Probability Questions and Answers

Answer: B

A product is odd only when both numbers are odd, so use the complement.

\[ P(\text{both odd}) = \frac{3}{6} \times \frac{3}{6} = \frac{1}{4} \]

\[ P(\text{product even}) = 1 - \frac{1}{4} = \frac{3}{4} \]

A single even factor is enough to make the product even, which is why the complement is so much easier to count.

Answer: A

Classical probability assumes every outcome in the sample space is equally likely, giving

\[ P(E) = \frac{\text{favourable outcomes}}{\text{total outcomes}} \]

It applies to fair dice, fair coins and well-shuffled cards. The relative-frequency approach estimates probability from observed data, and the subjective approach from personal judgement; neither assumes equal likelihood.

Answer: A

Choose one girl from 10 and two boys from 15, out of all ways to choose 3 from 25.

\[ P = \frac{{}^{10}C_{1} \times {}^{15}C_{2}}{{}^{25}C_{3}} = \frac{10 \times 105}{2300} = \frac{21}{46} \]

Answer: C

There are 20 balls in all.

\[ \text{total ways} = {}^{20}C_{2} = 190 \]

\[ \text{same colour} = {}^{10}C_{2} + {}^{10}C_{2} = 45 + 45 = 90 \]

\[ P = \frac{90}{190} = \frac{9}{19} \]

The complement - one of each colour - has probability \( \tfrac{100}{190} = \tfrac{10}{19} \), and the two add to 1.

Answer: A

Count each set, then remove the overlap.

\[ \text{multiples of 4} = \left\lfloor \tfrac{50}{4} \right\rfloor = 12, \qquad \text{multiples of 7} = \left\lfloor \tfrac{50}{7} \right\rfloor = 7 \]

\[ \text{multiples of 28} = \left\lfloor \tfrac{50}{28} \right\rfloor = 1 \]

\[ 12 + 7 - 1 = 18 \quad\Longrightarrow\quad P = \frac{18}{50} = \frac{9}{25} \]

28 is the only number counted twice, being a multiple of both 4 and 7.

Answer: D

Seat the five girls first, then place the boys in the gaps between them so that no two boys are adjacent.

\[ \text{total arrangements} = 7! = 5040 \]

Five girls create six gaps (including the ends). Choose two gaps and seat the boys in them:

\[ 5! \times {}^{6}C_{2} \times 2! = 120 \times 15 \times 2 = 3600 \]

\[ P = \frac{3600}{5040} = \frac{5}{7} \]

The gap method is the standard way to enforce a "no two together" condition.

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 \( B \).