The next step is to assign a cut-off probability value for decision-making. Based on that, the decision-maker will classify the observation as either class 1 or 0. If Pc is that cut-off value, the following condition occurs.

The following classification table is obtained if the cut-off value is 0.5.

| Launch T (F) | Actual Damage | Prediction (Pc = 0.5) |
| 66 | 0 | 0 |
| 70 | 1 | 0 |
| 69 | 0 | 0 |
| 80 | 0 | 0 |
| 68 | 0 | 0 |
| 67 | 0 | 0 |
| 72 | 0 | 0 |
| 73 | 0 | 0 |
| 70 | 0 | 0 |
| 57 | 1 | 1 |
| 63 | 1 | 1 |
| 70 | 1 | 0 |
| 78 | 0 | 0 |
| 67 | 0 | 0 |
| 53 | 1 | 1 |
| 67 | 0 | 0 |
| 75 | 0 | 0 |
| 70 | 0 | 0 |
| 81 | 0 | 0 |
| 76 | 0 | 0 |
| 79 | 0 | 0 |
| 75 | 1 | 0 |
| 76 | 0 | 0 |
| 58 | 1 | 1 |

The accuracy of the prediction may be obtained by counting the actual and predicted values or by using the function, ‘confusionMatrix’ of the ‘caret’ library.
True Positive (TP) = Damage = 1 & Prediction = 1
False Negative (FN) = Damage = 1 & Prediction = 0
False Positive (FP) = Damage = 0 & Prediction = 1
True Negative (TN) = Damage = 0 & Prediction = 0
Sensitivity = TP / (TP + FN) = 0.57 (57%)
Specificity = TN / (TN + FP) = 1 (100%)
overall accuracy = (TP + TN) / (TP + TN + FN + FP) = 0.875

Although the overall accuracy is at an impressive 87.5%, the true positive rate or the failure estimation rate is pretty average (57%). Considering the significance of the decision, one way to deal with it is to increase the sensitivity by reducing the cut-off probability to 0.2. That leads to the following.
Sensitivity = TP / (TP + FN) = 0.857 (85.7%)
Specificity = TN / (TN + FP) = 0.53 (53%)
overall accuracy = (TP + TN) / (TP + TN + FN + FP) = 0.625

Accuracy Paradox
As you can see, the sensitivity of the second case, the lower cut-off value, is higher, but the overall accuracy of the prediction is poorer. And this is a key step of decision making – choosing higher accuracy of predicting failures (positive values) over the overall classification accuracy.

