Compound Probability Practice Questions — Armed Services Vocational Aptitude Battery (ASVAB)

Question 1Question →

A quality control inspector is reviewing a batch of 12 electronic components, of which 8 are functioning properly and 4 are defective. If the inspector randomly selects 3 components from the batch one at a time without replacement, what is the probability that exactly 2 of the selected components are functioning properly and 1 is defective?

\frac{28}{55}

\frac{28}{165}

\frac{4}{9}

\frac{19}{33}

Show answer & explanation

Answer: $\frac{28}{55}$

Answer

The correct probability is

\frac{28}{55}

.

The correct answer accounts for both the dependent nature of the events (drawing without replacement) and the multiple combinations that satisfy the condition (3 different orders).

Step-by-Step Solution

1

Determine the probabilities of drawing 2 functioning (F) and 1 defective (D) component in one specific order, such as F, then F, then D.

P(F_1) = \frac{8}{12}

,

P(F_2) = \frac{7}{11}

, and

P(D_3) = \frac{4}{10}

.

Because the components are drawn without replacement, the total pool (denominator) and the pool of available specific components (numerator) decrease with each successful draw.

2

Calculate the joint probability of this specific sequence by multiplying the individual probabilities.

\frac{8}{12} \times \frac{7}{11} \times \frac{4}{10} = \frac{2}{3} \times \frac{7}{11} \times \frac{2}{5} = \frac{28}{165}

.

The probability of a sequence of dependent events occurring together is the product of their conditional probabilities.

3

Identify how many different sequences can result in exactly 2 functioning and 1 defective component.

There are 3 possible sequences: F-F-D, F-D-F, and D-F-F.

The single defective component can be the first, second, or third one drawn.

4

Calculate the total probability of achieving this outcome by multiplying the probability of a single sequence by the number of possible sequences.

3 \times \frac{28}{165} = \frac{84}{165}

.

Each of the 3 valid sequences is mutually exclusive and has the exact same probability of occurring, so we can sum their probabilities (or multiply by 3).

5

Simplify the final fraction.

\frac{84}{165}

simplifies to

\frac{28}{55}

by dividing the numerator and denominator by 3.

Fractions should always be presented in their simplest form.

Key Concept

Compound Probability of Dependent Events with Permutations

Question 2Question →

A secure military facility uses three independent detection systems to monitor its perimeter. The probability that the primary radar detects an incoming drone is 0.80. The probability that the infrared system detects it is 0.90. The probability that the acoustic sensors detect it is 0.75. If a drone approaches the perimeter, what is the probability that exactly two of the three systems detect it? Express your answer as a decimal.

Show answer & explanation

Answer: 0.375

Answer

0.375

The correct answer is found by calculating the compound probability for each of the three specific scenarios where exactly two systems succeed and one fails, and then adding those mutually exclusive probabilities together.

Step-by-Step Solution

1

Identify the mutually exclusive scenarios that result in exactly two systems detecting the drone.

Scenario 1: Radar and Infrared detect, Acoustic fails. Scenario 2: Radar and Acoustic detect, Infrared fails. Scenario 3: Infrared and Acoustic detect, Radar fails.

Because these three systems act independently, we must calculate the compound probability of each specific combination of two successes and one failure.

2

Calculate the probability of Scenario 1 (Radar and Infrared detect, Acoustic fails).

0.80 \times 0.90 \times (1 - 0.75) = 0.80 \times 0.90 \times 0.25 = 0.180

Multiply the probabilities of the two specific successful detections by the probability of the specific failure.

3

Calculate the probability of Scenario 2 (Radar and Acoustic detect, Infrared fails).

0.80 \times (1 - 0.90) \times 0.75 = 0.80 \times 0.10 \times 0.75 = 0.060

Multiply the probabilities of the two specific successful detections by the probability of the specific failure.

4

Calculate the probability of Scenario 3 (Infrared and Acoustic detect, Radar fails).

(1 - 0.80) \times 0.90 \times 0.75 = 0.20 \times 0.90 \times 0.75 = 0.135

Multiply the probabilities of the two specific successful detections by the probability of the specific failure.

5

Sum the probabilities of the three mutually exclusive scenarios.

0.180 + 0.060 + 0.135 = 0.375

To find the total probability of any of these mutually exclusive events occurring, add their individual joint probabilities together.

Key Concept

Calculating the probability of an exact number of successes among independent events by identifying all possible combinations of success and failure.

Estimated Time:2m 30s

Question 3Question →

A supply room contains 25 identical-looking computer mice, of which 10 are wireless and 15 are wired. If an IT technician randomly selects two mice from the room one after the other without replacement, what is the probability that both selected mice are wireless? (Express your answer as a decimal.)

Show answer & explanation

Answer: 0.15

Answer

0.15

The correct answer is found by multiplying the probability of the first event by the conditional probability of the second event. First draw:

\frac{10}{25}

. Second draw:

\frac{9}{24}

. Multiplying these gives

\frac{90}{600}

, which simplifies exactly to

0.15

.

Step-by-Step Solution

1

Determine the probability of the first event.

The probability of drawing the first wireless mouse is

\frac{10}{25}

.

There are 10 target items (wireless mice) out of 25 total items.

2

Determine the probability of the second event given the first event occurred.

The probability of drawing a second wireless mouse is

\frac{9}{24}

.

Because the selection is without replacement, one wireless mouse has been removed, leaving 9 wireless mice out of 24 total mice.

3

Calculate the compound probability.

\frac{10}{25} \times \frac{9}{24} = 0.15

For dependent events occurring in sequence, multiply the probability of the first event by the conditional probability of the second event.

Key Concept

Compound Probability (Dependent Events)

Estimated Time:1m 0s

Question 4Question →

A game at a school carnival involves drawing two wooden blocks from a bag without looking. The bag initially contains 9 red blocks, 7 yellow blocks, and 5 blue blocks. If a player draws two blocks one after the other without replacing the first block, what is the probability that the first block drawn is blue and the second block drawn is yellow?

\frac{1}{12}

\frac{5}{63}

\frac{4}{7}

\frac{12}{41}

Show answer & explanation

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

Answer

\frac{1}{12}

The correct answer is found by multiplying the probability of the first event by the probability of the second event, accounting for the reduced total pool. The probability of drawing a blue block first is

\frac{5}{21}

. Because this block is not replaced, only 20 blocks remain in the bag. The probability of then drawing a yellow block is

\frac{7}{20}

. Multiplying these probabilities gives

\frac{5}{21} \times \frac{7}{20} = \frac{35}{420}

, which simplifies to

\frac{1}{12}

.

Step-by-Step Solution

1

Calculate the total number of blocks initially in the bag.

9 \text{ (red)} + 7 \text{ (yellow)} + 5 \text{ (blue)} = 21 \text{ total blocks}

The denominator for the first probability requires the total count of all possible outcomes.

2

Determine the probability of the first event (drawing a blue block).

\frac{5}{21}

There are 5 blue blocks out of the 21 total blocks.

3

Determine the probability of the second event (drawing a yellow block) given that a blue block was already removed.

\frac{7}{20}

Because the first block was not replaced, there are now

21 - 1 = 20

total blocks. There are still 7 yellow blocks in the bag.

4

Multiply the probabilities of the two dependent events and simplify.

\frac{5}{21} \times \frac{7}{20} = \frac{35}{420} = \frac{1}{12}

To find the compound probability of both events occurring in sequence, you must multiply their individual probabilities.

Key Concept

Dependent Compound Probability

Question 5Question →

A navigation briefcase contains $15$ identical-looking USB drives. $9$ of the drives contain map data, and $6$ contain communication protocols. If a pilot randomly selects two drives from the briefcase one after the other without replacing the first, what is the probability that both selected drives contain map data? Express your answer as a simplified fraction.

Show answer & explanation

Answer: 12/35; 12 / 35

Answer

12/35

The correct answer is derived by multiplying the initial probability of drawing a map drive (

\frac{9}{15}

) by the updated probability of drawing a second map drive from the remaining inventory (

\frac{8}{14}

). Simplifying the fractions to

\frac{3}{5}

and

\frac{4}{7}

before multiplying gives the final result of

\frac{12}{35}

.

Step-by-Step Solution

1

Determine the probability of selecting a map data drive on the first draw.

P(First Map) =

9

map drives /

15

total drives =

\frac{9}{15} = \frac{3}{5}

.

The probability of a single event is the number of favorable outcomes divided by the total number of possible outcomes.

2

Determine the probability of selecting a map data drive on the second draw, given that one map drive has already been removed.

P(Second Map) =

8

map drives remaining /

14

total drives remaining =

\frac{8}{14} = \frac{4}{7}

.

Since the first drive is not replaced, both the number of map drives and the total number of drives decrease by one. This is a dependent event.

3

Calculate the compound probability of both dependent events occurring.

\frac{3}{5} \times \frac{4}{7} = \frac{12}{35}

.

To find the probability of two dependent events happening in sequence, you multiply the probability of the first event by the conditional probability of the second event.

Key Concept

Compound probability of dependent events (without replacement)

Question 6Question →

A supply cabinet contains $10$ identical safety helmets, of which $2$ are yellow and $8$ are white. A construction worker randomly selects a helmet, records its color, returns it to the cabinet, and then randomly selects a second helmet. What is the probability of selecting a yellow helmet both times? Express your answer as a decimal.

Show answer & explanation

Answer: 0.04

Answer

0.04

Since the helmet is replaced after the first draw, the two events are independent. The probability of drawing a yellow helmet is

\frac{2}{10}

(or

0.2

). To find the probability of drawing a yellow helmet twice in a row, multiply the probabilities of the two independent events:

0.2 \times 0.2 = 0.04

.

Step-by-Step Solution

1

Find the probability of the first event.

\frac{2}{10}

or

0.2

There are

2

yellow helmets out of

10

total.

2

Find the probability of the second event.

\frac{2}{10}

or

0.2

Because the helmet was replaced, the total number and the number of yellow helmets remain the same.

3

Calculate the compound probability.

0.04

For independent events, multiply the individual probabilities together:

0.2 \times 0.2 = 0.04

.

Key Concept

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

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

.

Question 7Question →

A motor pool dispatcher has a set of $6$ vehicle keys on a desk. Exactly $2$ of the keys belong to trucks, and $4$ belong to sedans. If the dispatcher blindly grabs two keys, one after the other without putting the first back, what is the probability that both keys belong to trucks?

\frac{1}{9}

\frac{2}{15}

\frac{1}{15}

\frac{3}{11}

Show answer & explanation

Answer: $\frac{1}{15}$

Answer

\frac{1}{15}

The correct answer accurately calculates the compound probability of dependent events. The chance of grabbing a truck key first is 2/6. Because that key is held and not replaced, there is only 1 truck key left among the 5 remaining keys, making the chance of the second draw 1/5. Multiplying these fractions (2/6 × 1/5) results in 2/30, which simplifies to 1/15.

Step-by-Step Solution

1

Determine the probability of selecting a truck key on the first draw.

\frac{2}{6}

There are

2

truck keys out of a total of

6

keys.

2

Determine the probability of selecting a truck key on the second draw, assuming one truck key has already been removed.

\frac{1}{5}

Since one truck key was taken and not replaced, there is now

1

truck key remaining out of a total of

5

keys.

3

Multiply the probabilities of the two dependent events to find the compound probability.

\frac{2}{6} \times \frac{1}{5} = \frac{2}{30} = \frac{1}{15}

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

Key Concept

Compound Probability (Dependent Events)

Compound Probability

Answer

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Common Mistakes

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Common Mistakes

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Common Mistakes

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Common Mistakes

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Common Mistakes

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Common Mistakes

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Common Mistakes