Study sets
English

Fundamental Information Technology Engineer Examination (FE) | Subject A Applied Fundamentals, Probability, Statistics, and Information Theory Problems 02

1 / 100.0s

Problem 1

Signed integers are represented in eight-bit two's complement. Positive 37 is 00100101.

Which bit string represents decimal -37?

View explanation

Inverting all bits of positive 37, 00100101, gives 11011010. Adding one gives 11011011. The inverted value without the final addition represents -38 in two's complement. Setting only the most significant bit follows a sign-magnitude idea, not two's complement.

Problem 2

A decimal floating-point system retains four significant digits and rounds after each operation. Let a = 1.000 × 10^4 and b = 1.000 × 10^0, and evaluate (a + b) - a from left to right.

Which combination gives the computed result and the main numerical error?

View explanation

Although a + b is 10001 exactly, rounding it to four significant digits gives 1.000 × 10^4, so the contribution of b disappears. Subtracting a then yields zero. Losing the smaller addend when numbers of very different magnitudes are added is loss of significance through absorption. Catastrophic cancellation instead concerns the loss of significant digits when nearly equal numbers are subtracted.

Problem 3

Access is denied when “the source is overseas OR multifactor authentication fails.” Access is allowed only when this denial condition is false.

Which permission condition correctly applies De Morgan's law?

View explanation

If the denial condition is A OR B, permission is NOT(A OR B). By De Morgan's law, this is (NOT A) AND (NOT B): a domestic source and successful multifactor authentication. Keeping OR would allow access even when one of the denial conditions is true.

Problem 4

A factory makes 60% of its products on machine A and 40% on machine B. A has a 1% defect rate, while B has a 3% defect rate. All products come from these two machines. A randomly selected product is found to be defective.

What is the conditional probability that the defective product was made by machine B?

View explanation

The joint probability of an A defect is 0.6 × 0.01 = 0.006, and that of a B defect is 0.4 × 0.03 = 0.012. Total defect probability is 0.018, so the probability of B given a defect is 0.012 / 0.018 = 2/3. Looking only at B's 3% defect rate ignores the production mix.

Problem 5

In a quality test, the null hypothesis H0 is “the lot's defect rate is within the standard.” A lot is classified as abnormal when H0 is rejected. Sample size and all other conditions remain unchanged while only the significance level is reduced.

Which statement correctly identifies the error and the general effect of reducing the significance level?

View explanation

Rejecting H0 when it is true—that is, labeling a conforming lot abnormal—is a Type I error. A smaller significance level makes rejection harder and reduces Type I error. With sample size and other conditions fixed, however, failure to reject a genuinely nonconforming lot, a Type II error, may become more likely.

Problem 6

Monthly data for a city show a strong positive correlation between cold-drink sales and water-accident counts. Both tend to rise in hotter months, but there is no direct evidence that drink sales cause accidents.

Which interpretation is most appropriate?

View explanation

Higher temperatures can increase both demand for cold drinks and participation in water activities, making temperature a plausible common cause. Correlation shows that two variables vary together but does not determine causal direction or rule out confounding. Additional data and a suitable experimental or analytical design are required to support causation.

Problem 7

Consider a stable M/M/1 model with Poisson arrivals, exponentially distributed service time, and one server. The average arrival rate lambda is 4 per hour and service rate mu is 6 per hour. Use utilization rho = lambda / mu and mean time in system W = 1 / (mu - lambda).

Which combination gives utilization and mean time in the system, including waiting and service?

View explanation

Utilization is 4 / 6 = 2/3. Mean time in the system is 1 / (6 - 4) = 0.5 hour, or 30 minutes. The value 1 / mu = 1/6 hour = 10 minutes is mean service time only and excludes waiting. Because lambda is less than mu, the model has a steady state.

Problem 8

Events A and B are independent, with P(A) = 1/4 and P(B) = 1/8. The self-information of event E is I(E) = -log2 P(E) bits.

How much self-information is conveyed by the joint occurrence of A and B?

View explanation

Independence gives a joint probability of 1/4 × 1/8 = 1/32. Therefore, self-information is -log2(1/32) = 5 bits. For independent events, multiplying probabilities means the individual information values, 2 and 3 bits, add.

Problem 9

Four symbols A, B, C, and D have probabilities 1/2, 1/4, 1/8, and 1/8, respectively. Construct a binary Huffman code and calculate average code length as the sum of each symbol's probability multiplied by its code length.

Which combination correctly gives the code lengths and average length?

View explanation

Merge the two least probable symbols C and D to form 1/4, merge that node with B's 1/4 to form 1/2, and finally merge it with A's 1/2. The resulting lengths are 1 for A, 2 for B, and 3 for C and D. The weighted average is 1/2 × 1 + 1/4 × 2 + 1/8 × 3 + 1/8 × 3 = 1.75 bits; it is not an unweighted average.

Problem 10

A finite automaton has initial state S0 and accepting state S2. It reads input from left to right, one symbol at a time, using this transition table: Current state | Input 0 | Input 1 S0 | S1 | S0 S1 | S1 | S2 S2 | S1 | S0 A string is accepted only if the state after all input is read is S2.

Which combination gives the final states for input strings 1001 and 01010?

View explanation

For 1001, the transitions are S0 → S0 → S1 → S1 → S2, so it is accepted. For 01010, they are S0 → S1 → S2 → S1 → S2 → S1, so the final state is S1 and the string is rejected. This automaton checks whether the input ends in 01 after all symbols are read, not whether it passed through S2 at any earlier point.