Probability Questions and Answers

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.

Answer: A

Each coin can land in 2 ways, and the ten coins are independent.

\[ 2^{10} = 1024 \]

The rule is \( 2^{n} \) for \( n \) coins - not \( 2n \), which would give only 20.

Answer: A

Four coins give \( 2^{4} = 16 \) outcomes. Choose which two of the four show heads:

\[ {}^{4}C_{2} = 6 \]

\[ P = \frac{6}{16} = \frac{3}{8} \]

The six arrangements are HHTT, HTHT, HTTH, THHT, THTH and TTHH.

Answer: B

Since there are 4 kings in a deck of 52, the probability of drawing a king is 4/52 = 1/13.

Hence, the probability of not picking a king is 1 - 1/13 = 12/13.

(∵ For an event E, P (E) =1-P (not E)).

Answer: A

Without replacement, the second draw sees one fewer sock of each relevant kind.

\[ P = \frac{5}{8} \times \frac{4}{7} = \frac{20}{56} = \frac{5}{14} \]

Using \( \tfrac{5}{8} \) twice would assume replacement and give \( \tfrac{25}{64} \).

Answer: A

The special rule of multiplication states that

\[ P(A \cap B) = P(A) \times P(B) \]

and this holds only when the events are independent - the occurrence of one leaving the probability of the other unchanged.

Mutual exclusivity belongs to the special rule of addition, \( P(A \cup B) = P(A) + P(B) \). The two conditions are quite different: mutually exclusive events with non-zero probabilities are in fact never independent, since one occurring makes the other impossible.

Answer: D

The sum is given to be 8, so the sample space shrinks to the outcomes that produce it.

\[ (2,6),\ (3,5),\ (4,4),\ (5,3),\ (6,2) \]

Five outcomes, of which exactly one starts with a 4.

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

Note that \( (4,4) \) is a single outcome, not two. Without the given sum, the probability of "first throw 4 and total 8" would be \( \tfrac{1}{36} \).

Answer: B

The favourable outcomes are 1 together with the even numbers 2, 4 and 6.

\[ \{1, 2, 4, 6\} \rightarrow 4 \text{ outcomes} \]

\[ P = \frac{4}{6} = \frac{2}{3} \]

1 is odd, so it does not overlap with the even numbers and nothing has to be subtracted - counting only three outcomes would leave 1 out.

Answer: C

The bag holds \( 4 + 5 + 6 = 15 \) balls, 5 of them red.

\[ P = \frac{{}^{5}C_{3}}{{}^{15}C_{3}} = \frac{10}{455} = \frac{2}{91} \]