A single die is rolled. The probability of getting 1 or an even number is
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.
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A bag contains 4 white, 5 red and 6 blue balls. Three balls are drawn at random from the bag. The probability that all of them are red, is:
Answer: C
The bag holds \( 4 + 5 + 6 = 15 \) balls, 5 of them red.
\[ P = \frac{{}^{5}C_{3}}{{}^{15}C_{3}} = \frac{10}{455} = \frac{2}{91} \]
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The sum of the probabilities of all elementary events of an experiment is p, then
Answer: C
The elementary events of an experiment are mutually exclusive and between them cover every possible outcome, so their probabilities must account for the whole.
\[ p = 1 \]
This is the normalisation axiom of probability. Any set of assignments summing to less or more than 1 is not a valid probability distribution.
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What is probability of drawing two clubs from a well shuffled pack of 52 cards?
Answer: B
There are 13 clubs in the pack. Draw two without replacement:
\[ P = \frac{13}{52} \times \frac{12}{51} = \frac{1}{4} \times \frac{12}{51} \]
\[ = \frac{12}{204} = \frac{1}{17} \]
Equivalently \( \dfrac{{}^{13}C_{2}}{{}^{52}C_{2}} = \dfrac{78}{1326} = \dfrac{1}{17} \). The second denominator is 51, not 52, because the first card is not replaced.
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Two dice are thrown simultaneously. What is the probability of getting two numbers whose product is even?
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.
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In which approach to probability the outcomes are equally likely to occur?
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.
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In a class, there are 15 boys and 10 girls. Three students are selected at random. The probability that 1 girl and 2 boys are selected, is:
Answer: A
Choose one girl from 10 and two boys from 15, out of all ways to choose 3 from 25.
\[ P = \frac{{}^{10}C_{1} \times {}^{15}C_{2}}{{}^{25}C_{3}} = \frac{10 \times 105}{2300} = \frac{21}{46} \]
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A box has 10 black and 10 white balls. What is the probability of getting two balls of the same color?
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.
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Tickets numbered 1 to 50 are mixed and one ticket is drawn at random. Find the probability that the ticket drawn has a number which is a multiple of 4 or 7?
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.
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What are the chances that no two boys are sitting together for a photograph if there are 5 girls and 2 boys?
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.
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