Probability Questions and Answers

Answer: B

A pot is chosen first, each with probability \( \tfrac{1}{2} \), and then a marble from it.

\[ P = \frac{1}{2} \times \frac{5}{8} + \frac{1}{2} \times \frac{4}{6} \]

\[ = \frac{5}{16} + \frac{1}{3} = \frac{15}{48} + \frac{16}{48} = \frac{31}{48} \]

Pooling all 14 marbles into one pot would give \( \tfrac{9}{14} \) - the first option - but that would treat the two pots as equally stocked, which they are not.

Answer: C

When three coins are tossed simultaneously, there are 8 possible outcomes, which are (HHH), (HHT), (HTH), (THH), (HTT), (THT), (TTH), (TTT), where H is represents the head and T represents the tail.

Favorable outcomes of getting all heads or all tails are HHH and TTT.

We know that, Probability of an event E, P (E) =Number of favorable outcomes /Total number of outcomes

Hence, the required probability = 2/8 = 1/4

Answer: A

The two cases cannot both happen, so add their counts.

\[ \text{both white} = {}^{2}C_{2} = 1, \qquad \text{both blue} = {}^{3}C_{2} = 3 \]

\[ P = \frac{1 + 3}{{}^{15}C_{2}} = \frac{4}{105} \]

With only two white shirts there is exactly one way to take both, which is what keeps the numerator so small.

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: 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: B

A non-leap year has 365 days:

\[ 365 = 52 \times 7 + 1 \]

So there are 52 Thursdays for certain, plus one leftover day. A 53rd Thursday occurs only if that day is a Thursday, and it is equally likely to be any of the seven days.

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

A leap year has two leftover days and the probability rises to \( \tfrac{2}{7} \).

Answer: C

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

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

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

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: B

The total possible outcomes = 24

Numbers divisible by only 2 are 2, 4, 8, 10, 14, 16, 20, 22 (8 numbers) --- (1)

Numbers divisible only by 3 are 3, 9, 15, 21 (4 numbers) --- (2)

Numbers divisible by both 2 and 3 are 6, 12, 18, 24 (4 numbers) --- (3)

From (1), (2) and (3), we see that the number of favourable outcomes is 16 (i.e., 8 + 4 + 4).

We know that, Probability of an event E, P (E) =number of favourable outcomes/ total number of outcomes

=16/24

=2/3

Answer: C

There are 12 face cards - jack, queen and king in each of the four suits - so 40 cards are not face cards.

\[ P = \frac{40}{52} = \frac{10}{13} \]

Equivalently \( 1 - \tfrac{12}{52} = 1 - \tfrac{3}{13} = \tfrac{10}{13} \).