Probability Questions and Answers

Answer: D

A deterministic experiment gives the same result every time it is repeated under identical conditions - heating water to 100°C at sea level, for example.

A random experiment is the opposite: the conditions may be identical, yet the outcome varies from trial to trial. Probability theory is concerned with the second kind.

Answer: C

There are \( 10 + 25 = 35 \) tickets in the lottery.

\[ P = \frac{10}{35} = \frac{2}{7} \]

The denominator is the total number of tickets, not the number of blanks - using 25 would give \( \tfrac{2}{5} \).

Answer: C

Of the 30 cards, 17 are white and 13 green. Nine are marked IMPORTANT: 4 white and 5 green.

Adding the two counts directly would double-count the 5 green IMPORTANT cards, so use the addition rule:

\[ P(G \cup I) = P(G) + P(I) - P(G \cap I) \]

\[ = \frac{13}{30} + \frac{9}{30} - \frac{5}{30} = \frac{17}{30} \]

Equivalently, count the 13 green cards plus the 4 white IMPORTANT ones: \( \tfrac{17}{30} \).

Answer: C

Two coins give four outcomes: HH, HT, TH, TT. Exactly one head occurs in HT and TH.

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

HT and TH are distinct outcomes because the coins are distinguishable; treating them as one would give \( \tfrac{1}{3} \).

Answer: D

Among the first 20 natural numbers:

\[ \text{even} = 10, \qquad \text{prime} = 8 \ (2, 3, 5, 7, 11, 13, 17, 19) \]

Only 2 is both, so

\[ 10 + 8 - 1 = 17 \quad\Longrightarrow\quad P = \frac{17}{20} \]

2 is the only even prime, which is exactly why the overlap here is a single number.

Answer: D

There are four kings in the pack.

\[ P = \frac{{}^{4}C_{2}}{{}^{52}C_{2}} = \frac{6}{1326} = \frac{1}{221} \]

Equivalently \( \tfrac{4}{52} \times \tfrac{3}{51} = \tfrac{1}{221} \), the second fraction reflecting that the first card is not replaced.

Answer: C

There is exactly one queen of clubs and one king of hearts, and no card is both.

\[ P = \frac{1}{52} + \frac{1}{52} = \frac{2}{52} = \frac{1}{26} \]

The question names specific cards, not whole ranks, so the count is 2 rather than 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 \( \tfrac{3}{8} \).

Answer: B

An event is any subset of the sample space - one outcome or a collection of them.

Rolling an even number, for example, is the event \( \{2, 4, 6\} \).

Probability is the number assigned to an event, and a random variable is a function from outcomes to numbers - neither is the collection itself.

Answer: B

The colour of the transferred shirt is unknown, so split into the two cases and add them.

A black shirt is moved (probability \( \tfrac{5}{8} \)). The second box then has 4 black out of 9:

\[ \frac{5}{8} \times \frac{4}{9} = \frac{20}{72} \]

A green shirt is moved (probability \( \tfrac{3}{8} \)). The second box still has 3 black out of 9:

\[ \frac{3}{8} \times \frac{3}{9} = \frac{9}{72} \]

\[ P = \frac{20}{72} + \frac{9}{72} = \frac{29}{72} \]

The second box holds 9 shirts after the transfer, not 8 - the moved shirt is added to it.