Probability of Independent, Dependent, and Mutually Exclusive Events Practice Questions — GRE General Test — Page 2

Question 21Question →

During a quality assurance test of a dual-sensor monitoring device, Sensor $A$ operates independently of Sensor $B$ . The probability that Sensor $A$ detects a target signal during a test trial is $0.80$ , and the probability that Sensor $B$ detects the target signal during the same trial is $0.75$ . What is the probability that exactly one of the two sensors detects the target signal during a test trial?

Show answer & explanation

Answer: 0.35

Answer

0.35

To find the probability that exactly one sensor detects the signal, we must evaluate two disjoint scenarios: Sensor

A

detects while Sensor

B

fails (

0.80 \times 0.25 = 0.20

), or Sensor

A

fails while Sensor

B

detects (

0.20 \times 0.75 = 0.15

). Summing these two probabilities gives

0.20 + 0.15 = 0.35

. Alternatively, subtracting the probability that both sensors detect (

0.80 \times 0.75 = 0.60

) from the probability that at least one detects (

1 - 0.20 \times 0.25 = 0.95

) yields

0.95 - 0.60 = 0.35

.

Step-by-Step Solution

1

Determine the probabilities of individual event non-occurrences

P(A') = 0.20

and

P(B') = 0.25

The probability of an event not occurring is

1

minus the probability that it occurs.

2

Calculate joint probability for each mutually exclusive event outcome

P(A \text{ and } B') = 0.80 \times 0.25 = 0.20

and

P(A' \text{ and } B) = 0.20 \times 0.75 = 0.15

Since the sensors operate independently,

P(X \text{ and } Y) = P(X) \times P(Y)

.

3

Combine the independent outcomes that satisfy the condition

0.20 + 0.15 = 0.35

The scenarios (only

A

detects, or only

B

detects) are mutually exclusive, so their probabilities are added.

Key Concept

Probability of Independent and Mutually Exclusive Events

Question 22Question →

In a sample space of a random experiment, $A$ and $B$ are independent events such that $P(A) = 0.35$ and $P(A \cup B) = 0.74$ . Event $C$ is mutually exclusive with Event $A$ . If the conditional probability $P(C \mid B) = 0.20$ , what is the probability that Event $B$ occurs, but neither Event $A$ nor Event $C$ occurs?

0.19

0.27

0.39

0.48

0.51

Show answer & explanation

Answer: 0.27

Answer

0.27

To find the probability that Event

B

occurs without

A

or

C

, we must isolate the region of

B

that does not overlap with

A

or

C

. Since

A

and

B

are independent,

P(A \cup B) = P(A) + P(B) - P(A)P(B)

, which allows us to solve for

P(B) = 0.60

and

P(A \cap B) = 0.21

. Next, using the conditional probability

P(C \mid B) = 0.20

, we find

P(B \cap C) = 0.20 \times 0.60 = 0.12

. Because

A

and

C

are mutually exclusive, the intersections

A \cap B

and

B \cap C

do not overlap. Subtracting both intersection probabilities from

P(B)

yields

0.60 - 0.21 - 0.12 = 0.27

.

Step-by-Step Solution

1

Calculate the probability of Event

B

,

P(B)

, using the independence of

A

and

B

.

P(B) = 0.60

Since

A

and

B

are independent,

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

. Substituting into the union formula

P(A \cup B) = P(A) + P(B) - P(A)P(B)

gives

0.74 = 0.35 + P(B)(1 - 0.35)

, so

0.39 = 0.65 P(B)

, yielding

P(B) = 0.60

.

2

Find the joint probability

P(A \cap B)

.

P(A \cap B) = 0.21

By independence,

P(A \cap B) = P(A) \times P(B) = 0.35 \times 0.60 = 0.21

.

3

Calculate the joint probability

P(B \cap C)

using the conditional probability formula.

P(B \cap C) = 0.12

From the definition of conditional probability,

P(C \mid B) = \frac{P(B \cap C)}{P(B)}

, so

P(B \cap C) = P(C \mid B) \cdot P(B) = 0.20 \times 0.60 = 0.12

.

4

Determine the probability that

B

occurs but neither

A

nor

C

occurs.

P(B \cap A^c \cap C^c) = 0.27

Since

A

and

C

are mutually exclusive, events

(A \cap B)

and

(B \cap C)

are disjoint subsets of

B

. Therefore,

P(B \cap A^c \cap C^c) = P(B) - P(A \cap B) - P(B \cap C) = 0.60 - 0.21 - 0.12 = 0.27

.

Key Concept

Probability Rules for Independent, Dependent, and Mutually Exclusive Events

Question 23Question →

A jar contains $4$ red marbles, $6$ blue marbles, and $5$ green marbles. If two marbles are selected at random one after another without replacement, what is the probability that the first marble selected is red and the second marble selected is blue?

\frac{4}{35}

\frac{8}{75}

\frac{73}{105}

\frac{12}{55}

\frac{10}{29}

Show answer & explanation

Answer: $\frac{4}{35}$

Answer

The probability that the first marble selected is red and the second marble selected is blue is

\frac{4}{35}

.

To find the joint probability of two dependent sequential events, multiply the probability of the first event by the conditional probability of the second event. The probability of drawing a red marble first is

\frac{4}{15}

. Because the selection is made without replacement, there are

14

marbles remaining in the jar,

6

of which are blue. The probability of drawing a blue marble second is

\frac{6}{14}

. Multiplying these together yields

\frac{4}{15} \times \frac{6}{14} = \frac{24}{210} = \frac{4}{35}

.

Step-by-Step Solution

1

Calculate the total number of marbles in the jar initially.

Total marbles =

4 + 6 + 5 = 15

.

The sample space size for the first draw is the sum of all marbles.

2

Determine the probability of selecting a red marble on the first draw.

P(\text{Red}_1) = \frac{4}{15}

.

There are

4

favorable outcomes out of

15

total outcomes.

3

Determine the probability of selecting a blue marble on the second draw given that one red marble was removed without replacement.

P(\text{Blue}_2 \mid \text{Red}_1) = \frac{6}{14} = \frac{3}{7}

.

After removing one red marble, there are still

6

blue marbles, but only

14

total marbles remaining.

4

Apply the multiplication rule for dependent events.

P(\text{Red}_1 \text{ and } \text{Blue}_2) = \frac{4}{15} \times \frac{6}{14} = \frac{24}{210} = \frac{4}{35}

.

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

Key Concept

Conditional Probability and Dependent Events

Estimated Time:1m 30s

Question 24Question →

A reliability engineering test evaluates two independent components, Component X and Component Y, in a machine. The probability that Component X fails during operation is $0.20$ , and the probability that Component Y fails during operation is $0.30$ . What is the probability that at least one of the two components operates successfully during operation?

0.44

0.50

0.56

0.70

0.94

Show answer & explanation

Answer: $0.94$

Answer

The probability that at least one of the two components operates successfully is

0.94

.

The correct answer is

0.94

. The complement of the event 'at least one component operates successfully' is the event 'both components fail'. Because Component X and Component Y fail independently,

P(\text{both fail}) = P(\text{X fails}) \times P(\text{Y fails}) = 0.20 \times 0.30 = 0.06

. Subtracting this complementary probability from

1

yields

1 - 0.06 = 0.94

.

Step-by-Step Solution

1

Determine the probability that each component fails.

P(\text{X fails}) = 0.20

and

P(\text{Y fails}) = 0.30

.

These probabilities are explicitly given in the problem statement.

2

Calculate the joint probability that BOTH components fail simultaneously using the multiplication rule for independent events.

P(\text{both fail}) = P(\text{X fails}) \times P(\text{Y fails}) = 0.20 \times 0.30 = 0.06

.

Since the components fail independently, their joint failure probability is the product of their individual failure probabilities.

3

Apply the complement rule to find the probability that at least one component operates successfully.

P(\text{at least one succeeds}) = 1 - P(\text{both fail}) = 1 - 0.06 = 0.94

.

The event 'at least one component succeeds' is the exact complement of 'both components fail'.

Key Concept

Independent Events and Complement Probability Rule

Question 25Question →

An executive is monitoring two independent corporate projects, Project Alpha and Project Beta. Based on historical performance, the probability that Project Alpha meets its deadline is $\frac{4}{5}$ , and the probability that Project Beta meets its deadline is $\frac{3}{4}$ . What is the probability that exactly one of the two projects meets its deadline?

\frac{7}{20}

\frac{3}{5}

\frac{19}{20}

\frac{7}{9}

\frac{1}{20}

Show answer & explanation

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

Answer

The probability that exactly one project meets its deadline is

\frac{7}{20}

.

The event 'exactly one project meets its deadline' consists of two mutually exclusive scenarios: (1) Alpha meets its deadline and Beta does not, or (2) Alpha misses its deadline and Beta meets it. Using independence, the probability of Scenario 1 is

\frac{4}{5} \times \left(1 - \frac{3}{4}\right) = \frac{4}{5} \times \frac{1}{4} = \frac{4}{20}

. The probability of Scenario 2 is

\left(1 - \frac{4}{5}\right) \times \frac{3}{4} = \frac{1}{5} \times \frac{3}{4} = \frac{3}{20}

. Summing these mutually exclusive probabilities gives

\frac{4}{20} + \frac{3}{20} = \frac{7}{20}

.

Step-by-Step Solution

1

Determine the probabilities of individual events and their complements.

P(\text{Alpha meets}) = \frac{4}{5}

,

P(\text{Alpha misses}) = 1 - \frac{4}{5} = \frac{1}{5}

.

P(\text{Beta meets}) = \frac{3}{4}

,

P(\text{Beta misses}) = 1 - \frac{3}{4} = \frac{1}{4}

.

To find the probability of specific outcome combinations, the complementary probabilities for each independent event are required.

2

Identify the mutually exclusive cases that satisfy the condition 'exactly one project meets its deadline'.

Case 1: Alpha meets and Beta misses. Case 2: Alpha misses and Beta meets.

The event 'exactly one' consists of two distinct, non-overlapping scenarios.

3

Calculate the joint probability for each case using independence.

Case 1 probability:

\frac{4}{5} \times \frac{1}{4} = \frac{4}{20}

. Case 2 probability:

\frac{1}{5} \times \frac{3}{4} = \frac{3}{20}

.

Since the projects operate independently, joint probabilities are found by multiplying individual event probabilities.

4

Add the probabilities of the mutually exclusive cases.

\frac{4}{20} + \frac{3}{20} = \frac{7}{20}

.

For mutually exclusive events, the total probability of either case occurring is the sum of their individual probabilities.

Key Concept

Probability of Independent Events and Mutually Exclusive Cases

Question 26Question →

A laboratory tests two solar panels, Panel A and Panel B, under identical conditions. The probability that Panel A operates at peak efficiency on any given day is $0.75$ , and the probability that Panel B operates at peak efficiency on any given day is $0.60$ . The daily efficiency outcomes of the two panels are independent events. Which of the following statements must be true? Select all such statements.

Select all that apply

The probability that both panels operate at peak efficiency on a given day is

0.45

.

The probability that neither panel operates at peak efficiency on a given day is

0.35

.

The probability that at least one panel operates at peak efficiency on a given day is

0.90

.

The conditional probability that Panel A operates at peak efficiency, given that Panel B operates at peak efficiency, is

0.45

.

The probability that Panel A operates at peak efficiency and Panel B does not operate at peak efficiency on a given day is

0.30

.

Show answer & explanation

Answer: The probability that both panels operate at peak efficiency on a given day is $0.45$ .; The probability that at least one panel operates at peak efficiency on a given day is $0.90$ .; The probability that Panel A operates at peak efficiency and Panel B does not operate at peak efficiency on a given day is $0.30$ .

Answer

The correct statements are: the probability that both panels operate at peak efficiency is

0.45

; the probability that at least one panel operates at peak efficiency is

0.90

; and the probability that Panel A operates at peak efficiency while Panel B does not is

0.30

.

The statements confirming that both panels operate at peak efficiency (

0.45

), that at least one operates at peak efficiency (

0.90

), and that Panel A operates while Panel B does not (

0.30

) are mathematically sound applications of independent event rules.

Step-by-Step Solution

1

Identify given probabilities and independence condition

P(A) = 0.75

,

P(B) = 0.60

, and events

A

and

B

are independent.

Establishes the given parameter values.

2

Calculate joint probability of both events occurring

P(A \cap B) = P(A) \times P(B) = 0.75 \times 0.60 = 0.45

.

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

3

Determine complement probabilities and probability of neither event occurring

P(A^c) = 1 - 0.75 = 0.25

,

P(B^c) = 1 - 0.60 = 0.40

, so

P(A^c \cap B^c) = 0.25 \times 0.40 = 0.10

.

Complements of independent events are also independent.

4

Calculate the union probability (at least one panel at peak efficiency)

P(A \cup B) = 1 - P(A^c \cap B^c) = 1 - 0.10 = 0.90

.

The event 'at least one' is the logical complement of 'neither'.

5

Evaluate conditional probability

P(A \mid B)

and difference probability

P(A \cap B^c)

P(A \mid B) = P(A) = 0.75

and

P(A \cap B^c) = 0.75 \times 0.40 = 0.30

.

Independence implies

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

and

P(A \cap B^c) = P(A) P(B^c)

.

Key Concept

Probability rules for independent events, complement rule, and conditional probability definition

Estimated Time:1m 30s

Question 27Question →

Events $A$ and $B$ are mutually exclusive, with $P(A) = 0.25$ and $P(B) = 0.40$ . Event $C$ is independent of both event $A$ and event $B$ , with $P(C) = 0.50$ . What is the probability that event $C$ occurs and at least one of events $A$ or $B$ occurs?

0.050

0.125

0.325

0.650

0.825

Show answer & explanation

Answer: $0.325$

Answer

The probability that event

C

occurs and at least one of events

A

or

B

occurs is

0.325

.

The correct answer is

0.325

. First, since

A

and

B

are mutually exclusive events, the probability of at least one of them occurring is

P(A \text{ or } B) = P(A) + P(B) = 0.25 + 0.40 = 0.65

. Second, because event

C

is independent of both events, the probability that

C

occurs AND at least one of

A

or

B

occurs is given by the multiplication rule for independent events:

P(C) \times P(A \text{ or } B) = 0.50 \times 0.65 = 0.325

.

Step-by-Step Solution

1

Calculate the probability of the union of mutually exclusive events

A

and

B

.

P(A \text{ or } B) = P(A) + P(B) = 0.25 + 0.40 = 0.65

Since

A

and

B

are mutually exclusive,

P(A \cap B) = 0

, so their combined probability is simply the sum of their individual probabilities.

2

Calculate the joint probability of event

C

and event

(A \text{ or } B)

.

P(C \text{ and } (A \text{ or } B)) = P(C) \times P(A \text{ or } B) = 0.50 \times 0.65 = 0.325

Event

C

is independent of both

A

and

B

, which implies

C

is independent of

(A \text{ or } B)

. Therefore, the joint probability is found by multiplying their individual probabilities.

Key Concept

Probability rules for mutually exclusive events (addition rule) and independent events (multiplication rule).

Question 28Question →

Two software security tools, Tool X and Tool Y, operate independently to scan code repositories for vulnerabilities. The probability that Tool X detects a specific type of security flaw is $0.80$ , and the probability that Tool Y detects the same flaw is $0.75$ . What is the probability that exactly one of the two tools detects the flaw?

Show answer & explanation

Answer: 0.35

Answer

The probability that exactly one of the two tools detects the flaw is

0.35

.

The scenario requires finding the probability that exactly one tool detects the flaw. For independent events

X

and

Y

, 'exactly one' consists of two mutually exclusive events: (1) Tool X succeeds while Tool Y fails, which has probability

0.80 \times (1 - 0.75) = 0.80 \times 0.25 = 0.20

, and (2) Tool Y succeeds while Tool X fails, which has probability

0.75 \times (1 - 0.80) = 0.75 \times 0.20 = 0.15

. Summing these mutually exclusive probabilities gives

0.20 + 0.15 = 0.35

. Alternatively, one can subtract the probability of both tools succeeding (

0.80 \times 0.75 = 0.60

) from the probability of at least one tool succeeding (

0.80 + 0.75 - 0.60 = 0.95

), yielding

0.95 - 0.60 = 0.35

.

Step-by-Step Solution

1

Determine the complement probabilities for each independent tool failing to detect the flaw.

P(\text{Not X}) = 1 - 0.80 = 0.20

and

P(\text{Not Y}) = 1 - 0.75 = 0.25

.

The probability of an event not occurring is equal to 1 minus the probability that it occurs.

2

Calculate the joint probability of Tool X detecting the flaw and Tool Y failing to detect it.

P(\text{X and Not Y}) = 0.80 \times 0.25 = 0.20

.

Because the tools operate independently, the joint probability is the product of their individual probabilities.

3

Calculate the joint probability of Tool Y detecting the flaw and Tool X failing to detect it.

P(\text{Y and Not X}) = 0.75 \times 0.20 = 0.15

.

Tool independence allows multiplying the individual probabilities of detection and non-detection.

4

Sum the probabilities of the two mutually exclusive scenarios representing 'exactly one tool detects the flaw'.

P(\text{Exactly One}) = 0.20 + 0.15 = 0.35

.

The events 'X only' and 'Y only' cannot happen simultaneously, so their probabilities add directly.

Key Concept

Independence and Mutual Exclusivity Rules in Compound Probability

Probability of Independent, Dependent, and Mutually Exclusive Events

Answer

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