Answer :
To solve this problem, we can use the Hypergeometric distribution. The Hypergeometric distribution is used to calculate probabilities for draws without replacement. Here, we have:
- Total candles (N): 80
- Defective candles (K): 30
- Non-defective candles: 50 (80 - 30)
- Selected candles (n): 10
We'll calculate the probabilities for each part:
### a) Probability that all selected candles will be defective
We need all 10 selected candles to be defective. The probability for this scenario can be derived from the Hypergeometric distribution, which shows the likelihood of having a specific number of successes (in this case, defective candles) in a given number of draws.
The probability [tex]\( P(X = k) \)[/tex] where all [tex]\( k \)[/tex] selected candles are defective (here [tex]\( k = 10 \)[/tex]), is calculated as:
[tex]\[ P(X = 10) = \frac{\binom{K}{k} \times \binom{N-K}{n-k}}{\binom{N}{n}} \][/tex]
Substituting the values:
- [tex]\( K = 30 \)[/tex]
- [tex]\( k = 10 \)[/tex]
- [tex]\( N = 80 \)[/tex]
- [tex]\( n = 10 \)[/tex]
- [tex]\( N - K = 80 - 30 = 50 \)[/tex]
- [tex]\( n - k = 10 - 10 = 0 \)[/tex]
Following standard Hypergeometric computation principles, we get:
[tex]\[ P(X = 10) \approx 0.0000182479 \][/tex]
### b) Probability that 6 will be non-defective
We need 6 out of 10 selected candles to be non-defective, i.e., 4 defective and 6 non-defective.
The probability for this scenario can be derived again using the Hypergeometric distribution with:
[tex]\[ P(Y = m) = \frac{\binom{N-K}{m} \times \binom{K}{n-m}}{\binom{N}{n}} \][/tex]
Substituting the values:
- [tex]\( m = 6 \)[/tex] (non-defective)
- [tex]\( N - K = 50 \)[/tex]
- [tex]\( N = 80 \)[/tex]
- [tex]\( K = 30 \)[/tex]
- [tex]\( n = 10 \)[/tex]
- [tex]\( n - m = 10 - 6 = 4 \)[/tex]
Following standard Hypergeometric computation principles, we get:
[tex]\[ P(Y = 6) \approx 0.2644923901 \][/tex]
### c) Probability that all selected candles will be non-defective
We need all 10 selected candles to be non-defective. The probability for this scenario can also be derived from the Hypergeometric distribution:
[tex]\[ P(Z = q) = \frac{\binom{N-K}{q} \times \binom{K}{n-q}}{\binom{N}{n}} \][/tex]
Substituting the values:
- [tex]\( q = 10 \)[/tex]
- [tex]\( N - K = 50 \)[/tex]
- [tex]\( N = 80 \)[/tex]
- [tex]\( K = 30 \)[/tex]
- [tex]\( n = 10 \)[/tex]
- [tex]\( n - q = 10 - 10 = 0 \)[/tex]
Following standard Hypergeometric computation principles, we get:
[tex]\[ P(Z = 10) \approx 0.006238887 \][/tex]
So, summarizing the probability results:
a) The probability that all selected candles will be defective is approximately 0.0000182479.
b) The probability that 6 of the selected candles will be non-defective is approximately 0.2644923901.
c) The probability that all selected candles will be non-defective is approximately 0.006238887.
- Total candles (N): 80
- Defective candles (K): 30
- Non-defective candles: 50 (80 - 30)
- Selected candles (n): 10
We'll calculate the probabilities for each part:
### a) Probability that all selected candles will be defective
We need all 10 selected candles to be defective. The probability for this scenario can be derived from the Hypergeometric distribution, which shows the likelihood of having a specific number of successes (in this case, defective candles) in a given number of draws.
The probability [tex]\( P(X = k) \)[/tex] where all [tex]\( k \)[/tex] selected candles are defective (here [tex]\( k = 10 \)[/tex]), is calculated as:
[tex]\[ P(X = 10) = \frac{\binom{K}{k} \times \binom{N-K}{n-k}}{\binom{N}{n}} \][/tex]
Substituting the values:
- [tex]\( K = 30 \)[/tex]
- [tex]\( k = 10 \)[/tex]
- [tex]\( N = 80 \)[/tex]
- [tex]\( n = 10 \)[/tex]
- [tex]\( N - K = 80 - 30 = 50 \)[/tex]
- [tex]\( n - k = 10 - 10 = 0 \)[/tex]
Following standard Hypergeometric computation principles, we get:
[tex]\[ P(X = 10) \approx 0.0000182479 \][/tex]
### b) Probability that 6 will be non-defective
We need 6 out of 10 selected candles to be non-defective, i.e., 4 defective and 6 non-defective.
The probability for this scenario can be derived again using the Hypergeometric distribution with:
[tex]\[ P(Y = m) = \frac{\binom{N-K}{m} \times \binom{K}{n-m}}{\binom{N}{n}} \][/tex]
Substituting the values:
- [tex]\( m = 6 \)[/tex] (non-defective)
- [tex]\( N - K = 50 \)[/tex]
- [tex]\( N = 80 \)[/tex]
- [tex]\( K = 30 \)[/tex]
- [tex]\( n = 10 \)[/tex]
- [tex]\( n - m = 10 - 6 = 4 \)[/tex]
Following standard Hypergeometric computation principles, we get:
[tex]\[ P(Y = 6) \approx 0.2644923901 \][/tex]
### c) Probability that all selected candles will be non-defective
We need all 10 selected candles to be non-defective. The probability for this scenario can also be derived from the Hypergeometric distribution:
[tex]\[ P(Z = q) = \frac{\binom{N-K}{q} \times \binom{K}{n-q}}{\binom{N}{n}} \][/tex]
Substituting the values:
- [tex]\( q = 10 \)[/tex]
- [tex]\( N - K = 50 \)[/tex]
- [tex]\( N = 80 \)[/tex]
- [tex]\( K = 30 \)[/tex]
- [tex]\( n = 10 \)[/tex]
- [tex]\( n - q = 10 - 10 = 0 \)[/tex]
Following standard Hypergeometric computation principles, we get:
[tex]\[ P(Z = 10) \approx 0.006238887 \][/tex]
So, summarizing the probability results:
a) The probability that all selected candles will be defective is approximately 0.0000182479.
b) The probability that 6 of the selected candles will be non-defective is approximately 0.2644923901.
c) The probability that all selected candles will be non-defective is approximately 0.006238887.
