Question

In a canning industry, the total process time \( F_0 \) was calculated as 3 min. If each can contains 20 spores with decimal reduction time of 1.6 min, the probability of spoilage would be _________ \(\text{ in 100 cans. (round off to the nearest integer)}\)

Hint:

The probability of spoilage increases with the total process time and the number of spores. Use the formula \(P_{\text{spoilage}} = 1 - e^{-F_0/D}\) to calculate the spoilage probability.

Answer 1

Correct Answer: 25

The formula to calculate the probability of spoilage is based on the total process time and the decimal reduction time: \[ P_{\text{spoilage}} = 1 - e^{-\frac{F_0}{D}} \] Where: - \( F_0 = 3 \) min (total process time) - \( D = 1.6 \) min (decimal reduction time) First, calculate the probability for one spore:
\[ P_{\text{spoilage}} = 1 - e^{-\frac{3}{1.6}} \] \[ P_{\text{spoilage}} = 1 - e^{-1.875} = 1 - 0.153 \] \[ P_{\text{spoilage}} = 0.847 \] Since each can contains 20 spores, the total probability of spoilage in one can is:
\[ P_{\text{total}} = 1 - (1 - P_{\text{spoilage}})^{20} \] \[ P_{\text{total}} = 1 - (0.153)^{20} \] For 100 cans, multiply by 100 to get the spoilage rate:
\[ \text{Probability in 100 cans} = P_{\text{total}} \times 100 \] This gives approximately:
\[ \text{Probability in 100 cans} \approx 27% \] Rounded to the nearest integer:
\[ \boxed{27} \]