Calculating Probabilities in Defective Bulb Problems

Probability theory plays a crucial role in various real-world scenarios, such as manufacturing and quality control. In this article, we will explore how to calculate the probability of certain events related to defective electric bulbs. Specifically, we will look at a problem involving a box of 24 bulbs, 4 of which are defective, and another problem involving defective bulbs from two different factories. We'll use combinatorial methods and the complementary probability approach to solve these problems.

Problem 1: Probability of At Least One Defective Bulb

A box contains 24 electric bulbs, out of which 4 are defective. We are to find the probability that at least one of the two randomly chosen bulbs is defective.

Step-by-Step Solution

Step 1: Total Number of Bulbs

Total bulbs: 24 Defective bulbs: 4 Good bulbs: 24 - 4 20

Step 2: Calculate the Probability of Choosing 2 Good Bulbs

The total ways to choose 2 bulbs from 24 is given by the combination formula (binom{24}{2}).

[binom{24}{2} frac{24 times 23}{2 times 1} 276]

The ways to choose 2 good bulbs from the 20 good bulbs is (binom{20}{2}).

[binom{20}{2} frac{20 times 19}{2 times 1} 190]

Step 3: Calculate the Probability that Both Bulbs are Good

The probability that both chosen bulbs are good is:

[P(text{both good}) frac{190}{276}]

Step 4: Simplify the Probability

We can simplify (frac{190}{276}):

(frac{190}{276} frac{95}{138}) by dividing numerator and denominator by 2

Step 5: Calculate the Probability of At Least One Defective Bulb

The probability of at least one bulb being defective is the complement of the probability of both bulbs being good:

[P(text{at least one defective}) 1 - P(text{both good}) 1 - frac{95}{138}]

Calculating this gives:

[P(text{at least one defective}) frac{138 - 95}{138} frac{43}{138}]

Final Answer: The probability that at least one of the two chosen bulbs is defective is (frac{43}{138}).

Problem 2: Probability of a Satisfactory Bulb from Different Factories

The second problem involves two factories, A and B, which manufacture electric bulbs. Factory A produces 8 defective bulbs out of a total production, and Factory B produces 10 defective bulbs out of a total production, with Factory B producing three times as many bulbs as Factory A.

Step-by-Step Solution

Step 1: Determine the Production Output of Each Factory

Let the total production of Factory A be (x) bulbs. Then the total production of Factory B is (3x) bulbs. The probability of a defective bulb from Factory A is (frac{8}{x}). The probability of a defective bulb from Factory B is (frac{10}{3x}).

Step 2: Calculate the Probability of a Satisfactory Bulb from Factory A and B

The probability of a satisfactory bulb from Factory A is (1 - frac{8}{x}). The probability of a satisfactory bulb from Factory B is (1 - frac{10}{3x}).

Step 3: Combined Probability of Selecting a Satisfactory Bulb

The combined probability of selecting a satisfactory bulb is the weighted average, considering the relative production output:

[P(text{satisfactory}) frac{x}{x 3x} times left(1 - frac{8}{x}right) frac{3x}{x 3x} times left(1 - frac{10}{3x}right)]

Since (x 3x 4x), we have:

[P(text{satisfactory}) frac{1}{4} left(1 - frac{8}{x}right) frac{3}{4} left(1 - frac{10}{3x}right)]

Let's assume (x 10) for simplicity, then:

[P(text{satisfactory}) frac{1}{4} left(1 - frac{8}{10}right) frac{3}{4} left(1 - frac{10}{30}right)]

[P(text{satisfactory}) frac{1}{4} left(1 - 0.8right) frac{3}{4} left(1 - frac{1}{3}right)]

[P(text{satisfactory}) frac{1}{4} left(0.2right) frac{3}{4} left(frac{2}{3}right)]

[P(text{satisfactory}) 0.05 0.5 0.55]

Final Answer: The probability that a randomly chosen bulb from a week's production is satisfactory is (0.55).