Find the probability that atmost 5 defective fuses will be found in a box of 200 fuses if experience shows that 2 percent of such fuses are defective (e^-4= 0.0183)
The Correct Answer and Explanation is:
We are given:
- A box contains 200 fuses.
- The probability that any individual fuse is defective is 2% or 0.02.
- We are to find the probability that at most 5 defective fuses are found in the box.
Step 1: Define the distribution
Let XX be the number of defective fuses in the box of 200.
Since we have a fixed number of trials (200), a constant probability of success (0.02), and independent outcomes, X∼Binomial(n=200,p=0.02)X \sim \text{Binomial}(n = 200, p = 0.02).
However, calculating binomial probabilities for large nn can be complex. Since nn is large and pp is small, we use the Poisson approximation to the binomial distribution.
Step 2: Use Poisson approximation
If X∼Binomial(n,p)X \sim \text{Binomial}(n, p), then for large nn, small pp, and λ=np\lambda = np, we can approximate: X∼Poisson(λ)X \sim \text{Poisson}(\lambda)
Here, λ=np=200×0.02=4\lambda = np = 200 \times 0.02 = 4
So X∼Poisson(4)X \sim \text{Poisson}(4)
We want P(X≤5)P(X \leq 5), the probability of at most 5 defective fuses.
Step 3: Calculate P(X≤5)P(X \leq 5) using Poisson probabilities
Poisson probability mass function: P(X=k)=e−λλkk!P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!}
Given e−4=0.0183e^{-4} = 0.0183, compute: P(X≤5)=∑k=054ke−4k!P(X \leq 5) = \sum_{k=0}^{5} \frac{4^k e^{-4}}{k!}
Now compute each term:
- P(0)=40⋅0.01830!=1⋅0.01831=0.0183P(0) = \frac{4^0 \cdot 0.0183}{0!} = \frac{1 \cdot 0.0183}{1} = 0.0183
- P(1)=41⋅0.01831!=4⋅0.01831=0.0732P(1) = \frac{4^1 \cdot 0.0183}{1!} = \frac{4 \cdot 0.0183}{1} = 0.0732
- P(2)=16⋅0.01832=0.1464P(2) = \frac{16 \cdot 0.0183}{2} = 0.1464
- P(3)=64⋅0.01836=0.1952P(3) = \frac{64 \cdot 0.0183}{6} = 0.1952
- P(4)=256⋅0.018324=0.1955P(4) = \frac{256 \cdot 0.0183}{24} = 0.1955
- P(5)=1024⋅0.0183120=0.1569P(5) = \frac{1024 \cdot 0.0183}{120} = 0.1569
Add them: P(X≤5)≈0.0183+0.0732+0.1464+0.1952+0.1955+0.1569=0.7855P(X \leq 5) \approx 0.0183 + 0.0732 + 0.1464 + 0.1952 + 0.1955 + 0.1569 = \boxed{0.7855}
Explanation
This problem requires calculating the probability that at most 5 fuses are defective in a batch of 200, given that 2% are usually defective. The number of defective fuses in the box can be modeled by a binomial distribution: each fuse has a 2% chance of being defective, and there are 200 independent fuses. This results in a binomial distribution with parameters n=200n = 200 and p=0.02p = 0.02.
However, since nn is large and pp is small, the Poisson distribution is a good approximation. The Poisson approximation to the binomial distribution is suitable when npnp (the expected number of defects) is moderate. Here, np=200×0.02=4np = 200 \times 0.02 = 4, which meets the condition.
We approximate the binomial with a Poisson distribution where the mean λ=4\lambda = 4. The Poisson distribution describes the probability of a given number of events happening in a fixed interval and is often used for modeling rare events.
To find the probability of at most 5 defective fuses, we calculate the sum of the Poisson probabilities from 0 to 5. Using the provided value e−4=0.0183e^{-4} = 0.0183, we apply the Poisson probability formula to each k=0k = 0 through 55. After computing each term, we sum them to obtain the cumulative probability.
The final result, P(X≤5)≈0.7855P(X \leq 5) \approx 0.7855, means that there’s approximately a 78.55% chance that 5 or fewer fuses in the box are defective, based on the given defect rate. This helps in quality control decisions and assessing the likelihood of acceptable batches.
