LLM Primer

Solutions

MATH-C07-C-001

There are three odd outcomes: 1, 3, and 5.

There are six total outcomes.

P(odd)=3/6=0.5P(\text{odd}) = 3/6 = 0.5

MATH-C07-C-002

P(not A)=1−P(A)=1−0.65=0.35P(\text{not } A) = 1 - P(A) = 1 - 0.65 = 0.35

MATH-C07-C-003

No.

The probabilities sum to:

0.2+0.3+0.6=1.10.2 + 0.3 + 0.6 = 1.1

A valid distribution over exclusive outcomes must sum to 1.

MATH-C07-C-004

E[X]=2(0.25)+6(0.75)=0.5+4.5=5E[X] = 2(0.25) + 6(0.75) = 0.5 + 4.5 = 5

MATH-C07-C-005

Standard deviation is the square root of variance:

16=4\sqrt{16} = 4

MATH-C07-C-006

The flips are independent:

P(HH)=P(H)P(H)=0.5⋅0.5=0.25P(HH) = P(H)P(H) = 0.5 \cdot 0.5 = 0.25

MATH-C07-C-007

P(A∣B)=P(A∩B)P(B)=0.180.6=0.3P(A \mid B) = \frac{P(A \cap B)}{P(B)} = \frac{0.18}{0.6} = 0.3

MATH-C07-C-008

P(A∣B)=P(B∣A)P(A)P(B)P(A \mid B) = \frac{P(B \mid A)P(A)}{P(B)}

So:

P(A∣B)=0.9⋅0.10.3=0.090.3=0.3P(A \mid B) = \frac{0.9 \cdot 0.1}{0.3} = \frac{0.09}{0.3} = 0.3

MATH-C07-C-009

A Bernoulli distribution models one binary yes/no outcome.

MATH-C07-C-010

No.

Positive covariance means two variables tend to move together. It does not prove that one causes the other.

MATH-C07-C-011

The probabilities are not equal, so do not count two labels out of three. Add the assigned probabilities:

0.3+0.1=0.40.3 + 0.1 = 0.4

MATH-C07-C-012

Use the overlap rule:

P(A∪B)=P(A)+P(B)−P(A∩B)P(A \cup B) = P(A) + P(B) - P(A \cap B)

So:

0.5+0.4−0.1=0.80.5 + 0.4 - 0.1 = 0.8

MATH-C07-C-013

A mini-batch average loss is computed from sampled examples.

It is usually an estimate of expected loss, not the exact expected loss over the whole data-generating process.

MATH-C07-C-014

Bayes' rule is:

P(A∣B)=P(B∣A)P(A)P(B)P(A \mid B) = \frac{P(B \mid A)P(A)}{P(B)}

The numerator is P(A and B). Dividing by P(B) restricts attention to cases where the evidence happened, producing P(A | B).