If 5 of a Company’s 10 delivery trucks do not meet emission standards and 3 of them are chosen for inspection, then what is the probability that none of the trucks chosen will meet emission standards?
Understanding the Probability Problem
The problem asks for the probability that none of the 3 delivery trucks chosen for inspection will meet emission standards. We are given that a company has 10 delivery trucks in total, and 5 of these 10 trucks do not meet emission standards. This means the remaining 10 - 5 = 5 trucks do meet emission standards.
We are choosing 3 trucks out of the total 10. The selection is done without replacement, and the order in which the trucks are chosen does not matter. This type of problem involves combinations.
To find the probability, we need two values:
- The total number of ways to choose 3 trucks from the 10 available trucks.
- The number of ways to choose 3 trucks such that none of them meet emission standards (meaning all 3 are from the group that does not meet standards).
Calculating Total Ways to Choose Trucks
The total number of ways to choose 3 trucks from 10 is given by the combination formula \(\binom{n}{k} = \frac{n!}{k!(n-k)!}\), where \(n\) is the total number of items, and \(k\) is the number of items to choose.
Here, \(n = 10\) (total trucks) and \(k = 3\) (trucks chosen for inspection).
Total combinations = \(\binom{10}{3} = \frac{10!}{3!(10-3)!} = \frac{10!}{3!7!}\)
Let's expand the factorials:
\(\frac{10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{(3 \times 2 \times 1)(7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1)}\)
Cancel out the 7! terms:
\(\frac{10 \times 9 \times 8}{3 \times 2 \times 1}\)
Calculate the denominator: \(3 \times 2 \times 1 = 6\)
Calculate the numerator: \(10 \times 9 \times 8 = 720\)
Total combinations = \(\frac{720}{6} = 120\)
There are 120 different ways to choose 3 trucks from the 10 trucks.
Calculating Ways to Choose Non-Compliant Trucks
We want the probability that none of the chosen trucks meet emission standards. This means all 3 chosen trucks must come from the group of trucks that do not meet emission standards. There are 5 trucks that do not meet emission standards.
We need to find the number of ways to choose 3 trucks from these 5 non-compliant trucks. Again, we use the combination formula.
Here, \(n = 5\) (non-compliant trucks) and \(k = 3\) (trucks chosen).
Number of ways to choose 3 non-compliant trucks = \(\binom{5}{3} = \frac{5!}{3!(5-3)!} = \frac{5!}{3!2!}\)
Let's expand the factorials:
\(\frac{5 \times 4 \times 3 \times 2 \times 1}{(3 \times 2 \times 1)(2 \times 1)}\)
Cancel out the 3! terms:
\(\frac{5 \times 4}{2 \times 1}\)
Calculate the denominator: \(2 \times 1 = 2\)
Calculate the numerator: \(5 \times 4 = 20\)
Number of ways to choose 3 non-compliant trucks = \(\frac{20}{2} = 10\)
There are 10 different ways to choose 3 trucks such that none of them meet emission standards.
Calculating the Probability
The probability of an event is calculated as:
Probability = (Number of favorable outcomes) / (Total number of possible outcomes)
In this case:
- Favorable outcome: Choosing 3 trucks that do not meet emission standards. The number of ways is 10.
- Total possible outcomes: Choosing any 3 trucks from the 10. The number of ways is 120.
Probability (none meet standards) = \(\frac{\text{Number of ways to choose 3 non-compliant trucks}}{\text{Total number of ways to choose 3 trucks}}\)
Probability = \(\frac{10}{120}\)
Simplify the fraction by dividing both numerator and denominator by their greatest common divisor, which is 10:
\(\frac{10 \div 10}{120 \div 10} = \frac{1}{12}\)
The probability that none of the trucks chosen will meet emission standards is \(\frac{1}{12}\).
Summary of Probability Calculation
We can summarize the calculation using combination notation directly:
Total number of ways to choose 3 trucks from 10: \(\binom{10}{3} = 120\)
Number of ways to choose 3 trucks from the 5 non-compliant trucks: \(\binom{5}{3} = 10\)
Probability (none meet standards) = \(\frac{\binom{5}{3}}{\binom{10}{3}} = \frac{10}{120} = \frac{1}{12}\)
Revision Table: Key Concepts in Probability
| Concept | Description | Formula/Notation |
|---|---|---|
| Probability | Measure of the likelihood of an event occurring. | \(P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total outcomes}}\) |
| Combination | Selecting items from a set where order does not matter. Used when sampling without replacement. | \(\binom{n}{k} = C(n, k) = \frac{n!}{k!(n-k)!}\) |
| Sampling without Replacement | Once an item is selected, it is not returned to the pool for subsequent selections. | Applies to problems like drawing cards or selecting items from a finite group. |
| Event | A specific outcome or set of outcomes of a random process. | E.g., choosing 3 trucks that fail emission standards. |
Additional Information: Understanding Combinations
Combinations are fundamental in probability when dealing with selections from a group. The key difference between combinations and permutations is whether the order of selection matters.
- Combinations: Order does NOT matter. Choosing trucks A, B, C is the same as choosing B, C, A. Used when we are interested in the composition of the chosen group.
- Permutations: Order DOES matter. Arranging items in a sequence. Choosing A then B then C is different from choosing B then A then C. Used when arrangement or order is important.
In this problem, the question is simply about the group of 3 trucks chosen for inspection, not the sequence in which they were picked. Therefore, combinations are the correct tool.
The formula \(\binom{n}{k}\) counts how many unique subsets of size \(k\) can be formed from a set of size \(n\).
In our case, \(n=10\) (total trucks) and \(k=3\) (chosen trucks) for the total possible outcomes. For the favorable outcomes, \(n=5\) (non-compliant trucks) and \(k=3\) (chosen from this group).
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