Probability Questions and Answers

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

Q: The probability that a prime number selected at random from the numbers (1,2,3, ..........35) is :

The primes up to 35 are\[ 2,\ 3,\ 5,\ 7,\ 11,\ 13,\ 17,\ 19,\ 23,\ 29,\ 31 \]which is 11 numbers out of 35.\[ P = \frac{11}{35} \]1 is not prime - counting it would give 12 and the first distractor.

Q: The following probabilities are given; choose the correct answer for that which is not possible.

Every probability lies between 0 and 1 inclusive.\[ 0 \leq P(E) \leq 1 \]\[ 0.15 \ \checkmark \qquad \frac{2}{7} \approx 0.286 \ \checkmark \qquad \frac{7}{5} = 1.4 \ \times \]\( \tfrac{7}{5} \) exceeds 1, so it cannot be a probability - nothing can be more certain than certain.

Q: If three coins are tossed simultaneously, than the probability of getting at least two heads, is

Three coins give \( 2^{3} = 8 \) equally likely outcomes.Those with at least two heads are\[ \text{HHT},\ \text{HTH},\ \text{THH},\ \text{HHH} \]\[ P = \frac{4}{8} = \frac{1}{2} \]"At least two" includes all three, so the HHH case must not be left out; counting only exactly two would give \( \tfrac{3}{8} \).

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51.

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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52.

A bag contains 3 red and 2 blue marbles. A marble is drawn at random. The probability of drawing a black ball is :

A.3/5
B.2/5
C.0/5
D.1/5

Answer: C

The bag holds only red and blue marbles, so a black one cannot be drawn at all.

\[ P(\text{black}) = \frac{\text{number of black marbles}}{\text{total marbles}} = \frac{0}{5} = 0 \]

An event that cannot happen is an impossible event, and its probability is exactly zero.

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53.

The probability that it will rain tomorrow is 0.85. What is the probability that it will not rain tomorrow

A.0.25
B.0.145
C.3/20
D.none of these

Answer: C

An event and its complement together account for every possibility, so their probabilities add to 1.

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

\[ 0.15 = \frac{15}{100} = \frac{3}{20} \]

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54.

What is the probability that a number selected from the numbers (1, 2, 3,..........,15) is a multiple of 4?

A.1/5
B.4/5
C.2/15
D.1/3

Answer: A

The multiples of 4 between 1 and 15 are 4, 8 and 12 - three of them.

\[ P = \frac{3}{15} = \frac{1}{5} \]

16 is the next multiple but lies outside the range, which is what keeps the count at three.

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55.

What are the total outcomes when we throw three coins?

Answer: C

Each coin lands in one of two ways, and the three coins are independent, so the outcomes multiply.

\[ 2 \times 2 \times 2 = 2^{3} = 8 \]

Listing them: HHH, HHT, HTH, HTT, THH, THT, TTH, TTT. In general \( n \) coins give \( 2^{n} \) outcomes.

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56.

The probability that a prime number selected at random from the numbers (1,2,3, ..........35) is :

A.12/35
B.11/35
C.13/35
D.none of these

Answer: B

The primes up to 35 are

\[ 2,\ 3,\ 5,\ 7,\ 11,\ 13,\ 17,\ 19,\ 23,\ 29,\ 31 \]

which is 11 numbers out of 35.

\[ P = \frac{11}{35} \]

1 is not prime - counting it would give 12 and the first distractor.

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57.

The sum of the probability of an event and non event is :

A.2
B.1
C.none of these.

Answer: B

An event and its complement are exhaustive: between them they cover every possible outcome, and they cannot both happen.

\[ P(E) + P(\bar{E}) = 1 \]

This is why \( P(\bar{E}) = 1 - P(E) \), the form used whenever the complement is easier to count than the event itself.

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58.

The following probabilities are given; choose the correct answer for that which is not possible.

A.0.15
B.2/7
C.7/5
D.none of these.

Answer: C

Every probability lies between 0 and 1 inclusive.

\[ 0 \leq P(E) \leq 1 \]

\[ 0.15 \ \checkmark \qquad \frac{2}{7} \approx 0.286 \ \checkmark \qquad \frac{7}{5} = 1.4 \ \times \]

\( \tfrac{7}{5} \) exceeds 1, so it cannot be a probability - nothing can be more certain than certain.

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59.

If three coins are tossed simultaneously, than the probability of getting at least two heads, is

A.1/4
B.3/8
C.1/2
D.1/8

Answer: C

Three coins give \( 2^{3} = 8 \) equally likely outcomes.

Those with at least two heads are

\[ \text{HHT},\ \text{HTH},\ \text{THH},\ \text{HHH} \]

\[ P = \frac{4}{8} = \frac{1}{2} \]

"At least two" includes all three, so the HHH case must not be left out; counting only exactly two would give \( \tfrac{3}{8} \).

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60.

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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