Complementary Probability and At-Least-One Scenarios Alıştırma Soruları — GMAT

Soru 1Soru →

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}

Cevabı ve açıklamayı göster

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

Cevap

\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}

.

Adım Adım Çözüm

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.

Anahtar Kavram

Complementary Probability:

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

Soru 2Soru →

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 $0.992$ .

Doğru

Yanlış

Cevabı ve açıklamayı göster

Cevap: True

Cevap

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

.

Adım Adım Çözüm

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'.

Anahtar Kavram

Complementary Probability and At-Least-One Scenarios

Soru 3Soru →

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}

Cevabı ve açıklamayı göster

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

Cevap

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}

.

Adım Adım Çözüm

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.

Anahtar Kavram

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

Soru 4Soru →

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 $N$ ?

Cevabı ve açıklamayı göster

Cevap: 7

Cevap

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}

.

Adım Adım Çözüm

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.

Anahtar Kavram

Complementary Probability and Dependent Sampling (Without Replacement)

Soru 5Soru →

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}

Cevabı ve açıklamayı göster

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

Cevap

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}

.

Adım Adım Çözüm

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'.

Anahtar Kavram

Complementary Probability

Tahmini Süre:2m 0s

Soru 6Soru →

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 $\frac{8}{15}$ .

Doğru

Yanlış

Cevabı ve açıklamayı göster

Cevap: True

Cevap

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}

.

Adım Adım Çözüm

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.

Anahtar Kavram

Complementary Probability and At-Least-One Scenarios without Replacement

Soru 7Soru →

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 $a + b$ ?

Cevabı ve açıklamayı göster

Cevap: 59

Cevap

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

.

Adım Adım Çözüm

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.

Anahtar Kavram

Complementary Probability and At-Least-One Scenarios

Complementary Probability and At-Least-One Scenarios

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar