Probability Questions and Answers

Answer: D

A probability distribution pairs each possible outcome of an experiment with its probability - for a single die, the six faces each with \( \tfrac{1}{6} \).

A random variable is the quantity being measured, and a contingency table cross-tabulates two categorical variables; neither is the listing of outcomes with their probabilities.

Answer: A

Choose 2 of the 4 red shirts and 1 of the 3 blue, out of all ways to choose 3 from 15.

\[ P = \frac{{}^{4}C_{2} \times {}^{3}C_{1}}{{}^{15}C_{3}} = \frac{6 \times 3}{455} \]

\[ = \frac{18}{455} \]

The two selections are made together, so the counts multiply; the remaining colours play no part.

Answer: A

Of the faces 1 to 6, the composite numbers are 4 and 6.

\[ P = \frac{2}{6} = \frac{1}{3} \]

2, 3 and 5 are prime, and 1 is neither prime nor composite - including it would give \( \tfrac{1}{2} \).

Answer: D

Let \( A \) be "at least one die shows 4" and \( B \) be "the sum is 7", over the 36 outcomes.

\[ |A| = 11 \quad (\text{six with the first a 4, six with the second, less the double-counted } (4,4)) \]

\[ |B| = 6: \ (1,6),(2,5),(3,4),(4,3),(5,2),(6,1) \]

\[ |A \cap B| = 2: \ (3,4),(4,3) \]

\[ |A \cup B| = 11 + 6 - 2 = 15 \quad\Longrightarrow\quad P = \frac{15}{36} \]

Both corrections matter: \( (4,4) \) within \( A \), and the two outcomes shared by \( A \) and \( B \).

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 \( \tfrac{2}{7} \).