Probability Questions and Answers

Answer: B

ASSASSINATION has 13 letters: A, S, S, A, S, S, I, N, A, T, I, O, N.

\[ \text{vowels: A, A, A, I, I, O} \rightarrow 6 \]

\[ \text{consonants: S, S, S, S, N, N, T} \rightarrow 7 \]

\[ P(\text{consonant}) = \frac{7}{13} \]

The two counts must add to 13, which is the check that neither has been miscounted.

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: C

The bucket holds \( 10 + 8 + 12 = 30 \) balls.

\[ P = \frac{10}{30} = \frac{1}{3} \]

The decimal 0.33 is only an approximation of \( \tfrac{1}{3} \), which is the exact value.

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: C

Complementary events cover every possibility between them, so their probabilities add to 1.

\[ P(\text{not } A) = 1 - P(A) = 1 - 0.15 = 0.85 \]

Answer: D

Count each set, then remove the overlap.

\[ \text{multiples of 3} = 6, \qquad \text{multiples of 5} = 4, \qquad \text{multiples of 15} = 1 \]

\[ 6 + 4 - 1 = 9 \quad\Longrightarrow\quad P = \frac{9}{20} \]

15 is a multiple of both and would otherwise be counted twice.

Answer: B

Each suit has three face cards - jack, queen and king - so there are \( 3 \times 4 = 12 \) in the pack.

\[ P = \frac{12}{52} = \frac{3}{13} \]

The ace is not a face card here, as the question makes explicit; counting it would give \( \tfrac{16}{52} \).

Answer: D

Each die shows one of 6 faces, and the three dice are independent.

\[ 6^{3} = 216 \]

The rule is \( 6^{n} \) for \( n \) dice; \( 3^{4} = 81 \) and \( 3 \times 6 = 18 \) are the shapes of the two nearest distractors.

Answer: C

A leap year has 366 days, which is

\[ 366 = 52 \times 7 + 2 \]

So it holds 52 complete weeks - and therefore 52 Sundays for certain - plus two extra days. A 53rd Sunday appears only if one of those two is a Sunday.

The two leftover days are one of seven equally likely consecutive pairs:

\[ \{\text{Sun,Mon}\},\ \{\text{Mon,Tue}\},\ \{\text{Tue,Wed}\},\ \{\text{Wed,Thu}\},\ \{\text{Thu,Fri}\},\ \{\text{Fri,Sat}\},\ \{\text{Sat,Sun}\} \]

Two of the seven contain a Sunday, so

\[ P = \frac{2}{7} \]

In a non-leap year there is only one leftover day and the probability drops to \( \tfrac{1}{7} \).

Answer: D

\[ P = \frac{{}^{6}C_{2}}{{}^{15}C_{2}} = \frac{15}{105} = \frac{1}{7} \]

Equivalently, drawing one after the other: \( \tfrac{6}{15} \times \tfrac{5}{14} = \tfrac{30}{210} = \tfrac{1}{7} \). The \( \tfrac{6}{15} \) option ignores the second draw entirely.