Probability and Statistics

59 questions

Question 21Question →

A box contains a mixture of red, blue, and green tokens. There are exactly 1616 red tokens in the box. The probability of randomly selecting a red token is 25\frac{2}{5}. If there are 1414 blue tokens in the box, what is the probability of randomly selecting a green token on a single draw?

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Answer: 14\frac{1}{4}

Answer

The probability of randomly selecting a green token is 14\frac{1}{4}.
To find the probability of drawing a green token, you must first know the total number of tokens. Because the 1616 red tokens make up 25\frac{2}{5} of the total, the total number of tokens is 16÷25=4016 \div \frac{2}{5} = 40. The box contains 1616 red and 1414 blue tokens, accounting for 3030 tokens. This leaves 1010 green tokens (40−3040 - 30). The probability of selecting a green token is the number of green tokens divided by the total number of tokens: 1040\frac{10}{40}, which reduces to 14\frac{1}{4}.

Step-by-Step Solution

1
Determine the total number of tokens in the box.
Total tokens = 4040
Since 1616 red tokens represent a probability of 25\frac{2}{5}, we can find the total by dividing the part by its fraction: 16÷25=16×52=4016 \div \frac{2}{5} = 16 \times \frac{5}{2} = 40.
2
Calculate the number of green tokens.
Green tokens = 1010
Subtract the known red and blue tokens from the total: 40−16 (red)−14 (blue)=10 green tokens40 - 16 \text{ (red)} - 14 \text{ (blue)} = 10 \text{ green tokens}.
3
Calculate the probability of drawing a green token.
Probability = 14\frac{1}{4}
Probability is the number of favorable outcomes divided by total outcomes: 1040\frac{10}{40}, which simplifies to 14\frac{1}{4}.

Key Concept

Single Event Probability and Total Outcomes
Question 22Question →

A logistics officer is analyzing data from various base operations. Match each statistical description on the left with its correctly calculated value on the right.

Click a left item, then click its matching right item

Items

The mean weight, in pounds, of five crates weighing 4545, 6060, 5555, 4040, and 5050.
The median of six daily supply requests recorded as 1212, 1818, 1414, 2222, 1515, and 2020.
The target value needed as a fifth shipment to achieve a mean of 2525, given prior shipments of 2020, 2424, 2828, and 2222.
The mode of vehicle repair times, in hours, logged as 44, 33, 66, 44, 55, 88, and 44.

Matches

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Answer

The mean crate weight matches with 50, the median supply request matches with 16.5, the missing shipment for a target mean matches with 31, and the mode of repair times matches with 4.
Each pairing correctly applies the fundamental rules of mean, median, mode, and algebraic representation of statistical targets. The calculations strictly follow the rules for measures of central tendency.

Step-by-Step Solution

1
Calculate the mean for the first description.
(45+60+55+40+50)/5=50(45 + 60 + 55 + 40 + 50) / 5 = 50
The mean is the arithmetic average of a data set.
2
Calculate the median for the second description.
Sorted list: 12,14,15,18,20,2212, 14, 15, 18, 20, 22. Average of middle two: (15+18)/2=16.5(15 + 18) / 2 = 16.5
The median is the middle value of an ordered set, or the average of the two middle values for an even number of items.
3
Calculate the missing value for the third description.
125−(20+24+28+22)=31125 - (20 + 24 + 28 + 22) = 31
The sum of all values must equal the target mean multiplied by the total number of values.
4
Identify the mode for the fourth description.
The number 44 appears three times.
The mode is the most frequent value in a data set.

Key Concept

Measures of Central Tendency
Question 23Question →

A maintenance kit contains 1212 metric wrenches, 1515 standard wrenches, 99 screwdrivers, and an unknown number of pliers. If the probability of randomly selecting a screwdriver from the entire kit is 16\frac{1}{6}, what is the probability of randomly selecting a standard wrench? Express your answer as a fraction in simplest form.

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Answer: 5/18;518;5/185/18; \frac{5}{18}; 5 / 18

Answer

5/18
By using the given probability of selecting a screwdriver (1/6) and the known count of screwdrivers (9), we determine that the total number of items in the kit is 54 (because 9 is 1/6 of 54). The probability of randomly selecting a standard wrench is then the number of standard wrenches (15) divided by the total (54). Simplifying the fraction 15/54 by dividing both numbers by 3 yields 5/18.

Step-by-Step Solution

1
Set up an equation to find the total number of items in the kit.
9Total=16\frac{9}{\text{Total}} = \frac{1}{6}
The probability of an event is the number of favorable outcomes divided by the total number of possible outcomes. Here, we know the probability and the exact count of screwdrivers.
2
Solve for the total number of items.
\text{Total} = 9 \times 6 = 54
Cross-multiplying isolates the total variable, revealing there are 54 items in the entire kit.
3
Calculate the probability of selecting a standard wrench.
Probability=1554\text{Probability} = \frac{15}{54}
There are 15 standard wrenches out of the 54 total items calculated.
4
Simplify the fraction.
1554=518\frac{15}{54} = \frac{5}{18}
Both 15 and 54 are divisible by 3. Dividing the numerator and denominator by 3 gives the simplest form.

Key Concept

Calculating probabilities by determining the total sample size from a known probability.
Question 24Question →

An artillery spotter is recording the impact distances (in meters) from a target center for five test rounds. The distances are 3434, 4141, 2828, 5252, and 4040. A sixth round is fired. If the mean distance for all six rounds is exactly 4141 meters, what was the impact distance of the sixth round?

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Answer: 5151

Answer

The correct impact distance of the sixth round is 5151 meters.
To find a missing value when the mean is known, you must work backward using the total sum. Since the mean of 66 rounds is 4141, their total sum must be 41×6=24641 \times 6 = 246. The sum of the first five rounds is 195195. Subtracting the current sum from the required total sum (246−195246 - 195) reveals that the sixth round must be 5151.

Step-by-Step Solution

1
Calculate the required sum of the distances for all six rounds.
41×6=24641 \times 6 = 246
The mean of a data set is its total sum divided by the number of items. To find the required total sum, multiply the target mean (4141) by the total number of rounds (66).
2
Calculate the sum of the distances for the first five rounds.
34+41+28+52+40=19534 + 41 + 28 + 52 + 40 = 195
This establishes the total distance that has already been accounted for by the initial test rounds.
3
Subtract the sum of the first five rounds from the required total sum to find the sixth round.
246−195=51246 - 195 = 51
The difference between the total needed for all six rounds and the total already recorded gives the exact value required for the final round.

Key Concept

Solving for a missing value given a target mean
Question 25Question →

A military supply depot receives a shipment of 400400 uniform jackets. Of these jackets, 25%25\% are size Small, 38\frac{3}{8} are size Medium, and the remaining jackets are divided equally between size Large and size Extra-Large. If a quartermaster randomly selects one jacket from the entire shipment, what is the probability that it is a size Extra-Large jacket?

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Answer: 316\frac{3}{16}

Answer

316\frac{3}{16}
The probability of a single event is the number of favorable outcomes divided by the total number of possible outcomes. First, determine the number of Extra-Large jackets. There are 100100 Small jackets (25%25\% of 400400) and 150150 Medium jackets (38\frac{3}{8} of 400400). This leaves 150150 jackets (400−100−150=150400 - 100 - 150 = 150). Since these remaining jackets are divided equally between Large and Extra-Large, there are exactly 7575 Extra-Large jackets. The probability of randomly selecting an Extra-Large jacket is the number of Extra-Large jackets (7575) over the total number of jackets (400400). Simplifying the fraction 75400\frac{75}{400} by dividing the numerator and denominator by 2525 yields 316\frac{3}{16}.

Step-by-Step Solution

1
Calculate the number of Small jackets in the shipment.
25%25\% of 400400 is 100100 jackets.
To find the remaining jackets, you first need to determine the exact counts of the known sizes.
2
Calculate the number of Medium jackets in the shipment.
38\frac{3}{8} of 400=3×50=150400 = 3 \times 50 = 150 jackets.
Converting the fraction to an exact count is required to subtract from the total.
3
Determine the number of jackets remaining for the Large and Extra-Large sizes.
400−(100+150)=400−250=150400 - (100 + 150) = 400 - 250 = 150 remaining jackets.
Subtracting the known sizes from the total gives the combined count of the remaining two sizes.
4
Calculate the number of Extra-Large jackets.
150÷2=75150 \div 2 = 75 Extra-Large jackets.
The problem states the remaining jackets are divided equally between the final two sizes.
5
Calculate the probability of selecting an Extra-Large jacket.
Probability =75400= \frac{75}{400}.
Single event probability is the number of favorable outcomes divided by the total number of possible outcomes.
6
Simplify the probability fraction.
75400=316\frac{75}{400} = \frac{3}{16}.
Standardized tests require probability fractions to be in simplest form.

Key Concept

Single Event Probability
Question 26Question →

A structural mechanic is testing the tensile strength of a batch of aluminum rivets. The mean tensile strength of the first 44 rivets tested is 410410 MPa. If the quality standard requires the mean strength of 55 rivets to be exactly 415415 MPa, what tensile strength must the 55 th rivet have?

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Answer: 435

Answer

The 5th rivet must have a tensile strength of 435 MPa.
The correct answer is derived by finding the total required sum for 5 rivets (2075) and subtracting the existing sum of the first 4 rivets (1640).

Step-by-Step Solution

1
Calculate the total tensile strength of the first 4 rivets.
1640 MPa
The sum of a data set is equal to its mean multiplied by the number of values.
2
Calculate the required total tensile strength for all 5 rivets.
2075 MPa
To achieve a target mean of 415 MPa with 5 items, the overall sum must equal the new mean multiplied by the new count.
3
Find the difference between the target total and the current total.
435 MPa
Subtracting the sum of the first 4 items from the required sum of all 5 items isolates the value needed for the 5th item.

Key Concept

Calculating a missing data value given an initial mean and a target mean.

Alternative Method

Use the deviation method: The first 4 rivets have a mean of 410 MPa, which is 5 MPa below the target mean of 415 MPa. This creates a total deficit of 4 × 5 = 20 MPa. The 5th rivet must provide enough strength to cover its own requirement (415 MPa) plus the 20 MPa deficit from the others. Therefore, 415 + 20 = 435 MPa.
Estimated Time:1m 0s
Question 27Question →

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.

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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).
0.80×0.90×(1−0.75)=0.80×0.90×0.25=0.1800.80 \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×(1−0.90)×0.75=0.80×0.10×0.75=0.0600.80 \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)×0.90×0.75=0.20×0.90×0.75=0.135(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=0.3750.180 + 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 28Question →

A quartermaster is recording the number of uniform items issued per day over an 88-day period. The recorded numbers are 2424, 1919, 2828, 1717, 3333, 1616, 2020, and 2727. What is the median number of uniform items issued?

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Answer: 22

Answer

The median number of uniform items issued is 22.
The correct answer accurately sorts the dataset into numerical order (16,17,19,20,24,27,28,3316, 17, 19, 20, 24, 27, 28, 33) and calculates the average of the two middlemost numbers, 2020 and 2424, which results in 2222.

Step-by-Step Solution

1
Sort the recorded numbers in ascending numerical order.
The sorted list is: 16,17,19,20,24,27,28,3316, 17, 19, 20, 24, 27, 28, 33.
The median is the middle value of an ordered dataset, so ordering is always the first step.
2
Identify the middle value(s).
Since there are 88 data points (an even number), there are two middle values: the 4th and 5th numbers, which are 2020 and 2424.
An even number of data points does not have a single exact middle number.
3
Calculate the average (mean) of the two middle values.
20+242=442=22\frac{20 + 24}{2} = \frac{44}{2} = 22
The median for an even number of data points is found by taking the mean of the two middlemost values.

Key Concept

To find the median of a dataset with an even number of values, the data must first be ordered from least to greatest, and then the mean of the two central values must be calculated.
Question 29Question →

A damage control officer is reviewing the completion times (in minutes) for various emergency response drills. Match each statistical description on the left with its correctly calculated value on the right.

Click a left item, then click its matching right item

Items

Mean of the drill times 12,18,15,11,12, 18, 15, 11, and 1414 minutes
Median of the drill times 9,14,8,13,16,9, 14, 8, 13, 16, and 1212 minutes
Mode of the drill times 10,15,12,10,18,10,10, 15, 12, 10, 18, 10, and 1414 minutes
Fourth drill time needed to achieve a mean of 1515 minutes, if the first three times are 16,12,16, 12, and 1414 minutes

Matches

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Answer

Mean pairs with 14; Median pairs with 12.5; Mode pairs with 10; Missing drill time pairs with 18.
Each statistical operation maps strictly to one correct outcome. Calculating the arithmetic mean gives 14, evaluating the sorted middle yields 12.5, counting frequencies identifies 10 as the mode, and algebraic substitution for a target mean resolves to 18.

Step-by-Step Solution

1
Calculate the mean of the first data set.
12+18+15+11+14=7012 + 18 + 15 + 11 + 14 = 70. 70÷5=1470 \div 5 = 14.
The mean requires summing all given values and dividing by the count of the values.
2
Calculate the median of the second data set.
Sorted list: 8,9,12,13,14,168, 9, 12, 13, 14, 16. The median is (12+13)÷2=12.5(12 + 13) \div 2 = 12.5.
The median is the exact middle value. For an even number of data points, it is the average of the two middle numbers in a sorted list.
3
Identify the mode of the third data set.
The number 1010 appears three times, which is more than any other number.
The mode is the value that occurs most frequently in the distribution.
4
Calculate the missing value for the fourth scenario.
Total needed: 15×4=6015 \times 4 = 60. Current sum: 16+12+14=4216 + 12 + 14 = 42. Missing value: 60−42=1860 - 42 = 18.
To achieve a specific mean, the total sum must equal the mean multiplied by the number of events. Subtracting known events yields the missing event.

Key Concept

Measures of Central Tendency

Alternative Method

To find the missing drill time for a target mean of 15, you can use the deviation method: 16 is (+1), 12 is (-3), and 14 is (-1). The sum of these deviations is -3. To balance the average, the missing value must counteract this with an opposite deviation of +3. Therefore, the missing value is 15 + 3 = 18.
Estimated Time:1m 30s
Question 30Question →

A 5-character security passcode must be created using the digits 0 through 9. The passcode must begin with an odd digit, end with an even digit, and no digit may be repeated anywhere in the passcode. How many different passcodes can be created?

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Answer: 8,400

Answer

8,400
To find the total number of possible passcodes, we evaluate the restricted positions first. The first digit must be odd (5 options: 1, 3, 5, 7, 9). The last digit must be even (5 options: 0, 2, 4, 6, 8). Because odd and even sets do not overlap, exactly one odd and one even digit are removed from the total pool of 10 digits. Because repetition is not allowed, 10−2=810 - 2 = 8 digits remain available for the middle three positions. The number of ways to fill the middle three spots is a permutation without repetition: 8×7×6=3368 \times 7 \times 6 = 336. Multiplying the independent choices for all positions gives 5×336×5=8,4005 \times 336 \times 5 = 8,400.

Step-by-Step Solution

1
Determine the number of options for the first character.
5 options
The first digit must be odd, and there are 5 odd digits (1, 3, 5, 7, 9) in the standard set of digits from 0-9.
2
Determine the number of options for the last character.
5 options
The last digit must be even, and there are 5 even digits (0, 2, 4, 6, 8) in the standard set.
3
Determine the number of available digits for the remaining three middle positions.
8 digits
There are 10 digits in total. Since no digit may be repeated and 2 distinct digits have already been used for the first and last positions, exactly 8 digits remain available.
4
Calculate the number of ways to arrange the middle three characters.
336 ways
Using permutations for the 3 middle spots from the 8 remaining digits: 8 × 7 × 6 = 336.
5
Calculate the total number of possible passcodes.
8,400 passcodes
Using the Fundamental Counting Principle, multiply the number of options for each stage: 5 × 336 × 5 = 8,400.

Key Concept

Fundamental Counting Principle with Restrictions
Question 31Question →

A flight engineer is monitoring the fuel consumption, in gallons per hour (GPH), of a cargo aircraft during 44 different flight legs. The consumption for the first 33 legs was 520520, 480480, and 550550 GPH.

If the target mean fuel consumption for all 44 legs is 510510 GPH, what must the fuel consumption be for the 44 th leg?

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Answer: 490

Answer

490
To find the missing value when given the mean, first determine the total sum required for all data points. Since the mean for 44 legs is 510510 GPH, the total fuel consumption must be 4×510=20404 \times 510 = 2040 GPH. The sum of the first three legs is 520+480+550=1550520 + 480 + 550 = 1550 GPH. Subtracting this sum from the required total gives the fuel consumption required for the fourth leg: 2040−1550=4902040 - 1550 = 490 GPH.

Step-by-Step Solution

1
Set up the formula for the mean of 44 flight legs.
520+480+550+x4=510\frac{520 + 480 + 550 + x}{4} = 510
The mean is calculated by dividing the sum of all values by the total number of values.
2
Calculate the total target fuel consumption required for all 44 legs.
510×4=2040510 \times 4 = 2040
Multiplying the target mean by the total number of legs gives the required overall sum.
3
Calculate the sum of the fuel consumption for the first 33 legs.
520+480+550=1550520 + 480 + 550 = 1550
This represents the total fuel consumed so far.
4
Subtract the sum of the first 33 legs from the total target consumption to find the missing value.
2040−1550=4902040 - 1550 = 490
The difference between the required total and the current total must be the consumption of the 44 th leg.

Key Concept

Calculating a missing value when given the mean
Question 32Question →

A scientist records the temperature changes, in degrees Celsius, of a chemical solution during five different phases of an experiment. The recorded changes are: −2.5-2.5, 3.03.0, −5.5-5.5, 4.54.5, and −1.0-1.0. What is the range of these temperature changes?

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Answer: 10.010.0

Answer

10.010.0
The correct answer is found by taking the highest value in the dataset (4.54.5) and subtracting the lowest value (−5.5-5.5). Using the rules for subtracting negative numbers, 4.5−(−5.5)4.5 - (-5.5) becomes 4.5+5.54.5 + 5.5, which equals 10.010.0.

Step-by-Step Solution

1
Order the dataset from least to greatest.
−5.5,−2.5,−1.0,3.0,4.5-5.5, -2.5, -1.0, 3.0, 4.5
Sorting the data is necessary to correctly identify the minimum and maximum values in the set.
2
Identify the maximum and minimum values.
Maximum = 4.54.5, Minimum = −5.5-5.5
The range formula requires the absolute highest and lowest numbers in the dataset.
3
Subtract the minimum value from the maximum value.
4.5−(−5.5)=4.5+5.5=10.04.5 - (-5.5) = 4.5 + 5.5 = 10.0
Range is defined as the difference between the highest and lowest values, measuring the total spread of the data.

Key Concept

Range is calculated by finding the difference between the maximum and minimum values in a dataset.
Question 33Question →

A maintenance technician is tracking the number of work orders completed each day. Over a 55-day work week, the technician wants to achieve a mean of 1414 completed work orders per day. If the technician completed 1212, 1616, 1111, and 1313 work orders on the first four days, how many work orders must be completed on the fifth day to exactly meet the target mean?

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Answer: 18

Answer

The technician must complete 18 work orders on the fifth day.
To find a missing value when the mean is known, first determine the total sum required by multiplying the target mean (14) by the total number of data points (5), which equals 70. Next, add the known values from the first four days (12 + 16 + 11 + 13 = 52). Finally, subtract the sum of the known values from the total required sum (70 - 52 = 18). Therefore, 18 work orders must be completed on the fifth day.

Step-by-Step Solution

1
Determine the total sum required for all 5 days.
14×5=7014 \times 5 = 70
The mean is the total sum divided by the number of values. To find the required total sum, multiply the target mean by the total number of days.
2
Calculate the sum of the work orders completed so far.
12 + 16 + 11 + 13 = 52
This gives the total number of work orders already completed in the first four days.
3
Find the difference between the required total and the completed total.
70 - 52 = 18
Subtracting the sum of the first four days from the required 5-day total yields the exact value needed for the fifth day.

Key Concept

Calculating a missing value from a given mean requires multiplying the mean by the total number of items to find the required sum, then subtracting the known values.
Question 34Question →

Match each data set to its corresponding statistical range.

Click a left item, then click its matching right item

Items

Data Set: {15,8,22,14,9}\{15, 8, 22, 14, 9\}
Data Set: {−3,5,12,−7,4}\{-3, 5, 12, -7, 4\}
Data Set: {4.5,7.2,3.1,8.9,6.0}\{4.5, 7.2, 3.1, 8.9, 6.0\}
Data Set: {100,100,100,100,100}\{100, 100, 100, 100, 100\}

Matches

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Answer

Match the first data set to 1414, the second to 1919, the third to 5.85.8, and the fourth to 00.
Each data set is correctly matched to its range by finding the maximum and minimum values in the set, and then subtracting the minimum from the maximum.

Step-by-Step Solution

1
Calculate the range for the first data set: {15,8,22,14,9}\{15, 8, 22, 14, 9\}.
Maximum = 2222, Minimum = 88. Range = 22−8=1422 - 8 = 14.
The range is the difference between the largest and smallest values in the set.
2
Calculate the range for the second data set: {−3,5,12,−7,4}\{-3, 5, 12, -7, 4\}.
Maximum = 1212, Minimum = −7-7. Range = 12−(−7)=1912 - (-7) = 19.
When subtracting a negative number, it is equivalent to adding its absolute value.
3
Calculate the range for the third data set: {4.5,7.2,3.1,8.9,6.0}\{4.5, 7.2, 3.1, 8.9, 6.0\}.
Maximum = 8.98.9, Minimum = 3.13.1. Range = 8.9−3.1=5.88.9 - 3.1 = 5.8.
Identify the extreme decimal values to find the overall spread of the data.
4
Calculate the range for the fourth data set: {100,100,100,100,100}\{100, 100, 100, 100, 100\}.
Maximum = 100100, Minimum = 100100. Range = 100−100=0100 - 100 = 0.
If all values in a data set are identical, there is no dispersion, resulting in a range of 00.

Key Concept

The range is a measure of dispersion calculated as the difference between the highest (maximum) and lowest (minimum) values in a given data set.
Question 35Question →

An armory stores a mixture of lithium, alkaline, and nickel-metal hydride (NiMH) batteries. The probability of randomly selecting a lithium battery from the storage is 25\frac{2}{5}. If there are exactly 4848 lithium batteries in the armory, what is the total number of batteries stored?

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Answer: 120120

Answer

The total number of batteries stored is 120.
The correct answer is found by setting up the proportion derived from the probability definition: 2/5 = 48/x. Cross-multiplying gives 2x = 240, and dividing by 2 results in 120 total batteries.

Step-by-Step Solution

1
Set up the probability formula using the given information.
Probability=Number of LithiumTotal Batteries\text{Probability} = \frac{\text{Number of Lithium}}{\text{Total Batteries}}, so 25=48x\frac{2}{5} = \frac{48}{x}, where xx is the total number of batteries.
Single event probability is the ratio of favorable outcomes to the total number of possible outcomes.
2
Solve the proportion for xx by cross-multiplying.
2x=5×482x = 5 \times 48
Cross-multiplying allows us to solve an equation with fractions and isolate the variable terms.
3
Simplify the equation and isolate xx.
2x=2402x = 240, so x=2402=120x = \frac{240}{2} = 120.
Dividing both sides by 2 isolates the variable, revealing the total number of batteries.

Key Concept

Calculating the total number of outcomes given the probability of a single event and the number of favorable outcomes.
Question 36Question →

A quartermaster is organizing a drawer of 6060 uniform patches. The drawer contains 1515 rank insignia patches, 2121 unit patches, and the remaining patches are flag patches. If one patch is randomly selected from the drawer, what is the probability of selecting a flag patch? Express your answer as a decimal.

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Answer: 0.4

Answer

0.4
The probability is found by dividing the number of flag patches (2424) by the total number of patches (6060). The fraction 2460\frac{24}{60} simplifies to 25\frac{2}{5}, which is 0.40.4 when expressed as a decimal.

Step-by-Step Solution

1
Calculate the number of flag patches in the drawer.
60−(15+21)=60−36=2460 - (15 + 21) = 60 - 36 = 24 flag patches.
To determine the probability, the exact number of the desired item (flag patches) must first be known.
2
Set up the probability fraction.
2460\frac{24}{60}
The probability of a single event is the number of favorable outcomes divided by the total number of possible outcomes.
3
Convert the fraction to a decimal.
2460=25=0.4\frac{24}{60} = \frac{2}{5} = 0.4
The question explicitly specifies that the final answer must be expressed as a decimal.

Key Concept

Single Event Probability
Question 37Question →

A motor pool inventory shows that 14\frac{1}{4} of the vehicles are heavy trucks, 13\frac{1}{3} are light tactical vehicles, and the remaining vehicles are armored personnel carriers. If a vehicle is randomly selected for an inspection, what is the probability that it is an armored personnel carrier?

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Answer: 512\frac{5}{12}

Answer

512\frac{5}{12}
The total probability of all mutually exclusive events must sum to 1. First, find the combined probability of heavy trucks and light tactical vehicles by using a common denominator of 12: 14\frac{1}{4} becomes 312\frac{3}{12} and 13\frac{1}{3} becomes 412\frac{4}{12}. Their sum is 712\frac{7}{12}. Since the remaining vehicles are armored personnel carriers, subtract this sum from the total probability of 1 (or 1212\frac{12}{12}). The result is 512\frac{5}{12}.

Step-by-Step Solution

1
Find the combined probability of selecting a heavy truck or a light tactical vehicle.
14+13\frac{1}{4} + \frac{1}{3}
To determine the portion of the fleet that is not armored personnel carriers, the given fractions must be added.
2
Add the fractions by finding a common denominator.
312+412=712\frac{3}{12} + \frac{4}{12} = \frac{7}{12}
Fractions must have the same denominator (12) before their numerators can be added.
3
Subtract the combined fraction from 1 (the total probability).
1−712=1212−712=5121 - \frac{7}{12} = \frac{12}{12} - \frac{7}{12} = \frac{5}{12}
The sum of probabilities for all possible events must equal exactly 1. The remainder represents the armored personnel carriers.

Key Concept

Single Event Probability from Complementary Events
Estimated Time:1m 0s
Question 38Question →

An electrician's toolbox contains 88 red wire connectors, 1414 blue wire connectors, and 1010 green wire connectors. If the electrician randomly selects one connector without looking, what is the probability that it is NOT a blue connector?

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Answer: 916\frac{9}{16}

Answer

916\frac{9}{16}
The probability of an event is calculated by dividing the number of favorable outcomes by the total number of possible outcomes. There are 32 total connectors in the toolbox. The number of connectors that are NOT blue is 18 (8 red + 10 green). Therefore, the probability is 18/32, which simplifies to 9/16.

Step-by-Step Solution

1
Calculate the total number of wire connectors in the toolbox.
8+14+10=328 + 14 + 10 = 32 total connectors.
The denominator of a probability fraction represents the total number of possible outcomes.
2
Determine the number of favorable outcomes (connectors that are NOT blue).
88 (red) +10+ 10 (green) =18= 18 connectors that are not blue. Alternatively, 3232 (total) −14- 14 (blue) =18= 18.
The question asks for the probability of selecting a connector that is NOT blue.
3
Set up the probability fraction and simplify.
1832=18÷232÷2=916\frac{18}{32} = \frac{18 \div 2}{32 \div 2} = \frac{9}{16}
Probability is favorable outcomes over total outcomes, and must always be expressed in simplest form.

Key Concept

Single Event Probability
Question 39Question →

A supply sergeant is organizing a crate of 150150 signal flares. The crate contains 5050 red flares, 4040 green flares, 3636 yellow flares, and the remaining flares are white. If a recruit randomly selects one flare from the crate without looking, what is the probability that they select a white flare? Express your answer as a decimal.

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Answer: 0.16

Answer

0.16
The correct probability is found by first determining the number of white flares (150−126=24150 - 126 = 24), and then dividing that number by the total number of flares to find the probability of a single event (24÷150=0.1624 \div 150 = 0.16).

Step-by-Step Solution

1
Calculate the total number of red, green, and yellow flares.
50+40+36=12650 + 40 + 36 = 126
This is required to find out how many flares are in the known color categories.
2
Calculate the number of white flares.
150−126=24150 - 126 = 24
Subtracting the other flares from the total gives the remaining white flares.
3
Calculate the probability of selecting a white flare.
24150\frac{24}{150}
Probability is the number of favorable outcomes (white flares) divided by the total number of possible outcomes (total flares).
4
Convert the fraction to a decimal.
24150=425=0.16\frac{24}{150} = \frac{4}{25} = 0.16
The question specifically asks for the answer to be expressed as a decimal.

Key Concept

Single Event Probability
Estimated Time:1m 0s
Question 40Question →

A dining facility manager is planning a special event and will randomly select one pie from a large frozen delivery. According to the shipping manifest, 13\frac{1}{3} of the pies are apple, 14\frac{1}{4} are cherry, 15\frac{1}{5} are blueberry, and the rest are pecan. What is the probability that the randomly selected pie is a pecan pie?

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Answer: 1360\frac{13}{60}

Answer

The probability of selecting a pecan pie is 1360\frac{13}{60}.
The sum of the probabilities of all possible outcomes in this scenario must equal 1. By finding a common denominator (60), we determine that apple, cherry, and blueberry pies account for 4760\frac{47}{60} of the delivery. The remaining fraction, 1360\frac{13}{60}, represents the probability of selecting a pecan pie.

Step-by-Step Solution

1
Find a common denominator to add the fractions of the known pie types (13\frac{1}{3}, 14\frac{1}{4}, and 15\frac{1}{5}).
The least common multiple of 3, 4, and 5 is 60.
Fractions must have the same denominator before they can be added.
2
Convert each fraction to an equivalent fraction with a denominator of 60.
13=2060\frac{1}{3} = \frac{20}{60}, 14=1560\frac{1}{4} = \frac{15}{60}, and 15=1260\frac{1}{5} = \frac{12}{60}.
This allows for straight addition of the numerators.
3
Add the equivalent fractions to find the total portion of apple, cherry, and blueberry pies.
2060+1560+1260=4760\frac{20}{60} + \frac{15}{60} + \frac{12}{60} = \frac{47}{60}.
This sum represents all the pies in the delivery that are NOT pecan.
4
Subtract this sum from 1 (the whole delivery) to find the probability of selecting a pecan pie.
1−4760=6060−4760=13601 - \frac{47}{60} = \frac{60}{60} - \frac{47}{60} = \frac{13}{60}.
The sum of all possible mutually exclusive outcomes must equal 1.

Key Concept

Single Event Probability and Fraction Operations
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