Light bulbs manufactured at a certain factory have a 3% probability of being defective. What is the probability that 2 out of a shipment of 30 will be defective?
A. 0.25
B. 0.0016
C. 0.0035
D. 0.167
Answer (2)
Use the binomial probability formula: P ( X = k ) = ( k n ) p k ( 1 − p ) ( n − k ) .
Identify the parameters: n = 30 , k = 2 , p = 0.03 .
Calculate the probability: P ( X = 2 ) = ( 2 30 ) ( 0.03 ) 2 ( 0.97 ) 28 ≈ 0.1712 .
The closest answer is 0.167 .
Explanation
Analyze the problem Let's analyze the problem. We are given that light bulbs have a 3% chance of being defective, and we want to find the probability that exactly 2 out of a shipment of 30 are defective. This is a binomial probability problem, where we have a fixed number of trials (30 light bulbs), each trial is independent, there are only two outcomes (defective or not defective), and the probability of success (a light bulb being defective) is constant (3%).
Introduce the binomial probability formula The binomial probability formula is: P ( X = k ) = ( k n ) p k ( 1 − p ) n − k , where:
n is the number of trials (light bulbs)
k is the number of successes (defective light bulbs)
p is the probability of success on a single trial (probability of a light bulb being defective)
( k n ) is the binomial coefficient, which represents the number of ways to choose k successes from n trials.
Identify the parameters In our case, we have:
n = 30 (number of light bulbs)
k = 2 (number of defective light bulbs)
p = 0.03 (probability of a light bulb being defective)
Calculate the binomial coefficient First, let's calculate the binomial coefficient: ( 2 30 ) = 2 ! ( 30 − 2 )! 30 ! = 2 ! 28 ! 30 ! = 2 × 1 30 × 29 = 15 × 29 = 435
Calculate p^k Next, let's calculate p k : p k = ( 0.03 ) 2 = 0.0009
Calculate (1-p)^(n-k) Now, let's calculate ( 1 − p ) n − k : ( 1 − p ) n − k = ( 1 − 0.03 ) 30 − 2 = ( 0.97 ) 28 ≈ 0.4371
Calculate the final probability Finally, let's calculate the probability: P ( X = 2 ) = ( 2 30 ) p k ( 1 − p ) n − k = 435 × 0.0009 × 0.4371 ≈ 0.1712
Compare and choose the closest option Comparing the calculated probability (0.1712) with the given options: a. 0.25 b. 0.0016 c. 0.0035 d. 0.167
The closest option is d. 0.167.
State the final answer Therefore, the probability that exactly 2 out of 30 light bulbs are defective is approximately 0.167.
Examples
Binomial probability is useful in many real-world scenarios. For example, it can be used to determine the probability of a certain number of products being defective in a manufacturing process, the probability of a certain number of customers making a purchase after receiving an advertisement, or the probability of a certain number of voters supporting a particular candidate in an election. Understanding binomial probability helps in making informed decisions in various fields such as quality control, marketing, and political science. Let's say a marketing team sends out 1000 emails, and they know that on average, 5% of recipients will click on the link. Using binomial probability, they can calculate the likelihood of exactly 50 people clicking the link, or the probability of at least 30 people clicking the link, helping them assess the effectiveness of their campaign.
The probability that exactly 2 out of 30 light bulbs are defective is calculated using the binomial probability formula, resulting in approximately 0.1712. The closest multiple-choice option is D. 0.167. This shows the practical application of binomial probabilities in real-world scenarios such as quality control.
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