Question

There are four calculators and it is known that exactly two of them are defective. They are tested one by one, in a random order till both the defective calculators are identified. Then the probability that only two tests are required is: (a) \(\frac{5}{6}\) (b) \(\frac{1}{2}\) (c) \(\frac{1}{6}\) (d) \(\frac{1}{3}\)

Short answer

Answer: The probability is 1/4 or 25%.

Step-by-step solution

Calculate Total Outcomes

To calculate the total outcomes, we find the number of different ways the calculators can be tested. Since there are 4 calculators, each with 2 ways(correct or defective) to be selected, there are a total of \(4!\) (4 factorial) ways to test them, which can be calculated as: \(4! = 4 \times 3 \times 2 \times 1 = 24.\) There are 24 possible ways to test the calculators.

Calculate Favorable Outcomes

For the number of favorable outcomes (i.e., finding both defective calculators in only two tests), we need to calculate the combinations of choosing the defective calculators in two tests. The favorable outcomes are the number of pairs that can be made with 2 defective calculators out of 4 calculators. This can be calculated using the combination formula: C(n, r) = \({n!}/{(n-r)! r!}\), where n = the total number of items and r = the number of items to choose. In this case, n = 4 and r = 2, so we calculate C(4, 2) = \({4!}/{(4-2)! 2!}\) = \({4!}/{2! 2!}\) = \({24}/{(2 \times 1) (2 \times 1)}\) = \({24}/{4}\) = 6. There are 6 favorable outcomes.

Calculate the Probability

Now that we have the total outcomes and favorable outcomes, we can find the probability that only two tests are required to identify both defective calculators using the formula: probability = favorable outcomes / total outcomes. The probability = 6/24 = 1/4, which can be simplified as 0.25 or 25%. Thus, the probability that only two tests are required to identify both defective calculators is 1/4, and the correct answer is (b) \(\frac{1}{2}\).

Key concepts

Combinatorics

Combinatorics is a branch of mathematics that deals with counting, arranging, and grouping objects. It allows us to determine the number of ways something can occur. This is particularly useful in probability when we want to know how many possible outcomes there are for an event. In our exercise, we look at how different calculators can be tested in various sequences. We use combinations and permutations to get the count of both favorable and total outcomes.

Permutations involve arrangements where the order matters. For example, choosing which calculator to test first, second, and so on. Combinations, on the other hand, involve selecting items where the order does not matter, such as selecting defective calculators. In our context, combinatorics helped us find that there are 24 different sequences to test four calculators, calculated using factorials.

Defective Items

A defective item is one that does not function as intended. Identifying defective items in a set, like calculators, depends on how they behave when tested. In our exercise, we know there are two defective calculators out of four. The challenge is to find these two defective ones through testing.

• The problem requires calculating the probability of finding both defective ones in only two tests.
• All calculators are tested in random order.
• Recognizing defective items requires methods like combinations, which involves choosing sets of items where the sequence does not matter.

By understanding the concept of defective items, we explore the scenarios where solutions reflect real-world situations, like product testing and quality control.

Factorial Calculation

Factorial calculation is the process of multiplying a series of descending natural numbers. It's denoted by an exclamation mark, like \(4!\). A factorial gives us the total number of ways to arrange a set number of items.

• Factorials are widely used in permutations to calculate the number of different sequences possible.
• In our exercise, \(4!\) means \(4 \times 3 \times 2 \times 1 = 24\), reflecting the total sequences to test four calculators.

Factorials play a key role in combinatorics, helping to solve complex problems in calculations involving arrangements and selections. It is fundamental in understanding how large sets can be organized in different ways. Factorial calculations simplify our task of determining probabilities by giving us a reliable way to count possibilities.