Probability Questions and Answers

Answer: D

A probability distribution pairs each possible outcome of an experiment with its probability - for a single die, the six faces each with \( \tfrac{1}{6} \).

A random variable is the quantity being measured, and a contingency table cross-tabulates two categorical variables; neither is the listing of outcomes with their probabilities.

Answer: A

Joint probability is the probability that two events occur together, written \( P(A \cap B) \).

The second option describes conditional probability, \( P(A \mid B) \); mutual exclusivity would make the joint probability zero; and prior probability is the belief held before new evidence, a term from Bayes' theorem.

Answer: B

The special rule of addition is

\[ P(A \cup B) = P(A) + P(B) \]

and it holds only when the events are mutually exclusive - they cannot both occur, so \( P(A \cap B) = 0 \) and there is no overlap to subtract.

Independence is the condition for the special rule of multiplication. The general addition rule, \( P(A \cup B) = P(A) + P(B) - P(A \cap B) \), needs no condition at all.

Answer: C

A binomial experiment requires exactly three things: each trial has only two outcomes, the probability of success is the same on every trial, and the trials are independent.

The number of trials \( n \) may be any positive integer - even one - so "must have at least 3 trials" is not a condition.

Answer: A

Classical probability assumes every outcome in the sample space is equally likely, giving

\[ P(E) = \frac{\text{favourable outcomes}}{\text{total outcomes}} \]

It applies to fair dice, fair coins and well-shuffled cards. The relative-frequency approach estimates probability from observed data, and the subjective approach from personal judgement; neither assumes equal likelihood.

Answer: A

For a Poisson distribution with parameter \( \lambda \),

\[ \text{mean} = \lambda, \qquad \text{variance} = \lambda \]

so the two are always equal - a property unique to the Poisson among the standard discrete distributions, and the usual quick test of whether count data are Poisson.

Answer: A

The bag holds \( 2 + 3 + 2 = 7 \) balls, of which 5 are not blue.

\[ P = \frac{{}^{5}C_{2}}{{}^{7}C_{2}} = \frac{10}{21} \]

Counting the non-blue balls directly is quicker than working through the red and green cases separately.

Answer: A

"Neither red nor green" leaves only the blue balls.

\[ \text{total} = 8 + 7 + 6 = 21 \]

\[ P = \frac{7}{21} = \frac{1}{3} \]

Recognising the complement as a single colour turns a two-case count into one division.

Answer: C

Two throws give 36 outcomes. Those summing to 9 are

\[ (3,6),\ (4,5),\ (5,4),\ (6,3) \]

\[ P = \frac{4}{36} = \frac{1}{9} \]

Order matters, so \( (3,6) \) and \( (6,3) \) count separately.

Answer: D

"At most two heads" means everything except three heads, so use the complement.

\[ P(\text{3 heads}) = \frac{1}{8} \]

\[ P(\text{at most 2}) = 1 - \frac{1}{8} = \frac{7}{8} \]

Counting zero, one and two heads directly gives \( 1 + 3 + 3 = 7 \) of the 8 outcomes - the same answer for more work.