Independent and Dependent Events Alıştırma Soruları — GMAT

Soru 1Soru →

A quality assurance technician randomly selects 2 microprocessors from a batch of $16$ microprocessors, of which $6$ were manufactured at Facility A and $10$ were manufactured at Facility B. The selections are made one after another without replacement. What is the probability that both selected microprocessors were manufactured at Facility A?

Cevabı ve açıklamayı göster

Cevap: 0.125

Cevap

The probability that both selected microprocessors were manufactured at Facility A is 0.125 (or 1/8).

Because the selections are made without replacement, the outcome of the first draw changes the pool of available microprocessors for the second draw. The probability of choosing a Facility A microprocessor first is

\frac{6}{16} = \frac{3}{8}

. Following that selection,

5

Facility A microprocessors remain among

15

total microprocessors, giving a conditional probability of

\frac{5}{15} = \frac{1}{3}

for the second selection. Multiplying these dependent probabilities yields

\frac{3}{8} \times \frac{1}{3} = \frac{1}{8} = 0.125

.

Adım Adım Çözüm

1

Determine the probability of selecting a Facility A microprocessor on the first draw.

P(\text{First is Facility A}) = \frac{6}{16} = \frac{3}{8}

There are 6 microprocessors from Facility A out of 16 total microprocessors.

2

Determine the conditional probability of selecting a Facility A microprocessor on the second draw.

P(\text{Second is Facility A} \mid \text{First is Facility A}) = \frac{5}{15} = \frac{1}{3}

Since the selection is made without replacement, 5 Facility A microprocessors remain out of a reduced total of 15 microprocessors.

3

Multiply the dependent probabilities to find the combined probability.

P(\text{Both are Facility A}) = \frac{3}{8} \times \frac{1}{3} = \frac{3}{24} = \frac{1}{8} = 0.125

The joint probability of sequential dependent events is the product of the initial probability and the conditional probability.

Anahtar Kavram

Probability of Dependent Events (Sampling without Replacement)

Tahmini Süre:1m 30s

Soru 2Soru →

A financial analyst chooses two investment assets at random, one after another without replacement, from a portfolio consisting of $3$ technology stocks and $2$ utility stocks. The probability of achieving a target return depends on the composition of the selected assets:
- If both selected assets are technology stocks, the probability of achieving the target return is $0.80$ .
- If exactly one selected asset is a technology stock, the probability of achieving the target return is $0.50$ .
- If neither selected asset is a technology stock, the probability of achieving the target return is $0.10$ .

What is the probability, expressed as a decimal, that the analyst achieves the target return?

Cevabı ve açıklamayı göster

Cevap: 0.55

Cevap

The probability that the analyst achieves the target return is 0.55.

To find the overall probability of achieving the target return, calculate the probability of each mutually exclusive composition of assets selected without replacement, and weight each by its conditional probability of success. The probability of selecting two technology stocks is (3/5) * (2/4) = 0.30. The probability of selecting one technology stock and one utility stock in either order is (3/5)*(2/4) + (2/5)*(3/4) = 0.60. The probability of selecting two utility stocks is (2/5) * (1/4) = 0.10. Multiplying each by its respective success rate (0.80, 0.50, and 0.10) yields 0.24 + 0.30 + 0.01 = 0.55.

Adım Adım Çözüm

1

Determine the probabilities of the three mutually exclusive selection scenarios without replacement.

P(TT) = 0.30, P(1T) = 0.60, P(UU) = 0.10.

Since selections are made without replacement, the outcome of the first draw affects the total count and available assets for the second draw.

2

Calculate the joint probability of achieving the target return for each scenario.

P(TT and Target) = 0.24, P(1T and Target) = 0.30, P(UU and Target) = 0.01.

Multiply the probability of drawing each combination by the conditional probability of success for that specific combination.

3

Sum the joint probabilities across all possible scenarios.

Total Probability = 0.24 + 0.30 + 0.01 = 0.55.

The scenarios are mutually exclusive and exhaustive, so their sum gives the total overall probability of achieving the target return.

Anahtar Kavram

Probability of Dependent Events and Law of Total Probability

Tahmini Süre:2m 0s

Soru 3Soru →

A quality control engineer inspects a shipment of 25 solar panels, of which 9 are premium grade and 16 are standard grade. The engineer randomly selects 2 panels from the shipment, one after another without replacement. What is the probability, expressed as a decimal, that at least one of the two selected panels is premium grade?

Cevabı ve açıklamayı göster

Cevap: 0.6

Cevap

The probability that at least one of the two selected panels is premium grade is 0.6.

Selection without replacement creates dependent events because the outcome of the first draw alters the pool available for the second draw. Out of 25 panels, 16 are standard grade. The probability that the first drawn panel is standard grade is 16/25. If the first panel is standard grade, 15 standard panels remain out of 24 total panels, giving a probability of 15/24 for the second draw. The probability of selecting two standard panels is (16/25) × (15/24) = 0.40. Using complementary probability, the probability that at least one panel is premium grade is 1 - 0.40 = 0.60.

Adım Adım Çözüm

1

Determine the initial counts of total, premium, and standard grade panels.

Total panels = 25, premium grade panels = 9, standard grade panels = 16.

Establishing the sample space composition is necessary to calculate draw probabilities.

2

Calculate the probability that neither panel selected is premium grade (i.e., both are standard grade).

P(both standard) = (16 / 25) * (15 / 24) = (16 / 25) * (5 / 8) = 0.40.

Because selection is done without replacement, the total pool size and remaining standard panels each decrease by 1 for the second draw.

3

Apply complementary probability to determine the probability of selecting at least one premium panel.

P(at least one premium) = 1 - P(both standard) = 1 - 0.40 = 0.60.

The scenario of selecting at least one premium panel is the exact complementary event of selecting zero premium panels.

Anahtar Kavram

Dependent Events and Complementary Probability

Alternatif Yöntem

Sum the probabilities of mutually exclusive favorable outcomes: P(1st premium, 2nd standard) + P(1st standard, 2nd premium) + P(both premium) = (9/25)(16/24) + (16/25)(9/24) + (9/25)(8/24) = 0.24 + 0.24 + 0.12 = 0.60.

Tahmini Süre:1m 30s

Soru 4Soru →

A jar contains $3$ green marbles and $2$ red marbles. If two marbles are selected at random one after another without replacement, what is the probability that both selected marbles are green?

Cevabı ve açıklamayı göster

Cevap: 0.3

Cevap

The probability that both selected marbles are green is

0.3

.

Because selection is done without replacement, the outcome of the first draw changes the composition of the jar for the second draw. The probability of selecting a green marble first is

3/5

. With one green marble removed,

2

green marbles remain among

4

total marbles, so the probability of selecting green on the second draw is

2/4

. Multiplying these sequential probabilities gives

(3/5) \times (2/4) = 6/20 = 0.3

.

Adım Adım Çözüm

1

Calculate the probability of drawing a green marble on the first selection.

P(\text{First Green}) = \frac{3}{5}

There are

3

green marbles out of

5

total marbles in the jar.

2

Calculate the conditional probability of drawing a green marble on the second selection.

P(\text{Second Green} \mid \text{First Green}) = \frac{2}{4} = \frac{1}{2}

Because the first marble was not replaced,

2

green marbles remain out of a total of

4

remaining marbles.

3

Multiply the probabilities of the dependent sequential events.

P(\text{Both Green}) = \frac{3}{5} \times \frac{1}{2} = \frac{3}{10} = 0.3

The probability of both dependent events occurring is the product of the first event's probability and the conditional probability of the second event.

Anahtar Kavram

Probability of Dependent Events Without Replacement

Soru 5Soru →

A container holds $4$ red tokens and $6$ blue tokens. If two tokens are selected at random one after another without replacement, what is the probability that both selected tokens are red?

\frac{4}{25}

\frac{8}{45}

\frac{2}{15}

\frac{2}{5}

\frac{1}{3}

Cevabı ve açıklamayı göster

Cevap: $\frac{2}{15}$

Cevap

\frac{2}{15}

The option stating

\frac{2}{15}

is correct because selecting without replacement makes the second event dependent on the first. The first probability is

\frac{4}{10}

and the second probability is

\frac{3}{9}

. Multiplying these gives

\frac{4}{10} \times \frac{3}{9} = \frac{12}{90} = \frac{2}{15}

.

Adım Adım Çözüm

1

Calculate the probability of selecting a red token on the first draw.

P(\text{First Red}) = \frac{4}{10} = \frac{2}{5}

There are

4

red tokens out of a total of

10

tokens.

2

Calculate the conditional probability of selecting a red token on the second draw given the first was red.

P(\text{Second Red} \mid \text{First Red}) = \frac{3}{9} = \frac{1}{3}

Because the selection is without replacement,

1

red token and

1

total token have been removed, leaving

3

red tokens out of

9

remaining tokens.

3

Multiply the sequential probabilities to find the combined probability.

P(\text{Both Red}) = \frac{2}{5} \times \frac{1}{3} = \frac{2}{15}

The probability of both dependent events occurring sequentially is the product of their individual conditional probabilities.

Anahtar Kavram

Probability of Dependent Events Without Replacement

Soru 6Soru →

A quality control inspector tests items from a batch containing $n$ high-grade items and $4$ low-grade items, where $n > 1$ . Two items are selected sequentially. If the first item selected is high-grade, it is kept out of the batch for testing, and additionally $1$ low-grade item is also removed from the batch before the second selection. If the probability of selecting two high-grade items in succession is $\frac{3}{8}$ , what is the value of $n$ ?

Cevabı ve açıklamayı göster

Cevap: 6

Cevap

The initial number of high-grade items

n

is

6

.

Because the outcome of the first draw alters both the count of high-grade items and the count of low-grade items, the two draws are dependent events. The probability of drawing a high-grade item on the first draw is

\frac{n}{n+4}

. Following this successful outcome, the remaining pool consists of

n-1

high-grade items and

3

low-grade items, yielding a total of

n+2

items. The probability of a second high-grade selection is therefore

\frac{n-1}{n+2}

. Multiplying these probabilities gives

\frac{n(n-1)}{(n+4)(n+2)} = \frac{3}{8}

. Solving the resulting quadratic equation

5n^2 - 26n - 24 = 0

gives

n = 6

(since

n

must be a positive integer).

Adım Adım Çözüm

1

Calculate the probability of selecting a high-grade item on the first draw.

P(H_1) = \frac{n}{n+4}

The initial batch contains

n

high-grade items out of a total of

n+4

items.

2

Determine the conditional sample space for the second draw after event

H_1

occurs.

High-grade items remaining =

n-1

; low-grade items remaining =

4-1=3

; total remaining items =

(n-1)+3 = n+2

.

Drawing a high-grade item removes

1

high-grade item and

1

additional low-grade item from the batch.

3

Calculate the conditional probability of selecting a high-grade item on the second draw.

P(H_2 | H_1) = \frac{n-1}{n+2}

There are now

n-1

high-grade items out of

n+2

total remaining items.

4

Set up and solve the algebraic probability equation.

\frac{n(n-1)}{(n+4)(n+2)} = \frac{3}{8} \implies 8n^2 - 8n = 3(n^2 + 6n + 8) \implies 5n^2 - 26n - 24 = 0 \implies (5n+4)(n-6) = 0 \implies n = 6

.

Because

n > 1

is required, the negative fraction root

-\frac{4}{5}

is discarded.

Anahtar Kavram

Probability of dependent events with non-standard sample space modifications

Soru 7Soru →

A box contains $x$ blue spheres and $8$ yellow spheres, where $x$ is a positive integer. If two spheres are selected at random one after another without replacement, the probability that both selected spheres are blue is $\frac{5}{17}$ . What is the total number of spheres in the box initially?

10

15

18

20

25

Cevabı ve açıklamayı göster

Cevap: 18

Cevap

The total number of spheres in the box initially is 18.

The probability of drawing two blue spheres sequentially without replacement is given by

\frac{x}{x+8} \times \frac{x-1}{x+7} = \frac{5}{17}

. Expanding and rearranging the equation gives

3x^2 - 23x - 70 = 0

, which factors into

(3x + 7)(x - 10) = 0

. Since

x

must be a positive integer,

x = 10

. The total number of spheres in the box initially is

x + 8 = 10 + 8 = 18

.

Adım Adım Çözüm

1

Set up the probability expression for dependent sequential events.

P(\text{both blue}) = \frac{x}{x+8} \times \frac{x-1}{x+7}

Because the draws occur without replacement, the total count decreases from

x+8

to

x+7

and the number of blue spheres decreases from

x

to

x-1

for the second draw.

2

Equate to the given probability and clear denominators to form a quadratic equation.

\frac{x(x-1)}{(x+8)(x+7)} = \frac{5}{17} \implies 17(x^2 - x) = 5(x^2 + 15x + 56) \implies 3x^2 - 23x - 70 = 0

Cross-multiplying converts the rational probability equation into a standard quadratic equation.

3

Factor the quadratic equation to find the positive integer root.

(3x + 7)(x - 10) = 0 \implies x = 10

Since the count of spheres must be a positive integer,

x = 10

is the only valid solution for the number of blue spheres.

4

Calculate the total initial number of spheres.

\text{Total} = x + 8 = 10 + 8 = 18

The question asks for the total initial number of spheres, which includes both the

x

blue spheres and the 8 yellow spheres.

Anahtar Kavram

Probability of Dependent Events Without Replacement

Tahmini Süre:2m 0s

Soru 8Soru →

A research laboratory's secure vault contains $N$ data drives, where $N > 4$ . Exactly $3$ of the drives contain encrypted security keys, while the remaining $N - 3$ drives contain standard operational data. If two data drives are selected at random one after another without replacement, the probability that at least one of the selected drives contains an encrypted security key is $\frac{8}{15}$ . What is the value of $N$ ?

8

9

10

12

15

Cevabı ve açıklamayı göster

Cevap: 10

Cevap

10

The correct option is 10. The probability of choosing at least one encrypted security key drive in two draws without replacement is

1 - P(\text{no security key drives})

. Since there are

N - 3

standard drives out of

N

total drives,

P(\text{no security key drives}) = \frac{N - 3}{N} \times \frac{N - 4}{N - 1}

. Setting

1 - \frac{(N - 3)(N - 4)}{N(N - 1)} = \frac{8}{15}

gives

\frac{(N - 3)(N - 4)}{N(N - 1)} = \frac{7}{15}

. Solving

15(N^2 - 7N + 12) = 7(N^2 - N)

leads to

8N^2 - 98N + 180 = 0

, or

4N^2 - 49N + 90 = 0

, which factors as

(4N - 9)(N - 10) = 0

. Because

N

must be an integer,

N = 10

.

Adım Adım Çözüm

1

Express the complementary probability event

The probability of at least one drive having an encrypted key is

1 - P(\text{both drives are standard data drives})

. Thus,

P(\text{both standard}) = 1 - \frac{8}{15} = \frac{7}{15}

.

Using the complementary probability rule is more efficient than calculating the individual probabilities of key-then-standard, standard-then-key, and key-then-key.

2

Set up the probability expression for dependent sequential selection without replacement

For the first draw, the probability of selecting a standard drive is

\frac{N - 3}{N}

. For the second draw, since one standard drive has been removed, the probability is

\frac{N - 4}{N - 1}

. Therefore,

\frac{(N - 3)(N - 4)}{N(N - 1)} = \frac{7}{15}

.

The events are dependent because the selection is made without replacement.

3

Solve the algebraic equation for N

Expanding both sides gives

15(N^2 - 7N + 12) = 7(N^2 - N)

, which simplifies to

15N^2 - 105N + 180 = 7N^2 - 7N \implies 8N^2 - 98N + 180 = 0

. Dividing by 2 yields

4N^2 - 49N + 90 = 0

. Factoring gives

(4N - 9)(N - 10) = 0

.

Since

N

must be a positive integer greater than 4,

N = 10

.

Anahtar Kavram

Dependent probability without replacement and complementary probability

Tahmini Süre:2m 0s

Soru 9Soru →

A glass bowl contains 6 red candies and $N$ green candies, where $N$ is a positive integer. If two candies are selected at random one after another without replacement, the probability that both candies selected are red is $\frac{1}{3}$ . What is the value of $N$ ?

Cevabı ve açıklamayı göster

Cevap: 4

Cevap

4

The correct value of

N

is 4. Because the selection is performed without replacement, the probability of drawing the first red candy is

\frac{6}{N+6}

and the second red candy is

\frac{5}{N+5}

. Setting their product equal to

\frac{1}{3}

yields

\frac{30}{(N+6)(N+5)} = \frac{1}{3}

, which expands to

N^2 + 11N - 60 = 0

. Factoring gives

(N-4)(N+15) = 0

, yielding

N = 4

as the only positive solution.

Adım Adım Çözüm

1

Define total outcomes and event probabilities for dependent selection

Initial total candies =

N + 6

. Probability of first red =

\frac{6}{N+6}

. Probability of second red given first is red =

\frac{5}{N+5}

.

Since selection is without replacement, the total count and red count both decrease by 1 after the first draw, making the events dependent.

2

Set up the probability multiplication equation

P(\text{both red}) = \frac{6}{N+6} \times \frac{5}{N+5} = \frac{30}{(N+6)(N+5)} = \frac{1}{3}

.

The probability of joint dependent events

A

and

B

occurring is

P(A \cap B) = P(A) \times P(B|A)

.

3

Solve the equation for

N

(N+6)(N+5) = 90 \Rightarrow N^2 + 11N - 60 = 0 \Rightarrow (N-4)(N+15) = 0

.

Cross-multiplying gives a quadratic equation in terms of

N

.

4

Select the valid positive integer root

N = 4

.

The count of candies must be a positive number, so

N = -15

is discarded.

Anahtar Kavram

Probability of Dependent Events (Sampling without Replacement)

Soru 10Soru →

A charity organization's steering board consists of $6$ donors and $4$ volunteers. Two board members are selected at random, one after another without replacement, to attend a national conference. What is the probability that the first member selected is a donor and the second member selected is a volunteer?

\frac{6}{25}

\frac{4}{15}

\frac{1}{3}

\frac{8}{15}

\frac{12}{25}

Cevabı ve açıklamayı göster

Cevap: $\frac{4}{15}$

Cevap

The probability that the first member selected is a donor and the second member selected is a volunteer is

\frac{4}{15}

.

The correct probability is calculated by multiplying the probability of the first event by the conditional probability of the second event given that the first event occurred. The probability of choosing a donor first is

\frac{6}{10}

. Since the selection is made without replacement, there are

9

total members remaining for the second draw,

4

of whom are volunteers. Therefore, the probability of selecting a volunteer second is

\frac{4}{9}

. The overall probability is

\frac{6}{10} \times \frac{4}{9} = \frac{24}{90} = \frac{4}{15}

.

Adım Adım Çözüm

1

Calculate the probability of selecting a donor on the first choice.

Since there are

6

donors out of

10

total board members,

P(\text{1st Donor}) = \frac{6}{10} = \frac{3}{5}

.

The sample space initially contains

10

members,

6

of whom are donors.

2

Calculate the conditional probability of selecting a volunteer on the second choice given that a donor was chosen first.

After one donor is selected,

9

members remain,

4

of whom are volunteers. Thus,

P(\text{2nd Volunteer} \mid \text{1st Donor}) = \frac{4}{9}

.

The selection is made without replacement, reducing both the total number of members in the pool and the sample space size.

3

Multiply the sequential probabilities for dependent events.

P(\text{1st Donor and 2nd Volunteer}) = \frac{6}{10} \times \frac{4}{9} = \frac{24}{90} = \frac{4}{15}

.

For dependent events

A

and

B

, the joint probability is

P(A \cap B) = P(A) \times P(B \mid A)

.

Anahtar Kavram

Probability of Dependent Events (Sampling Without Replacement)

Soru 11Soru →

A corporate compliance archive contains $10$ confidential files and $5$ public files. An auditor randomly selects two files from the archive, one after another without replacement. What is the probability that at least one of the selected files is public?

\frac{2}{21}

\frac{10}{21}

\frac{11}{21}

\frac{5}{9}

\frac{4}{7}

Cevabı ve açıklamayı göster

Cevap: $\frac{4}{7}$

Cevap

The probability that at least one of the selected files is public is

\frac{4}{7}

.

To find the probability that at least one file is public, subtract the probability that no public files are selected (meaning both files are confidential) from 1. The probability of choosing a confidential file on the first draw is

\frac{10}{15} = \frac{2}{3}

. Since the selection is without replacement,

9

confidential files remain out of

14

total files. The probability of choosing a confidential file on the second draw is

\frac{9}{14}

. The joint probability of drawing two confidential files is

\frac{2}{3} \times \frac{9}{14} = \frac{3}{7}

. Subtracting this from 1 yields

1 - \frac{3}{7} = \frac{4}{7}

.

Adım Adım Çözüm

1

Identify the complementary event

The complementary event to 'at least one selected file is public' is 'both selected files are confidential'.

Calculating the probability of the complement is simpler than summing the probabilities of all positive individual outcomes.

2

Calculate the probability of drawing two confidential files sequentially without replacement

The probability of drawing a confidential file on the first selection is

\frac{10}{15} = \frac{2}{3}

. After removing one confidential file, 9 confidential files remain out of 14 total files. The probability on the second selection is

\frac{9}{14}

. Thus,

P(\text{both confidential}) = \frac{10}{15} \times \frac{9}{14} = \frac{2}{3} \times \frac{9}{14} = \frac{3}{7}

.

Because the drawing is done without replacement, the outcome of the first draw affects the total sample space and event space for the second draw.

3

Subtract the complementary probability from 1

P(\text{at least one public}) = 1 - P(\text{both confidential}) = 1 - \frac{3}{7} = \frac{4}{7}

.

The sum of the probabilities of an event and its complement equals 1.

Anahtar Kavram

Probability of Dependent Events and Complementary Probability

Tahmini Süre:1m 30s

Independent and Dependent Events

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

Alternatif Yöntem

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

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar