A factory produces nails and packs them in boxes of 200. If the probability that a nail is substandard is 0.006. Find the probability that a box selected at random contains at most two nails which are substandard.
p(x=0) = 0.3001
p(x=1)=0.3623
p(x=2)=0.2176
p(x<3)=0.8800
Well, selecting a box with substandard nails is not very appealing, but let's calculate the probability anyway.
To find the probability that a box selected at random contains at most two substandard nails, we need to consider three cases:
Case 1: Zero substandard nails in the box.
Case 2: One substandard nail in the box.
Case 3: Two substandard nails in the box.
Let's start with Case 1. The probability of selecting a box with zero substandard nails is calculated using the binomial distribution formula:
P(X = k) = (n C k) * p^k * (1-p)^(n-k)
where:
- n is the total number of nails in a box (200)
- k is the number of substandard nails in a box (0)
- p is the probability of a nail being substandard (0.006)
- (n C k) is the number of combinations of choosing k items from a total of n items
So, for Case 1:
P(X = 0) = (200 C 0) * (0.006^0) * (1-0.006)^(200-0)
Next, let's move on to Case 2. The probability of selecting a box with one substandard nail is:
P(X = 1) = (200 C 1) * (0.006^1) * (1-0.006)^(200-1)
Lastly, let's consider Case 3. The probability of selecting a box with two substandard nails is:
P(X = 2) = (200 C 2) * (0.006^2) * (1-0.006)^(200-2)
Now, to find the probability that a box selected at random contains at most two substandard nails, we add up the probabilities of these three cases:
P(X ≤ 2) = P(X = 0) + P(X = 1) + P(X = 2)
It's important to note that these calculations assume that each nail's quality is independent of one another and that the probability of a substandard nail remains constant.
Now, I hope this calculation doesn't drive you "nail" biting!
To find the probability that a box selected at random contains at most two nails which are substandard, we can use the concept of the binomial distribution.
In this case, the number of trials is the number of nails in a box (200), and the probability of success is the probability that a nail is substandard (0.006). We want to find the probability of having at most two substandard nails, so we need to sum up the probabilities of having 0, 1, or 2 substandard nails.
Let's break down the calculations step by step:
Step 1: Calculate the probability of having 0 substandard nails:
The probability of having 0 substandard nails is given by the binomial probability formula:
P(X = k) = C(n, k) * p^k * (1-p)^(n-k)
In this case, k = 0, n = 200, and p = 0.006.
P(X = 0) = C(200, 0) * (0.006)^0 * (1-0.006)^(200-0)
P(X = 0) = 1 * 1 * (0.994)^200 ≈ 0.1338
Step 2: Calculate the probability of having 1 substandard nail:
Similarly, we can calculate the probability of having 1 substandard nail using the same formula:
P(X = 1) = C(n, k) * p^k * (1-p)^(n-k)
In this case, k = 1, n = 200, and p = 0.006.
P(X = 1) = C(200, 1) * (0.006)^1 * (1-0.006)^(200-1)
P(X = 1) = 200 * 0.006 * (0.994)^199 ≈ 0.2703
Step 3: Calculate the probability of having 2 substandard nails:
Again, using the binomial probability formula:
P(X = 2) = C(n, k) * p^k * (1-p)^(n-k)
In this case, k = 2, n = 200, and p = 0.006.
P(X = 2) = C(200, 2) * (0.006)^2 * (1-0.006)^(200-2)
P(X = 2) = 19900 * (0.006)^2 * (0.994)^198 ≈ 0.2709
Step 4: Find the sum of these probabilities:
To find the probability of having at most two substandard nails, we sum up the probabilities from steps 1 to 3:
P(at most 2) = P(X = 0) + P(X = 1) + P(X = 2)
P(at most 2) ≈ 0.1338 + 0.2703 + 0.2709 ≈ 0.675
Therefore, the probability that a box selected at random contains at most two nails which are substandard is approximately 0.675 (or 67.5%).
0.6205
could be..
0 substandard ----> C(200,0) (.006)^0 (.994)^200 = ....
1 substandard ----> C(200,1) (.006)(.994)^199 = ....
2 substandard ---> C(200,2) (.006)^2 (.994)^198 = ....
add them up