There are 2% defective nuts in a large bulk of nuts Write the probability distribution for getting defective nuts in a random sample of 10 nuts
.02 * 10 = .2
Cannot show a probability distribution on these posts.
To calculate the probability distribution for getting defective nuts in a random sample of 10 nuts, we can use the binomial probability distribution formula:
P(x) = C(n, x) * p^x * (1-p)^(n-x)
Where:
P(x) is the probability of getting x defective nuts,
C(n, x) is the number of combinations of n items taken x at a time,
p is the probability of getting a defective nut (2% or 0.02 in this case),
and n is the sample size (10 nuts).
Now let's calculate the probability distribution for x = 0, 1, 2, ..., 10:
For x = 0:
P(0) = C(10, 0) * (0.02)^0 * (1-0.02)^(10-0) = 1 * 1 * 0.98^10 ≈ 0.817
For x = 1:
P(1) = C(10, 1) * (0.02)^1 * (1-0.02)^(10-1) = 10 * 0.02 * 0.98^9 ≈ 0.302
For x = 2:
P(2) = C(10, 2) * (0.02)^2 * (1-0.02)^(10-2) = 45 * 0.02^2 * 0.98^8 ≈ 0.263
For x = 3:
P(3) = C(10, 3) * (0.02)^3 * (1-0.02)^(10-3) = 120 * 0.02^3 * 0.98^7 ≈ 0.153
For x = 4:
P(4) = C(10, 4) * (0.02)^4 * (1-0.02)^(10-4) = 210 * 0.02^4 * 0.98^6 ≈ 0.067
For x = 5:
P(5) = C(10, 5) * (0.02)^5 * (1-0.02)^(10-5) = 252 * 0.02^5 * 0.98^5 ≈ 0.021
For x = 6:
P(6) = C(10, 6) * (0.02)^6 * (1-0.02)^(10-6) = 210 * 0.02^6 * 0.98^4 ≈ 0.005
For x = 7:
P(7) = C(10, 7) * (0.02)^7 * (1-0.02)^(10-7) = 120 * 0.02^7 * 0.98^3 ≈ 0.001
For x = 8:
P(8) = C(10, 8) * (0.02)^8 * (1-0.02)^(10-8) = 45 * 0.02^8 * 0.98^2 ≈ 0.0001
For x = 9:
P(9) = C(10, 9) * (0.02)^9 * (1-0.02)^(10-9) = 10 * 0.02^9 * 0.98^1 ≈ 0.000006
For x = 10:
P(10) = C(10, 10) * (0.02)^10 * (1-0.02)^(10-10) = 1 * 0.02^10 * 0.98^0 ≈ 0.0000001
Hence, the probability distribution for getting defective nuts in a random sample of 10 nuts is approximately as follows:
P(0) ≈ 0.817
P(1) ≈ 0.302
P(2) ≈ 0.263
P(3) ≈ 0.153
P(4) ≈ 0.067
P(5) ≈ 0.021
P(6) ≈ 0.005
P(7) ≈ 0.001
P(8) ≈ 0.0001
P(9) ≈ 0.000006
P(10) ≈ 0.0000001
To determine the probability distribution for getting defective nuts in a random sample, we can use the concept of the binomial distribution.
The binomial distribution describes the probability of getting a certain number of successes (in this case, defective nuts) out of a fixed number of independent Bernoulli trials (in this case, randomly selecting nuts). Each trial has only two possible outcomes: success (defective nut) or failure (non-defective nut).
In this scenario, the probability of getting a defective nut is given as 2%, which can be expressed as 0.02. Let's calculate the probability distribution for different numbers of defective nuts in a random sample of 10 nuts.
The probability of getting exactly k defective nuts in a random sample of 10 nuts can be calculated using the formula for the binomial distribution:
P(X = k) = C(n, k) * p^k * (1-p)^(n-k)
Where:
- P(X = k) is the probability of getting exactly k defective nuts
- C(n, k) is the number of ways to choose k defective nuts among n total nuts, calculated using the combination formula: C(n, k) = n! / (k! * (n-k)!)
- p is the probability of getting a defective nut in a single trial
- n is the number of trials (in this case, the number of nuts in the sample)
Let's calculate the probability distribution for values k = 0 to 10:
P(X = 0) = C(10, 0) * (0.02)^0 * (1-0.02)^(10-0)
P(X = 1) = C(10, 1) * (0.02)^1 * (1-0.02)^(10-1)
P(X = 2) = C(10, 2) * (0.02)^2 * (1-0.02)^(10-2)
...
P(X = 10) = C(10, 10) * (0.02)^10 * (1-0.02)^(10-10)
By calculating these probabilities, we can obtain the probability distribution for getting defective nuts in a random sample of 10 nuts.