Answer

The statement is true because the probability of failing all

3

independent attempts is

(1 - 0.8)^3 = 0.008

, making the probability of at least one successful opening

1 - 0.008 = 0.992

.

The statement accurately reflects the complementary probability principle

P(\text{at least one}) = 1 - P(\text{none})

. For

3

independent trials with single-trial failure probability

0.2

, the probability of all trials failing is

0.2^3 = 0.008

. Subtracting this from

1

yields

0.992

.

Step-by-Step Solution

1

Find the probability of failure on a single trial

P(\text{failure}) = 1 - 0.8 = 0.2

The failure of a trial is the complement of its success.

2

Calculate the probability of zero successes across 3 independent trials

P(\text{0 successes}) = (0.2)^3 = 0.008

For independent events, the joint probability is the product of individual probabilities.

3

Apply the complementary probability rule to find the probability of at least one success

P(\text{at least 1 success}) = 1 - 0.008 = 0.992

The event 'at least one success' is the exact complement of 'zero successes'.

Key Concept

Complementary Probability and At-Least-One Scenarios

Question 3Question →

A testing kit contains $8$ electronic sensors, of which exactly $3$ are uncalibrated and $5$ are fully calibrated. If $3$ sensors are chosen at random without replacement, what is the probability that at least $1$ of the chosen sensors is uncalibrated?

\frac{5}{28}

\frac{15}{28}

\frac{23}{28}

\frac{55}{56}

\frac{387}{512}

Show answer & explanation

Answer: $\frac{23}{28}$

Answer

The correct answer is

\frac{23}{28}

.

To find the probability that at least one sensor is uncalibrated, calculate 1 minus the probability that no selected sensors are uncalibrated. Out of 8 sensors, there are

\binom{8}{3} = 56

ways to pick 3. The number of ways to pick 3 calibrated sensors from the 5 available is

\binom{5}{3} = 10

. Thus,

P(\text{no uncalibrated}) = \frac{10}{56} = \frac{5}{28}

. Subtracting this from 1 gives

1 - \frac{5}{28} = \frac{23}{28}

.

Step-by-Step Solution

1

Calculate the total number of ways to choose 3 sensors out of 8.

\binom{8}{3} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56

total outcomes.

Selection is made without replacement and order does not matter.

2

Calculate the complementary event: selecting 0 uncalibrated sensors (all 3 selected sensors are calibrated).

\binom{5}{3} = \binom{5}{2} = \frac{5 \times 4}{2 \times 1} = 10

outcomes with all calibrated sensors.

There are 5 calibrated sensors available, and 3 must be selected from them.

3

Find the probability of the complementary event (no uncalibrated sensors selected).

P(\text{no uncalibrated}) = \frac{10}{56} = \frac{5}{28}

.

Divide the favorable outcomes for selecting all calibrated sensors by total possible outcomes.

4

Apply the complementary probability principle to find

P(\text{at least one uncalibrated})

.

P(\text{at least one uncalibrated}) = 1 - P(\text{no uncalibrated}) = 1 - \frac{5}{28} = \frac{23}{28}

.

The sum of probabilities of complementary events is 1.

Key Concept

Complementary Probability: P(at least one X) = 1 - P(no X)

Question 4Question →

A quality control inspector reviews a shipment containing a total of $N$ customized components, of which exactly 3 are defective. The inspector randomly selects 3 components from the shipment one by one without replacement. If the probability that at least one of the selected components is defective is equal to $\frac{31}{35}$ , what is the value of $N$ ?

Show answer & explanation

Answer: 7

Answer

The total number of components in the shipment,

N

, is 7.

To find

N

, apply the complementary probability formula

P(\text{at least 1 defective}) = 1 - P(\text{0 defective})

. Given that

P(\text{at least 1 defective}) = \frac{31}{35}

, the probability of drawing zero defective components is

1 - \frac{31}{35} = \frac{4}{35}

. Out of

N

total components,

N-3

are non-defective. Drawing 3 non-defective components without replacement gives

P(\text{0 defective}) = \frac{(N-3)(N-4)(N-5)}{N(N-1)(N-2)}

. Setting this equal to

\frac{4}{35}

and testing integer values starting at

N=6

yields

N=7

, since

\frac{4 \times 3 \times 2}{7 \times 6 \times 5} = \frac{24}{210} = \frac{4}{35}

.

Step-by-Step Solution

1

Calculate the probability of the complementary event (selecting no defective components).

P(\text{0 defective}) = 1 - \frac{31}{35} = \frac{4}{35}

.

Calculating 'at least one' directly requires summing three separate cases (1 defective, 2 defective, 3 defective), whereas using the complement

P(\text{at least 1}) = 1 - P(\text{none})

requires evaluating only one scenario.

2

Formulate the algebraic expression for picking 3 non-defective components without replacement.

P(\text{0 defective}) = \frac{\binom{N-3}{3}}{\binom{N}{3}} = \frac{(N-3)(N-4)(N-5)}{N(N-1)(N-2)}

.

There are

N-3

non-defective components out of

N

total components, and 3 are selected without replacement.

3

Equate the algebraic probability to the known complementary probability and solve for

N

.

\frac{(N-3)(N-4)(N-5)}{N(N-1)(N-2)} = \frac{4}{35} \implies N = 7.

Testing

N = 7

yields

\frac{4 \times 3 \times 2}{7 \times 6 \times 5} = \frac{24}{210} = \frac{4}{35}

. The function is strictly increasing for

N \ge 6

, making

N = 7

the unique integer solution.

Key Concept

Complementary Probability and Dependent Sampling (Without Replacement)

Question 5Question →

A financial firm's investment committee consists of 8 senior analysts and 4 junior analysts. If a project review panel of 3 members is chosen at random without replacement from this committee, what is the probability that the panel contains at least one junior analyst?

\frac{14}{55}

\frac{28}{55}

\frac{19}{27}

\frac{41}{55}

\frac{47}{55}

Show answer & explanation

Answer: $\frac{41}{55}$

Answer

The probability that the panel contains at least one junior analyst is

\frac{41}{55}

.

To find the probability of selecting at least one junior analyst, it is most efficient to use the complementary probability rule:

P(\text{at least one junior}) = 1 - P(\text{no junior})

. The total number of ways to choose 3 panel members from 12 committee members is

\binom{12}{3} = 220

. The number of ways to choose 3 senior analysts from 8 is

\binom{8}{3} = 56

. Thus, the probability of selecting no junior analysts is

\frac{56}{220} = \frac{14}{55}

. Subtracting this from 1 gives

1 - \frac{14}{55} = \frac{41}{55}

.

Step-by-Step Solution

1

Calculate total possible outcomes for choosing 3 members out of 12.

\binom{12}{3} = \frac{12 \times 11 \times 10}{3 \times 2 \times 1} = 220

total combinations.

Selection order does not matter, so combinations

nCr

are used.

2

Calculate the number of unfavorable outcomes where zero junior analysts are chosen (all 3 selected are senior analysts).

\binom{8}{3} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56

ways.

All 3 members must be selected exclusively from the 8 senior analysts.

3

Find the probability of selecting zero junior analysts.

P(\text{no junior analysts}) = \frac{56}{220} = \frac{14}{55}

.

Divide the unfavorable outcomes by the total outcomes.

4

Apply the complementary probability principle:

P(\text{at least 1 junior}) = 1 - P(\text{no junior})

.

P(\text{at least 1 junior}) = 1 - \frac{14}{55} = \frac{41}{55}

.

The event 'at least one junior analyst' is the logical complement of 'no junior analysts'.

Key Concept

Complementary Probability

Estimated Time:2m 0s

Question 6Question →

If 3 components are randomly selected without replacement from a batch of 10 components containing exactly 2 defective components, the probability that at least one selected component is defective is equal to $\frac{8}{15}$ .

True

False

Show answer & explanation

Answer: True

Answer

True. The probability of selecting at least one defective component is indeed

\frac{8}{15}

.

The statement correctly computes the probability of selecting at least one defective component using

1 - P(\text{no defective components}) = 1 - \frac{\binom{8}{3}}{\binom{10}{3}} = 1 - \frac{7}{15} = \frac{8}{15}

.

Step-by-Step Solution

1

Identify the complementary event

The complement of 'at least one defective component' is 'zero defective components' (all 3 selected components are non-defective).

Calculating

P(\text{at least one}) = 1 - P(\text{none})

avoids calculating multiple dependent individual outcomes.

2

Calculate the total combinations for choosing 3 components from 10

\binom{10}{3} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120

.

This establishes the sample space size.

3

Calculate the combinations for choosing 3 non-defective components from 8

\binom{8}{3} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56

.

There are 8 non-defective components in the batch.

4

Compute

P(\text{0 defective})

and subtract from 1

P(\text{0 defective}) = \frac{56}{120} = \frac{7}{15}

. Therefore,

P(\text{at least 1 defective}) = 1 - \frac{7}{15} = \frac{8}{15}

.

Completing the complementary calculation confirms the statement is correct.

Key Concept

Complementary Probability and At-Least-One Scenarios without Replacement

Question 7Question →

A financial firm has a team of 10 auditors consisting of 6 certified public accountants (CPAs) and 4 audit assistants. A special audit task force of 3 auditors is to be selected at random from the team. If the probability that the task force contains at least one CPA is expressed as a fraction $\frac{a}{b}$ in simplest form, what is the value of $a + b$ ?

Show answer & explanation

Answer: 59

Answer

59

To find the probability of selecting at least one CPA, it is most efficient to use the complementary probability formula:

P(\text{at least 1 CPA}) = 1 - P(\text{no CPAs})

. The total number of ways to select any 3 auditors from the team of 10 is given by

\binom{10}{3} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120

. The number of ways to select 3 auditors such that none are CPAs (i.e., all 3 are audit assistants) is

\binom{4}{3} = 4

. Therefore, the probability of choosing zero CPAs is

\frac{4}{120} = \frac{1}{30}

. Subtracting from 1 gives

1 - \frac{1}{30} = \frac{29}{30}

. The fraction

\frac{29}{30}

is in simplest form because 29 is a prime number and does not divide 30. Thus,

a = 29

and

b = 30

, making

a + b = 59

.

Step-by-Step Solution

1

Calculate total possible combinations for selecting 3 members from 10.

\binom{10}{3} = 120

Selection is made without replacement and order of selection does not matter.

2

Calculate combinations of selecting 3 members with no CPAs.

\binom{4}{3} = 4

All 3 selected members must come from the pool of 4 audit assistants.

3

Determine probability of the complement event (no CPAs).

P(\text{no CPA}) = \frac{4}{120} = \frac{1}{30}

Probability is favorable outcomes over total outcomes.

4

Compute probability of at least one CPA.

P(\text{at least 1 CPA}) = 1 - \frac{1}{30} = \frac{29}{30}

P(\text{at least one}) = 1 - P(\text{none}).

5

Find the sum of numerator a and denominator b in simplest form.

a + b = 29 + 30 = 59

29/30 cannot be simplified further as 29 is a prime number.

Key Concept

Complementary Probability and At-Least-One Scenarios

Complementary Probability and At-Least-One Scenarios

Answer

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