The problem requires finding the maximum number of exceptions without rejecting the VaR model using a binomial distribution. Here, the number of trials $N = 252$, and the probability of exception $p = 1 - 0.995 = 0.005$. The expected number of exceptions $\lambda = N \times p = 252 \times 0.005 = 1.26$. Using the Poisson approximation or exact binomial probabilities, we calculate the cumulative probabilities: P(X = 0) = 0.283 P(X = 1) = 0.358 P(X = 2) = 0.226 P(X = 3) = 0.095 P(X = 4) = 0.030 Summing these up: P(X $\le`$0) = 0.283 P(X $\le$1`) = 0.641 P(X$\le$2) = 0.867 P(X $\le`$3) = 0.962 P(X $\le$4`) = 0.992 To test at a 99$c$where$P(X \le c) \ge 0.99$. Since$P(X \le 4) \approx 0.992$, up to 4 exceptions are acceptable. If 5 or more exceptions occur, the model is rejected. Thus, the maximum acceptable number of losses exceeding the VaR is 4.