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

Question 1Question →

In a reliability study of a power station, Event $A$ is defined as an inverter system malfunction and Event $B$ is defined as a battery backup failure during peak hours. The probability that at least one of these two malfunctions occurs is $0.80$ , and the probability that an inverter system malfunction occurs without a battery backup failure is $0.35$ . If Event $A$ and Event $B$ are independent events, what is the probability of an inverter system malfunction, $P(A)$ ?

\frac{7}{20}

\frac{9}{20}

\frac{7}{11}

\frac{11}{20}

\frac{4}{5}

Show answer & explanation

Answer: $\frac{7}{11}$

Answer

The probability of an inverter system malfunction,

P(A)

, is

\frac{7}{11}

.

The probability of the union of two events can be decomposed as

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

. Given

P(A \cup B) = 0.80

and

P(A \cap B^c) = 0.35

, solving yields

P(B) = 0.45

. Because Event

A

and Event

B

are independent,

P(A \cap B) = P(A)P(B) = 0.45 P(A)

. Since

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

, we substitute to get

0.35 = P(A) - 0.45 P(A) = 0.55 P(A)

. Solving for

P(A)

gives

P(A) = \frac{0.35}{0.55} = \frac{7}{11}

.

Step-by-Step Solution

1

Relate the union probability to the difference event

A \cap B^c

and event

B

.

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

.

The union of two events

A

and

B

can be partitioned into the region where only

A

occurs (

A \cap B^c

) and all outcomes in

B

.

2

Calculate

P(B)

using the given values.

P(B) = P(A \cup B) - P(A \cap B^c) = 0.80 - 0.35 = 0.45 = \frac{9}{20}

.

Subtracting

P(A \cap B^c)

from

P(A \cup B)

yields

P(B)

.

3

Apply the definition of independent events to express

P(A \cap B)

in terms of

P(A)

.

P(A \cap B) = P(A) \cdot P(B) = 0.45 P(A)

.

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

4

Set up and solve the equation for

P(A)

.

P(A \cap B^c) = P(A) - P(A \cap B) \implies 0.35 = P(A) - 0.45 P(A) = 0.55 P(A) \implies P(A) = \frac{0.35}{0.55} = \frac{7}{11}

.

Dividing

0.35

by

0.55

yields the exact value of

P(A)

.

Key Concept

Probability rules for independent events and set operations on sample spaces

Estimated Time:2m 0s

Question 2Question →

A quality control inspector evaluates a batch of $16$ precision components. Exactly $10$ of the components meet all engineering specifications, while $6$ have minor surface defects. If the inspector randomly selects $2$ components from the batch one after another without replacement, what is the probability that both selected components meet all engineering specifications?

Show answer & explanation

Answer: 0.375

Answer

The probability that both selected components meet all engineering specifications is

0.375

(or

\frac{3}{8}

).

Since the components are selected without replacement, the outcome of the second draw depends on the outcome of the first draw. The probability of selecting a qualifying component first is

\frac{10}{16}

. Given that a qualifying component was drawn first,

9

qualifying components remain out of

15

total components. The joint probability of both events occurring is

\frac{10}{16} \times \frac{9}{15} = \frac{90}{240} = \frac{3}{8} = 0.375

.

Step-by-Step Solution

1

Determine the probability of selecting a component meeting specifications on the first draw.

P(E_1) = \frac{10}{16} = \frac{5}{8}

There are

10

qualifying components out of

16

total components.

2

Determine the conditional probability of selecting a second component meeting specifications, given that the first component selected also met specifications.

P(E_2 \mid E_1) = \frac{9}{15} = \frac{3}{5}

Because sampling is done without replacement,

9

qualifying components remain out of

15

total remaining components.

3

Apply the multiplication rule for dependent events to calculate the probability of both events occurring.

P(E_1 \cap E_2) = P(E_1) \times P(E_2 \mid E_1) = \frac{5}{8} \times \frac{3}{5} = \frac{3}{8} = 0.375

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

Key Concept

Probability of Dependent Events without Replacement

Estimated Time:1m 15s

Question 3Question →

Two events $A$ and $B$ within a sample space have probabilities $P(A) = 0.60$ and $P(B) = 0.30$ . Which of the following statements must be true? Select all that apply.

Select all that apply

If events

A

and

B

are independent, the probability that at least one of the two events occurs is

0.72

.

If events

A

and

B

are mutually exclusive, the joint probability

P(A \text{ and } B)

is

0

.

Events

A

and

B

can be both mutually exclusive and independent.

If events

A

and

B

are independent, the probability that both events occur is

0.90

.

If events

A

and

B

are mutually exclusive, the conditional probability

P(A \mid B)

is

0.30

.

Show answer & explanation

Answer: If events $A$ and $B$ are independent, the probability that at least one of the two events occurs is $0.72$ .; If events $A$ and $B$ are mutually exclusive, the joint probability $P(A \text{ and } B)$ is $0$ .

Answer

The correct statements are that if events are independent, the probability that at least one occurs is 0.72, and if events are mutually exclusive, their joint probability is 0.

The statement regarding independent events is correct because

P(A \cup B) = P(A) + P(B) - P(A)P(B) = 0.60 + 0.30 - 0.18 = 0.72

. The statement regarding mutually exclusive events is correct because by definition, mutually exclusive events cannot occur together, meaning

P(A \text{ and } B) = 0

.

Step-by-Step Solution

1

Analyze independence condition for

P(A \cup B)

P(A \cap B) = 0.60 \times 0.30 = 0.18

, so

P(A \cup B) = 0.60 + 0.30 - 0.18 = 0.72

.

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

2

Analyze mutual exclusivity definition

P(A \cap B) = 0

.

Mutually exclusive events cannot occur simultaneously.

3

Evaluate simultaneous independence and mutual exclusivity

Since

P(A)P(B) = 0.18 \neq 0

, the events cannot satisfy both conditions simultaneously.

Independence requires

P(A \cap B) = 0.18

, while mutual exclusivity requires

P(A \cap B) = 0

.

Key Concept

Probability rules for independent and mutually exclusive events

Question 4Question →

Events $E$ and $F$ are two events in a sample space such that $P(E) = 0.30$ and $P(F) = 0.40$ . Which of the following statements must be true? Select all that apply.

Select all that apply

If

E

and

F

are mutually exclusive, then

P(E \text{ and } F) = 0

.

If

E

and

F

are independent, then

P(E \text{ and } F) = 0.12

.

If

E

and

F

are independent, then

P(E \text{ or } F) = 0.70

.

If

E

and

F

are mutually exclusive, then

E

and

F

are independent.

If

E

and

F

are independent, then

P(E \text{ given } F) = 0.12

.

Show answer & explanation

Answer: If $E$ and $F$ are mutually exclusive, then $P(E \text{ and } F) = 0$ .; If $E$ and $F$ are independent, then $P(E \text{ and } F) = 0.12$ .

Answer

The correct statements are that if

E

and

F

are mutually exclusive, then

P(E \text{ and } F) = 0

, and if

E

and

F

are independent, then

P(E \text{ and } F) = 0.12

.

The statement asserting that

P(E \text{ and } F) = 0

for mutually exclusive events is correct because mutually exclusive events by definition cannot occur together. The statement asserting that

P(E \text{ and } F) = 0.12

for independent events is correct because independent events satisfy the multiplication rule

P(E \text{ and } F) = P(E) \times P(F) = 0.30 \times 0.40 = 0.12

.

Step-by-Step Solution

1

Evaluate the statement regarding mutually exclusive events.

P(E \text{ and } F) = 0

.

By definition, mutually exclusive events cannot both happen at the same time, so

P(E \text{ and } F) = 0

is true.

2

Evaluate the joint probability for independent events.

P(E \text{ and } F) = P(E) \times P(F) = 0.30 \times 0.40 = 0.12

.

The multiplication rule for independent events states that

P(E \text{ and } F) = P(E) \cdot P(F)

.

3

Evaluate the union probability

P(E \text{ or } F)

for independent events.

P(E \text{ or } F) = P(E) + P(F) - P(E \text{ and } F) = 0.30 + 0.40 - 0.12 = 0.58

.

Simply adding

P(E) + P(F) = 0.70

fails to subtract the intersection that is counted twice.

4

Evaluate conditional probability for independent events.

P(E \text{ given } F) = P(E) = 0.30

.

Independence implies that knowing event

F

occurred does not change the probability of event

E

.

Key Concept

Probability rules for mutually exclusive and independent events

Question 5Question →

A fair six-sided die with faces numbered $1$ through $6$ is rolled twice. What is the probability of rolling a $4$ on the first roll and an odd number on the second roll?

\frac{1}{12}

\frac{2}{3}

\frac{1}{3}

\frac{1}{4}

\frac{1}{36}

Show answer & explanation

Answer: $\frac{1}{12}$

Answer

\frac{1}{12}

Because the outcome of the first die roll does not affect the outcome of the second die roll, the two events are independent. The probability of rolling a

4

on the first roll is

\frac{1}{6}

, and the probability of rolling an odd number (

1, 3,

or

5

) on the second roll is

\frac{3}{6} = \frac{1}{2}

. Multiplying these individual probabilities together yields

\frac{1}{6} \times \frac{1}{2} = \frac{1}{12}

.

Step-by-Step Solution

1

Determine the probability of the first event (rolling a 4).

There is

1

favorable outcome out of

6

possible outcomes, so

P(\text{First roll is } 4) = \frac{1}{6}

.

Each face of a fair six-sided die is equally likely to land face up.

2

Determine the probability of the second event (rolling an odd number).

The odd numbers are

1, 3,

and

5

, giving

3

favorable outcomes out of

6

total outcomes, so

P(\text{Second roll is odd}) = \frac{3}{6} = \frac{1}{2}

.

Half of the outcomes on a standard six-sided die are odd.

3

Calculate the joint probability of both independent events occurring.

P(\text{4 on first AND odd on second}) = \frac{1}{6} \times \frac{1}{2} = \frac{1}{12}

.

For independent events

A

and

B

, the multiplication rule states that

P(A \text{ and } B) = P(A) \times P(B)

.

Key Concept

Probability of Independent Events

Estimated Time:1m 0s

Question 6Question →

A container holds 5 red marbles, 4 blue marbles, and 3 green marbles. Two marbles are drawn sequentially at random without replacement. Let event $A$ be the event that at least one of the drawn marbles is red, and let event $B$ be the event that the second marble drawn is green. What is the conditional probability $P(B \mid A)$ ?

\frac{1}{6}

\frac{1}{4}

\frac{3}{22}

\frac{3}{19}

\frac{5}{22}

Show answer & explanation

Answer: $\frac{1}{6}$

Answer

The conditional probability

P(B \mid A)

is

\frac{1}{6}

.

To find

P(B \mid A)

, we evaluate

\frac{P(A \cap B)}{P(A)}

. The probability of at least one red marble

P(A)

is

1 - P(\text{no red}) = 1 - \frac{7 \times 6}{12 \times 11} = \frac{90}{132}

. The intersection event

A \cap B

requires the second marble to be green and at least one marble to be red, which means the first marble must be red and the second green. The probability of this is

\frac{5 \times 3}{12 \times 11} = \frac{15}{132}

. Taking the ratio

\frac{15/132}{90/132}

yields

\frac{15}{90} = \frac{1}{6}

.

Step-by-Step Solution

1

Calculate the total number of ordered outcomes and the probability of event A using the complementary event.

Total outcomes drawing 2 marbles from 12 without replacement is

12 \times 11 = 132

. The complement

A^c

(no red marbles selected from the 7 non-red marbles) has

7 \times 6 = 42

outcomes. Thus,

P(A^c) = \frac{42}{132} = \frac{7}{22}

, which means

P(A) = 1 - \frac{7}{22} = \frac{15}{22} = \frac{90}{132}

.

Using the complement rule is the most efficient way to compute 'at least one' probabilities.

2

Determine the intersection event

A \cap B

and calculate its probability.

Event

B

specifies that the second marble is green. For event

A

(at least one marble is red) to also occur, the first marble must be red. Thus,

A \cap B

is equivalent to 'the first marble is red AND the second marble is green'. The number of favorable outcomes is

5 \times 3 = 15

. So

P(A \cap B) = \frac{15}{132}

.

Mutual exclusivity between red and green on the second draw simplifies the intersection logic.

3

Apply the conditional probability formula

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

.

P(B \mid A) = \frac{\frac{15}{132}}{\frac{90}{132}} = \frac{15}{90} = \frac{1}{6}

.

Evaluating the ratio yields the exact conditional probability requested.

Key Concept

Conditional Probability of Dependent Events

Estimated Time:3m 0s

Question 7Question →

A jar contains $3$ red marbles and $7$ blue marbles. A marble is drawn at random from the jar, its color is noted, and it is returned to the jar. A second marble is then drawn at random. What is the probability that both drawn marbles are red?

\frac{9}{100}

\frac{3}{10}

\frac{3}{5}

\frac{9}{49}

\frac{6}{7}

Show answer & explanation

Answer: $\frac{9}{100}$

Answer

The probability that both drawn marbles are red is

\frac{9}{100}

.

The option stating

\frac{9}{100}

is correct because the two draws are independent due to replacement. The probability of getting a red marble on any single draw is

\frac{3}{10}

. By the multiplication rule for independent events, the probability of both events occurring is

\frac{3}{10} \times \frac{3}{10} = \frac{9}{100}

.

Step-by-Step Solution

1

Calculate the total number of marbles in the jar.

3 \text{ red} + 7 \text{ blue} = 10 \text{ total marbles}

.

Probability requires knowing the size of the full sample space.

2

Find the probability of drawing a red marble on a single draw.

P(\text{Red}) = \frac{3}{10}

.

There are

3

favorable outcomes (red marbles) out of

10

total possible outcomes.

3

Apply the multiplication rule for independent events.

P(\text{Red}_1 \text{ and } \text{Red}_2) = \frac{3}{10} \times \frac{3}{10} = \frac{9}{100}

.

Because the first marble is replaced, the second draw is independent of the first, so joint probability is the product of individual probabilities.

Key Concept

Multiplication Rule for Independent Events:

P(A \text{ and } B) = P(A) \times P(B)

when events

A

and

B

are independent.

Estimated Time:45s

Question 8Question →

A machine operates using two independent components, Component $A$ and Component $B$ . The probability that Component $A$ functions properly on a given day is $0.90$ , and the probability that Component $B$ functions properly on that same day is $0.80$ . What is the probability that at least one of the components functions properly on a given day?

Show answer & explanation

Answer: 0.98

Answer

The probability that at least one component functions properly is

0.98

.

To determine the probability that at least one component functions properly, use the complement rule:

P(\text{at least one}) = 1 - P(\text{neither})

. Since Component

A

and Component

B

operate independently, the probability that

A

fails is

1 - 0.90 = 0.10

and the probability that

B

fails is

1 - 0.80 = 0.20

. The probability of both components failing simultaneously is

0.10 \times 0.20 = 0.02

. Subtracting this probability from

1

gives

1 - 0.02 = 0.98

.

Step-by-Step Solution

1

Find the probability of failure for each component.

P(A^c) = 1 - 0.90 = 0.10

and

P(B^c) = 1 - 0.80 = 0.20

The event that a component fails is the complement of the event that it functions properly.

2

Calculate the joint probability of both components failing.

P(A^c \text{ and } B^c) = 0.10 \times 0.20 = 0.02

Because the components operate independently, their failure events are independent, so their individual probabilities are multiplied.

3

Calculate the probability that at least one component functions properly.

P(\text{at least one functions}) = 1 - 0.02 = 0.98

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

Key Concept

Probability of Independent Events and Complement Rule

Estimated Time:45s

Question 9Question →

Let $A$ and $B$ be two events in a sample space such that $0 < P(A) < 1$ and $0 < P(B) < 1$ . Which of the following statements must be true? Select all that apply.

Select all that apply

If

A

and

B

are mutually exclusive, then

A

and

B

cannot be independent.

If

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

, then

A

and

B

are independent events.

If

A

and

B

are independent events, then their complements

A^c

and

B^c

must be mutually exclusive.

If

P(A|B) > P(A)

, then

P(B|A) > P(B)

.

If

A

and

B

are independent events, then

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

.

Show answer & explanation

Answer: If $A$ and $B$ are mutually exclusive, then $A$ and $B$ cannot be independent.; If $P(A \cup B) = P(A) + P(B) - P(A)P(B)$ , then $A$ and $B$ are independent events.; If $P(A|B) > P(A)$ , then $P(B|A) > P(B)$ .

Answer

The statements asserting that mutually exclusive non-impossible events cannot be independent, that

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

implies independence, and that

P(A|B) > P(A)

implies

P(B|A) > P(B)

are all true.

For events with probabilities strictly between 0 and 1: (1) Mutual exclusivity requires

P(A \cap B) = 0

, whereas independence requires

P(A \cap B) = P(A)P(B) > 0

, so mutually exclusive events cannot be independent. (2) Substituting

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

into the addition rule yields

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

, which defines independence. (3)

P(A|B) > P(A)

is mathematically equivalent to

P(A \cap B) > P(A)P(B)

, which in turn is equivalent to

P(B|A) > P(B)

.

Step-by-Step Solution

1

Analyze the relationship between mutual exclusivity and independence.

Mutually exclusive events satisfy

P(A \cap B) = 0

. For independent events,

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

. Since

P(A) > 0

and

P(B) > 0

,

P(A)P(B) > 0 \neq 0

. Thus, mutually exclusive non-impossible events can never be independent.

To evaluate structural compatibility between mutual exclusivity and independence.

2

Apply the addition rule of probability to check the union equation.

General addition rule:

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

. Comparing to the given

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

shows

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

, which is the exact condition for independence.

To verify if the given expression for union probability forces independence.

3

Evaluate the joint probability of complementary events

A^c

and

B^c

.

Independence of

A

and

B

implies independence of

A^c

and

B^c

. Thus

P(A^c \cap B^c) = (1-P(A))(1-P(B)) > 0

. Because the joint probability is positive, the complements are not mutually exclusive.

To test whether independence of events implies mutual exclusivity of their complements.

4

Examine the symmetry of conditional probability inequalities.

P(A|B) > P(A) \implies P(A \cap B) > P(A)P(B) \implies P(B|A) = \frac{P(A \cap B)}{P(A)} > P(B)

.

To evaluate directional dependence between conditional probabilities.

5

Check the simple addition rule for independent events.

Simple addition

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

applies only when

P(A \cap B) = 0

. Independent events have

P(A \cap B) = P(A)P(B) > 0

, so

P(A \cup B) < P(A) + P(B)

.

To distinguish between addition rules for mutually exclusive vs. independent events.

Key Concept

Theoretical relationships between independent, dependent, mutually exclusive, and conditional events.

Estimated Time:3m 0s

Question 10Question →

A box contains $4$ red blocks and $6$ yellow blocks. A block is selected at random from the box, its color is noted, and it is returned to the box. A second block is then selected at random from the box. What is the probability that both selected blocks are red?

\frac{4}{25}

\frac{2}{15}

\frac{2}{5}

\frac{4}{9}

\frac{4}{5}

Show answer & explanation

Answer: $\frac{4}{25}$

Answer

The probability that both selected blocks are red is

\frac{4}{25}

.

Because the first block is returned to the box before the second selection, the two draws are independent events. The probability of selecting a red block on any single draw is

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

. Applying the multiplication rule for independent events gives

P(\text{Both red}) = \frac{2}{5} \times \frac{2}{5} = \frac{4}{25}

.

Step-by-Step Solution

1

Determine the total number of blocks in the box.

The total number of blocks is

4 + 6 = 10

.

Probability requires finding the ratio of favorable outcomes to total possible outcomes.

2

Calculate the probability of drawing a red block on the first selection.

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

.

There are

4

red blocks out of

10

total blocks.

3

Calculate the probability of drawing a red block on the second selection.

Since the first block is returned to the box, the events are independent, so

P(\text{Second is red}) = \frac{4}{10} = \frac{2}{5}

.

Replacement preserves the original sample space composition.

4

Apply the multiplication rule for independent events.

P(\text{Both are red}) = P(\text{First is red}) \times P(\text{Second is red}) = \frac{2}{5} \times \frac{2}{5} = \frac{4}{25}

.

The probability of two independent events both occurring is the product of their individual probabilities.

Key Concept

Probability of Independent Events

Question 11Question →

A quality inspection bin contains $6$ components manufactured by Line 1 and $4$ components manufactured by Line 2. Two components are drawn randomly from the bin sequentially, without replacement. Let $B$ be the event that the second component drawn is manufactured by Line 1, and let $C$ be the event that at least one of the two components drawn is manufactured by Line 2. What is the conditional probability $P(B \mid C)$ ?

\frac{2}{5}

\frac{4}{15}

\frac{3}{5}

\frac{4}{9}

\frac{2}{3}

Show answer & explanation

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

Answer

The conditional probability

P(B \mid C)

is

\frac{2}{5}

.

To find

P(B \mid C)

, we evaluate the ratio

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

. The probability of event

C

(at least one component from Line 2) is most easily found by taking the complement of drawing two Line 1 components:

1 - \frac{6 \times 5}{10 \times 9} = \frac{60}{90}

. For event

B \cap C

to occur, the second component must be Line 1 and at least one component must be Line 2, meaning the sequence must be (Line 2, Line 1), which has probability

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

. Dividing

\frac{24}{90}

by

\frac{60}{90}

yields

\frac{24}{60} = \frac{2}{5}

.

Step-by-Step Solution

1

Calculate the total number of outcomes for drawing two components sequentially without replacement.

Total outcomes =

10 \times 9 = 90

.

There are 10 components available for the first selection and 9 remaining components for the second selection.

2

Determine the probability of event

C

(at least one component from Line 2) using the complement rule.

P(C) = 1 - P(\text{both from Line 1}) = 1 - \frac{6 \times 5}{90} = 1 - \frac{30}{90} = \frac{60}{90} = \frac{2}{3}

.

The complement of having at least one component from Line 2 is having both components drawn from Line 1.

3

Determine the probability of the joint event

B \cap C

.

P(B \cap C) = P(\text{first from Line 2 AND second from Line 1}) = \frac{4 \times 6}{90} = \frac{24}{90} = \frac{4}{15}

.

For event

B

(second is Line 1) and event

C

(at least one is Line 2) to occur simultaneously, the first component must be from Line 2 and the second from Line 1.

4

Apply the conditional probability formula

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

.

P(B \mid C) = \frac{24/90}{60/90} = \frac{24}{60} = \frac{2}{5}

.

The conditional probability isolates the probability of event

B

within the reduced sample space where event

C

has occurred.

Key Concept

Conditional Probability and Dependent Sequential Events

Estimated Time:2m 30s

Question 12Question →

A box contains $10$ cards: $4$ blue cards numbered $1, 2, 3, 5$ and $6$ red cards numbered $1, 2, 3, 4, 6, 8$ . Two cards are drawn sequentially at random without replacement from the box. Let $A$ be the event that the first card drawn is blue, and let $B$ be the event that the sum of the numbers on the two drawn cards is an even number. What is the value of the conditional probability $P(A \mid B)$ ?

Show answer & explanation

Answer: 0.4

Answer

0.4 (or 2/5)

The conditional probability

P(A \mid B)

represents the likelihood that the first card drawn was blue given that the sum of the two drawn cards is even. There are 40 total outcome pairs resulting in an even sum (20 where both are odd and 20 where both are even). Among these 40 outcomes, exactly 16 start with a blue card (12 starting with a blue odd card and 4 starting with a blue even card). Therefore,

P(A \mid B) = \frac{16}{40} = 0.4

.

Step-by-Step Solution

1

Classify the sample space of cards by color and number parity.

Blue cards consist of 3 odds (1, 3, 5) and 1 even (2). Red cards consist of 2 odds (1, 3) and 4 evens (2, 4, 6, 8). Across all 10 cards, there are 5 odd cards and 5 even cards.

Categorizing by parity is essential because the sum of two integers is even if and only if both numbers share the same parity (both odd or both even).

2

Calculate the total number of sequential draw outcomes belonging to event

B

(sum is even).

Number of (Odd, Odd) outcomes =

5 \times 4 = 20

. Number of (Even, Even) outcomes =

5 \times 4 = 20

. Total outcomes for event

B

,

N(B) = 20 + 20 = 40

.

Since draws are without replacement, drawing a card reduces the available count of that parity by 1 for the second draw.

3

Calculate the number of outcomes belonging to the joint event

A \cap B

(first card is blue AND sum is even).

Subcase 1 (Blue Odd 1st, Odd 2nd):

3 \times 4 = 12

outcomes. Subcase 2 (Blue Even 1st, Even 2nd):

1 \times 4 = 4

outcomes. Total outcomes for

A \cap B

,

N(A \cap B) = 12 + 4 = 16

.

To satisfy both event

A

(first card blue) and event

B

(even sum), the second card must match the parity of the selected blue card.

4

Compute the conditional probability

P(A \mid B)

.

P(A \mid B) = \frac{N(A \cap B)}{N(B)} = \frac{16}{40} = \frac{2}{5} = 0.4

.

By the definition of conditional probability,

P(A \mid B) = \frac{P(A \cap B)}{P(B)} = \frac{N(A \cap B)}{N(B)}

when all outcomes in the reduced sample space are equally likely.

Key Concept

Conditional Probability and Sequential Dependent Sampling

Question 13Question →

Two events $A$ and $B$ are defined on a sample space such that $P(A) = 0.60$ and $P(B) = 0.75$ . Which of the following statements must be true? Select all such statements.

Select all that apply

Events

A

and

B

cannot be mutually exclusive.

The probability that both events

A

and

B

occur,

P(A \cap B)

, is at least

0.35

.

Events

A

and

B

must be independent.

The conditional probability

P(A \mid B)

is at least

\frac{7}{15}

.

The maximum possible value of

P(A \text{ or } B)

is

0.90

.

Show answer & explanation

Answer: Events $A$ and $B$ cannot be mutually exclusive.; The probability that both events $A$ and $B$ occur, $P(A \cap B)$ , is at least $0.35$ .; The conditional probability $P(A \mid B)$ is at least $\frac{7}{15}$ .

Answer

The statements asserting that events

A

and

B

cannot be mutually exclusive, that the joint probability

P(A \cap B)

is at least

0.35

, and that the conditional probability

P(A \mid B)

is at least

\frac{7}{15}

are all correct.

The sum of the probabilities of events

A

and

B

(

1.35

) exceeds

1

, making mutual exclusivity impossible. The inclusion-exclusion principle dictates

P(A \cap B) \geq 0.60 + 0.75 - 1.00 = 0.35

. Consequently, the minimum conditional probability

P(A \mid B)

is

\frac{0.35}{0.75} = \frac{7}{15}

.

Step-by-Step Solution

1

Evaluate mutual exclusivity

If

A

and

B

were mutually exclusive,

P(A \cap B) = 0

, so

P(A \cup B) = P(A) + P(B) = 0.60 + 0.75 = 1.35

. Since probability cannot exceed

1

, the events cannot be mutually exclusive.

Verify if the sum of individual probabilities exceeds 1.

2

Determine the minimum joint probability

P(A \cap B)

Using

P(A \cup B) = P(A) + P(B) - P(A \cap B) \leq 1

, we have

0.60 + 0.75 - P(A \cap B) \leq 1 \implies P(A \cap B) \geq 0.35

.

Apply the inclusion-exclusion principle bounded by maximum total probability.

3

Test for required independence

Independence requires

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

. Since

P(A \cap B)

can legitimately range anywhere between

0.35

and

0.60

, independence is possible but not guaranteed.

Check whether joint probability is strictly fixed at the product of individual probabilities.

4

Calculate the lower bound for conditional probability

P(A \mid B)

P(A \mid B) = \frac{P(A \cap B)}{P(B)} \geq \frac{0.35}{0.75} = \frac{35}{75} = \frac{7}{15}

.

Substitute the minimum joint probability into the conditional probability formula.

Key Concept

Probability rules governing overlap, mutual exclusivity, joint probability bounds, and conditional probability.

Question 14Question →

A container holds $n$ spheres, exactly 5 of which are blue and the remaining $n - 5$ are green. Two spheres are drawn at random from the container one after another without replacement. If the probability that at least one of the selected spheres is green is $\frac{14}{15}$ , what is the total number of spheres $n$ in the container?

15

20

25

30

75

Show answer & explanation

Answer: 25

Answer

The total number of spheres

n

in the container is 25.

The probability of at least one green sphere is complementary to drawing zero green spheres (meaning both spheres drawn are blue). Subtracting

\frac{14}{15}

from 1 yields

P(\text{both blue}) = \frac{1}{15}

. Since the selection is without replacement,

P(\text{both blue}) = \frac{5}{n} \times \frac{4}{n-1} = \frac{20}{n(n-1)}

. Equating this to

\frac{1}{15}

gives

n(n-1) = 300

. Solving the quadratic equation

n^2 - n - 300 = 0

gives

n = 25

, which correctly represents the total number of spheres.

Step-by-Step Solution

1

Use the complement rule to determine the probability that both selected spheres are blue.

P(\text{both blue}) = 1 - P(\text{at least one green}) = 1 - \frac{14}{15} = \frac{1}{15}

.

The event that at least one sphere is green is the complement of the event that both drawn spheres are blue.

2

Set up the joint probability equation for drawing two blue spheres sequentially without replacement.

P(\text{both blue}) = \frac{5}{n} \times \frac{4}{n - 1} = \frac{20}{n(n - 1)}

.

There are 5 blue spheres initially out of

n

. After drawing one blue sphere, 4 blue spheres remain out of

n - 1

total spheres.

3

Equate the expressions and solve for

n

.

\begin{aligned} \frac{20}{n(n - 1)} &= \frac{1}{15} \\ n(n - 1) &= 300 \\ n^2 - n - 300 &= 0 \\ (n - 25)(n + 12) &= 0 \end{aligned}

Cross-multiplying gives a quadratic equation in terms of

n

.

4

Select the valid positive integer solution for

n

.

n = 25

(since

n > 0

).

The total number of spheres must be a positive integer.

Key Concept

Probability of Complementary Events and Dependent Sequential Events

Estimated Time:2m 0s

Question 15Question →

An automated risk-management system uses three independent algorithms—Algorithm X, Algorithm Y, and Algorithm Z—to detect fraudulent transactions. The probability that Algorithm X detects a given fraudulent transaction is $\frac{3}{5}$ , the probability that Algorithm Y detects it is $\frac{2}{3}$ , and the probability that Algorithm Z detects it is $\frac{3}{4}$ . If a fraudulent transaction occurs, what is the probability that it will be detected by at least two of these three algorithms?

Show answer & explanation

Answer: 0.75

Answer

The probability that the transaction is detected by at least two of the three algorithms is 0.75 (or 3/4).

Because the algorithms operate independently, the event 'at least two algorithms detect the transaction' consists of four mutually exclusive outcomes: exactly X and Y detect (probability 6/60 = 0.10), exactly X and Z detect (probability 9/60 = 0.15), exactly Y and Z detect (probability 12/60 = 0.20), and all three detect (probability 18/60 = 0.30). Summing these four probabilities gives 0.10 + 0.15 + 0.20 + 0.30 = 0.75.

Step-by-Step Solution

1

Determine the complementary probabilities of non-detection for each algorithm.

P(X does not detect) = 2/5, P(Y does not detect) = 1/3, and P(Z does not detect) = 1/4.

The probability of an event's complement is 1 minus the probability of the event.

2

Calculate the probability for each scenario where exactly two algorithms detect the transaction.

P(X and Y only) = 6/60, P(X and Z only) = 9/60, P(Y and Z only) = 12/60.

Since the algorithms operate independently, joint probabilities are calculated by multiplying individual probabilities.

3

Calculate the probability that all three algorithms detect the transaction.

P(X and Y and Z) = 18/60.

Multiplying the individual detection probabilities of all three independent algorithms.

4

Sum the probabilities of all qualifying mutually exclusive outcomes.

(6/60) + (9/60) + (12/60) + (18/60) = 45/60 = 0.75.

The events representing different combinations of detections are mutually exclusive, so their probabilities add directly.

Key Concept

Independent Events and Addition Rule for Mutually Exclusive Outcomes

Question 16Question →

A committee of $3$ members is to be selected at random without replacement from a group of $5$ data scientists and $5$ software engineers. What is the conditional probability that at least two data scientists are selected in total, given that the first person selected is a data scientist?

\frac{5}{18}

\frac{1}{2}

\frac{5}{9}

\frac{13}{18}

\frac{3}{4}

Show answer & explanation

Answer: $\frac{13}{18}$

Answer

The conditional probability is

\frac{13}{18}

.

The conditional probability is calculated by adjusting the sample space after the first draw. With 1 data scientist already selected, there are 4 data scientists and 5 software engineers left (9 total). Selecting at least 1 more data scientist in the next 2 draws has a probability complementary to selecting 0 more data scientists. The probability of selecting 0 additional data scientists is

\frac{\binom{5}{2}}{\binom{9}{2}} = \frac{10}{36} = \frac{5}{18}

. Therefore, the required conditional probability is

1 - \frac{5}{18} = \frac{13}{18}

.

Step-by-Step Solution

1

Determine the remaining pool of candidates after the first selection.

Since 1 data scientist is selected first, 4 data scientists and 5 software engineers remain (total of 9 candidates). Two more candidates must be selected.

The conditional statement fixes the first selection, changing the sample space for the remaining 2 selections.

2

Identify the condition required for the target event to occur.

Since 1 data scientist is already selected, having at least two data scientists in total means selecting at least 1 additional data scientist in the remaining 2 draws.

Total data scientists = 1 (first draw) + (number of data scientists in next 2 draws).

3

Calculate the complementary probability (selecting 0 additional data scientists).

The number of ways to pick 2 software engineers from 5 is

\binom{5}{2} = 10

. The total ways to pick 2 people from 9 is

\binom{9}{2} = 36

. The probability of 0 additional data scientists is

\frac{10}{36} = \frac{5}{18}

.

Selecting 0 additional data scientists is equivalent to selecting 2 software engineers from the remaining group.

4

Subtract the complementary probability from 1.

1 - \frac{5}{18} = \frac{13}{18}

.

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

Key Concept

Conditional Probability without Replacement

Question 17Question →

A quality control analyst evaluates a manufacturing process in which two specific types of flaws, Flaw $X$ and Flaw $Y$ , can occur on produced glass panels. The probability that a randomly selected panel has Flaw $X$ is $P(X) = 0.20$ , and the probability that it has Flaw $Y$ is $P(Y) = 0.30$ . The analyst confirms that Flaw $X$ and Flaw $Y$ are mutually exclusive events.

Which of the following statements MUST be true regarding these two flaw types? Select all such statements.

Select all that apply

The probability that a randomly selected panel has at least one of the two flaws is

0.50

.

The conditional probability of a panel having Flaw

X

given that it has Flaw

Y

,

P(X \mid Y)

, is equal to

0

.

The probability that a randomly selected panel has both Flaw

X

and Flaw

Y

is

0.06

.

Flaw

X

and Flaw

Y

are independent events.

The probability that a randomly selected panel has neither Flaw

X

nor Flaw

Y

is

0.94

.

Show answer & explanation

Answer: The probability that a randomly selected panel has at least one of the two flaws is $0.50$ .; The conditional probability of a panel having Flaw $X$ given that it has Flaw $Y$ , $P(X \mid Y)$ , is equal to $0$ .

Answer

The statement that the probability of having at least one flaw is 0.50 and the statement that the conditional probability P(X | Y) is 0 are both true.

Because Flaw

X

and Flaw

Y

are mutually exclusive, their intersection

P(X \cap Y) = 0

. By the addition rule, the probability of at least one flaw is

P(X \cup Y) = P(X) + P(Y) = 0.20 + 0.30 = 0.50

. Furthermore, the conditional probability

P(X \mid Y) = \frac{P(X \cap Y)}{P(Y)} = \frac{0}{0.30} = 0

. Thus, both the statement claiming the union probability is

0.50

and the statement claiming the conditional probability is

0

are correct.

Step-by-Step Solution

1

Analyze the definition of mutually exclusive events

Since Flaw

X

and Flaw

Y

are mutually exclusive, they cannot occur simultaneously on the same panel. Therefore,

P(X \cap Y) = 0

.

By definition, mutually exclusive events have an intersection probability of zero.

2

Calculate the union probability P(X or Y)

Using the addition rule for mutually exclusive events:

P(X \cup Y) = P(X) + P(Y) - P(X \cap Y) = 0.20 + 0.30 - 0 = 0.50

.

The probability of at least one event occurring is the sum of their individual probabilities when the intersection is zero.

3

Calculate the conditional probability P(X | Y)

P(X \mid Y) = \frac{P(X \cap Y)}{P(Y)} = \frac{0}{0.30} = 0

.

If Flaw

Y

is known to occur, Flaw

X

cannot occur due to mutual exclusivity.

4

Evaluate independence between the two events

For independence,

P(X \cap Y)

must equal

P(X) \times P(Y) = 0.20 \times 0.30 = 0.06

. Since

0 \neq 0.06

, the events are dependent.

Two events with non-zero probabilities that are mutually exclusive are always dependent because the occurrence of one guarantees the non-occurrence of the other.

Key Concept

Mutually Exclusive vs. Independent Events

Question 18Question →

A medical laboratory uses an automated analyzer to screen blood samples for two distinct markers, Marker A and Marker B. The probability that a randomly selected sample contains Marker A is $0.40$ , and the probability that it contains Marker B is $0.25$ . If the presence of Marker A and the presence of Marker B are independent events, what is the probability that a randomly selected sample contains at least one of these two markers?

0.10

0.45

0.55

0.65

0.70

Show answer & explanation

Answer: 0.55

Answer

The probability that a randomly selected sample contains at least one of the two markers is

0.55

.

To find the probability that a sample contains at least one marker, apply the general addition rule

P(\text{A or B}) = P(\text{A}) + P(\text{B}) - P(\text{A and B})

. Because the events are independent,

P(\text{A and B}) = P(\text{A}) \times P(\text{B}) = 0.40 \times 0.25 = 0.10

. Substituting the values gives

0.40 + 0.25 - 0.10 = 0.55

. Alternatively, using the complementary probability rule yields

1 - P(\text{neither}) = 1 - (1 - 0.40)(1 - 0.25) = 1 - (0.60 \times 0.75) = 1 - 0.45 = 0.55

.

Step-by-Step Solution

1

Identify the given probabilities and event relationship.

P(\text{A}) = 0.40

,

P(\text{B}) = 0.25

, and events A and B are independent.

Establishing the parameters is necessary to apply the appropriate probability formulas.

2

Calculate the joint probability

P(\text{A and B})

.

P(\text{A and B}) = P(\text{A}) \times P(\text{B}) = 0.40 \times 0.25 = 0.10

.

For independent events, the probability of both events occurring simultaneously is the product of their individual probabilities.

3

Apply the general addition rule for probability to find

P(\text{A or B})

.

P(\text{A or B}) = P(\text{A}) + P(\text{B}) - P(\text{A and B}) = 0.40 + 0.25 - 0.10 = 0.55

.

The probability of at least one event occurring requires subtracting the overlapping joint probability to avoid double-counting.

Key Concept

Probability of Independent Events and the General Addition Rule

Question 19Question →

Two events $E$ and $F$ in a sample space have probabilities $P(E) = 0.40$ and $P(F) = 0.50$ . Which of the following statements must be true? Select all such statements.

Select all that apply

If events

E

and

F

are mutually exclusive, then

P(E \text{ or } F) = 0.90

.

If events

E

and

F

are independent, then

P(E \text{ and } F) = 0.20

.

Events

E

and

F

cannot be both mutually exclusive and independent.

If events

E

and

F

are independent, then

P(E \text{ or } F) = 0.90

.

If events

E

and

F

are mutually exclusive, then

P(E \text{ and } F) = 0.90

.

Show answer & explanation

Answer: If events $E$ and $F$ are mutually exclusive, then $P(E \text{ or } F) = 0.90$ .; If events $E$ and $F$ are independent, then $P(E \text{ and } F) = 0.20$ .; Events $E$ and $F$ cannot be both mutually exclusive and independent.

Answer

The correct statements are: 'If events E and F are mutually exclusive, then P(E or F) = 0.90', 'If events E and F are independent, then P(E and F) = 0.20', and 'Events E and F cannot be both mutually exclusive and independent.'

The statement regarding mutually exclusive events correctly uses the addition rule P(E or F) = P(E) + P(F) = 0.90 because there is no overlap. The statement regarding independent events correctly applies the multiplication rule P(E and F) = P(E) * P(F) = 0.20. Finally, events with non-zero probabilities cannot be both mutually exclusive (requiring zero intersection) and independent (requiring positive intersection), making the impossibility statement correct.

Step-by-Step Solution

1

Evaluate the union probability for mutually exclusive events.

P(E or F) = P(E) + P(F) = 0.40 + 0.50 = 0.90.

By definition of mutually exclusive events, P(E and F) = 0.

2

Evaluate the joint probability for independent events.

P(E and F) = P(E) * P(F) = 0.40 * 0.50 = 0.20.

The multiplication rule applies directly to independent events.

3

Evaluate whether mutually exclusive events with non-zero probabilities can be independent.

They cannot be both mutually exclusive and independent.

Mutual exclusivity requires P(E and F) = 0, whereas independence requires P(E and F) = 0.20, which are contradictory.

4

Evaluate the union probability for independent events.

P(E or F) = 0.40 + 0.50 - 0.20 = 0.70.

The general addition rule P(E or F) = P(E) + P(F) - P(E and F) must account for the overlap.

Key Concept

Distinction between independent events and mutually exclusive events in probability

Question 20Question →

At an automated agricultural sorting facility, harvested apples are inspected for two independent quality conditions: surface bruising and internal core rot. The probability that a randomly chosen apple has surface bruising is $\frac{1}{6}$ , and the probability that it has internal core rot is $\frac{1}{5}$ . What is the probability that a randomly chosen apple has at least one of these two quality conditions?

\frac{1}{30}

\frac{2}{11}

\frac{1}{3}

\frac{11}{30}

\frac{2}{5}

Show answer & explanation

Answer: $\frac{1}{3}$

Answer

\frac{1}{3}

The probability of at least one of two independent events occurring is given by

P(B \cup R) = P(B) + P(R) - P(B \cap R)

. Since the events are independent,

P(B \cap R) = P(B) \times P(R) = \frac{1}{6} \times \frac{1}{5} = \frac{1}{30}

. Substituting these values gives

\frac{1}{6} + \frac{1}{5} - \frac{1}{30} = \frac{5}{30} + \frac{6}{30} - \frac{1}{30} = \frac{10}{30} = \frac{1}{3}

. Alternatively, using the complement rule,

P(\text{at least one}) = 1 - P(\text{neither}) = 1 - \left(1 - \frac{1}{6}\right)\left(1 - \frac{1}{5}\right) = 1 - \left(\frac{5}{6} \times \frac{4}{5}\right) = 1 - \frac{4}{6} = \frac{1}{3}

.

Step-by-Step Solution

1

Identify the given probabilities and event relationship.

Let

B

be the event that an apple has surface bruising, so

P(B) = \frac{1}{6}

. Let

R

be the event that an apple has internal core rot, so

P(R) = \frac{1}{5}

. The events are given as independent.

Establishing the mathematical definitions and given conditions is necessary before applying probability rules.

2

Calculate the joint probability of both events occurring,

P(B \cap R)

.

P(B \cap R) = P(B) \times P(R) = \frac{1}{6} \times \frac{1}{5} = \frac{1}{30}

.

For independent events, the probability that both occur simultaneously is the product of their individual probabilities.

3

Apply the addition rule of probability (inclusion-exclusion principle) to find

P(B \cup R)

.

P(B \cup R) = P(B) + P(R) - P(B \cap R) = \frac{1}{6} + \frac{1}{5} - \frac{1}{30} = \frac{5}{30} + \frac{6}{30} - \frac{1}{30} = \frac{10}{30} = \frac{1}{3}

.

The probability of at least one event occurring requires subtracting the intersection so that the overlapping outcome is not counted twice.

Key Concept

Probability of the Union of Independent Events

Estimated Time:1m 30s

Probability of Independent, Dependent, and Mutually Exclusive Events

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept