Question

In a lot of 50 light bulbs, there are 2 bad bulbs. An inspector examines five bulbs, which are selected at random and without replacement. (a) Find the probability of at least one defective bulb among the five. (b) How many bulbs should be examined so that the probability of finding at least one bad bulb exceeds \(\frac{1}{2}\) ?

Short answer

(a) The probability of finding at least one defective bulb in a sample of five is 1 minus the probability of picking five good bulbs from the total number of ways of picking five bulbs. (b) The least number of bulbs to be examined from the lot so that the probability of at least one defective bulb exceeds \(\frac{1}{2}\) can be found by setting the probability of finding only good bulbs in an examination of \(n\) bulbs less than \(\frac{1}{2}\) and then iterating \(n\) until the condition is met.

Step-by-step solution

Calculate the Probability for Scenario (a)

To find the probability of getting at least one defective bulb out of 5 bulbs, we can calculate the probability of getting no defective bulb and then subtract it from 1. First, finding the probability of selecting no defective bulb out of 5 is equivalent to picking from the 48 good ones. The total possible combinations of picking 5 bulbs out of 50 is \({{50}\choose{5}}\) and that of selecting 5 good bulbs is \({{48}\choose{5}}\). Hence, the probability of getting no defective bulbs \(P_0\) is given by:\[ P_0 = \frac{{48}\choose{5}}{{50}\choose{5}} \]To get the probability of getting at least one defective bulb, subtract \(P_0\) from 1:\[ P = 1 - P_0 \]

Determine the Sample Size for Scenario (b)

For scenario (b), we need to find the smallest number, say \(n\), of bulbs to be examined such that the probability of finding at least 1 bad bulb exceeds \(\frac{1}{2}\). To find this, we'll calculate the complementary probability for examining \(n\) bulbs without finding a bad one (i.e. finding only good ones) and set it to be less than \(\frac{1}{2}\). Hence, we get the condition:\[ \frac{{48}\choose{n}}{{50}\choose{n}} < \frac{1}{2} \]We increase \(n\) from 1 and check when the above condition first becomes true.

Calculation

In this step, we will carry out the calculation from the previous steps. For part (a) carry out the combination calculations and subtract from 1. For part (b), iterate from \(n=1\) to \(n=50\) and check when the condition from step 2 first becomes true. This will give us the least number of bulbs for which the required condition is met.

Key concepts

Combinatorics

Combinatorics is a branch of mathematics focusing on counting, both as a means and an end in obtaining results. It's crucial in a range of problems including those that involve probability. When you're dealing with scenarios where the order of selection doesn't matter, combinations are used—a key concept in combinatorics.

For example, if we want to select 5 light bulbs from a batch of 50, we use the combination formula, denoted as \( {n \choose k} \), where \( n \) is the total number of items to choose from, and \( k \) is the number of items to be chosen. This formula calculates the number of possible selections of \( k \) items from a set of \( n \) distinguishable items.

Now let's dive into a little formula explanation: \( {n \choose k} = \frac{n!}{k!(n - k)!} \), where '!' denotes factorial, the product of all positive integers up to that number. For instance, \( 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \). So, in our light bulb problem, the total possible combinations of picking 5 bulbs out of 50 is \( {50 \choose 5} \). The logic is applied similarly for picking the 5 good bulbs from the 48 good ones, \( {48 \choose 5} \).

Probability Without Replacement

The concept of 'probability without replacement' comes into play when the outcome of one event affects the outcome of following events, which is common in scenarios where objects are not returned to the pool after selection. This affects the total number of outcomes and thus the probability.

Consider our set of 50 bulbs which contains 2 bad ones. If we select one bulb, and do not replace it, before selecting the next one, we change the total number of bulbs to choose from and potentially the number of bad bulbs remaining. With each draw, the pool of bulbs is reduced by one, altering the probability with each selection.

This is why the probability of drawing a bad bulb changes after each bulb is taken out. In effect, the odds become increasingly in our favor or against us, depending on what we're drawing for. The math of this scenario requires us to adjust our calculations, factoring in the changing size of the bulb pool and the count of good vs. bad bulbs for each draw.

Hypergeometric Distribution

The hypergeometric distribution is a model that describes the probability of a specific number of successes in a series of draws without replacement from a finite population. This is exactly the kind of situation our light bulb problem presents.

This distribution is related to the concepts of combinatorics and probability without replacement. It calculates the probability of getting a certain number of successes (in our case, defective bulbs) in a certain number of draws from a population with a known amount of successes, without returning any draws back to the population.

The formula for the hypergeometric probability is: \[ P(X = k) = \frac{{{K \choose k} {{N-K} \choose {n-k}}}}{{N \choose n}} \] where:

• \( N \) is the population size (50 bulbs),
• \( n \) is the number of draws (5 bulbs),
• \( K \) is the total number of successful states in the population (2 defective bulbs),
• \( k \) is the number of successful states in the sample (the number of defective bulbs we are calculating for, at least one in our problem).

By using this approach, we avoid overestimating the probability by treating each draw as independent when it's not since the bulbs aren't replaced. It gives us a more precise and realistic probability calculation for scenarios akin to our exercise.