Probability Questions and Answers

Q: In a Poisson probability distribution

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.

Q: A bucket contains 10 brown balls, 8 green balls, and 12 red balls and you pick one randomly without looking. What is the probability that the ball will be brown?

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.

Q: The joint probability is

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.

Q: Tickets numbered 1 to 20 are mixed up and then a ticket is drawn at random. What is the probability that the ticket drawn has a number which is a multiple of 3 or 5?

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.

Q: The probability of a leap year selected at random contain 53 Sunday is:

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 \( \tfrac{1}{7} \).

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71.

A letter is chosen at random from the letters of the word ASSASSINATION. The probability that the letter chosen is a consonant is:

A.6/13
B.7/13
C.1
D.none of these.

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.

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72.

In a Poisson probability distribution

A.The mean and variance of the distribution are the same (equal)
B.The probability of success is always greater than 5
C.The number of trials is always less than 5
D.It always contains a contingency table

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.

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73.

A bucket contains 10 brown balls, 8 green balls, and 12 red balls and you pick one randomly without looking. What is the probability that the ball will be brown?

A.0.33
B.0.61
C.1/3
D.4/15

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.

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74.

The joint probability is

A.The likelihood of two events happening together
B.The likelihood of an event happening given that another event has already happened
C.Based on two mutually exclusive events
D.Also called Prior probability

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.

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75.

If P (A) and P (not A) are complementary events and P (A) = 0.15, then P (not A) =?

A.0.35
B.Cannot be determined
C.0.85
D.0.3

Answer: C

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

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

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76.

Tickets numbered 1 to 20 are mixed up and then a ticket is drawn at random. What is the probability that the ticket drawn has a number which is a multiple of 3 or 5?

A.1/2
B.2/5
C.8/15
D.9/20

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.

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77.

One card is drawn at random from a pack of 52 cards. What is the probability that the card drawn is a face card (Jack, Queen and King only)?

A.1/13
B.3/13
C.1/4
D.9/52

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 \( \tfrac{16}{52} \).

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78.

The total events to throw three dice simultaneously is

A.6
B.18
C.81
D.216

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.

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79.

The probability of a leap year selected at random contain 53 Sunday is:

A.53/ 366
B.1/7
C.2/7
D.53/365

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 \( \tfrac{1}{7} \).

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80.

A box has 6 black, 4 red, 2 white and 3 blue shirts. Find the probability of drawing 2 black shirts if they are picked randomly?

A.1/8
B.2/15
C.6/15
D.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.

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