Probability Fundamentals - IB Mathematics Analysis and Approaches

1. A fair four-sided die (with faces numbered 1, 2, 3, 4) and a fair six-sided die (with faces numbered 1, 2, 3, 4, 5, 6) are rolled. What is the probability that the sum of the outcomes is a prime number?
   Answer: 11/24 The total number of possible outcomes in the sample space is $4 \times 6 = 24$. The possible sums range from $1+1=2$ to $4+6=10$. The prime numbers in this range are 2, 3, 5, and 7. We need to find the number of ways to obtain each of these sums: - Sum = 2: (1,1) - 1 way - Sum = 3: (1,2), (2,1) - 2 ways - Sum = 5: (1,4), (2,3), (3,2), (4,1) - 4 ways - Sum = 7: (1,6), (2,5), (3,4), (4,3) - 4 ways The total number of favourable outcomes is $1 + 2 + 4 + 4 = 11$. The probability is the number of favourable outcomes divided by the total number of outcomes, which is $\frac{11}{24}$.
2. A student who is a member of the Art club is chosen at random. What is the probability that this student is also a member of the Band club?
   Answer: 1/3 The question asks for the probability of a student being in the Band club given that they are in the Art club. This means our sample space is restricted to the students in the Art club. There are 60 students in the Art club. This is our new total. Of these 60 students, we are told that 20 are also in the Band club. Therefore, the probability is the number of students in both clubs divided by the number of students in the Art club. Probability = $\frac{\text{Number in A and B}}{\text{Number in A}} = \frac{20}{60} = \frac{1}{3}$.
3. A fair coin is tossed, and a fair six-sided die is rolled. What is the probability that the coin shows heads or the die shows a number less than 3?
   Answer: 2/3 Let H be the event that the coin shows heads, and D be the event that the die shows a number less than 3. The total sample space has $2 \times 6 = 12$ outcomes. $P(H) = 1/2$. The numbers less than 3 on a six-sided die are 1 and 2. So there are 2 favourable outcomes out of 6. $P(D) = 2/6 = 1/3$. The events H and D are independent. The event 'H and D' means the coin is heads AND the die is a 1 or 2. The probability is $P(H \cap D) = P(H) \times P(D) = (1/2) \times (1/3) = 1/6$. We want to find $P(H \cup D)$. Using the addition rule: $P(H \cup D) = P(H) + P(D) - P(H \cap D) = 1/2 + 1/3 - 1/6$. To add these fractions, we find a common denominator, which is 6. $\frac{3}{6} + \frac{2}{6} - \frac{1}{6} = \frac{4}{6} = \frac{2}{3}$.
4. Find the value of $P_1 - P_2$.
   Answer: 3/50 There are a total of 5 marbles in the bag. First, calculate $P_1$ (with replacement). The probability of drawing a red marble is $3/5$. Since the marble is replaced, the probability of drawing a second red marble is also $3/5$. $P_1 = \frac{3}{5} \times \frac{3}{5} = \frac{9}{25}$. Next, calculate $P_2$ (without replacement). The probability of drawing a first red marble is $3/5$. After drawing one red marble, there are 2 red marbles left and a total of 4 marbles. The probability of drawing a second red marble is $2/4 = 1/2$. $P_2 = \frac{3}{5} \times \frac{2}{4} = \frac{6}{20} = \frac{3}{10}$. Finally, find the difference $P_1 - P_2$. $P_1 - P_2 = \frac{9}{25} - \frac{3}{10}$. The common denominator is 50. $\frac{9 \times 2}{25 \times 2} - \frac{3 \times 5}{10 \times 5} = \frac{18}{50} - \frac{15}{50} = \frac{3}{50}$.
5. Let A and B be two events such that $P(A) = 0.6$, $P(B) = 0.5$, and $P(A \cap B) = 0.2$. Find the value of $P(A' \cup B')$.
   Answer: 0.8 This question requires the use of De Morgan's Laws for probability. The event $A' \cup B'$ is the complement of the event $A \cap B$. Therefore, $P(A' \cup B') = P((A \cap B)')$. Using the complement rule, $P((A \cap B)') = 1 - P(A \cap B)$. Given that $P(A \cap B) = 0.2$, we have $P(A' \cup B') = 1 - 0.2 = 0.8$.
6. What is the probability that the two balls drawn (one from each bag in sequence) are of different colours?
   Answer: 23/50 We can solve this using a tree diagram. There are two scenarios where the balls have different colours: (Red from 1st, Blue from 2nd) or (Blue from 1st, Red from 2nd). Scenario 1: Draw Red from 1st bag, then Blue from 2nd bag. The probability of drawing a red ball from the first bag is $P(R_1) = \frac{3}{5}$. The second bag now contains 5 red and 5 blue balls (total 10). The probability of drawing a blue ball from the second bag is $P(B_2|R_1) = \frac{5}{10}$. The probability of this scenario is $\frac{3}{5} \times \frac{5}{10} = \frac{15}{50}$. Scenario 2: Draw Blue from 1st bag, then Red from 2nd bag. The probability of drawing a blue ball from the first bag is $P(B_1) = \frac{2}{5}$. The second bag now contains 4 red and 6 blue balls (total 10). The probability of drawing a red ball from the second bag is $P(R_2|B_1) = \frac{4}{10}$. The probability of this scenario is $\frac{2}{5} \times \frac{4}{10} = \frac{8}{50}$. The total probability of drawing balls of different colours is the sum of the probabilities of these two mutually exclusive scenarios: $\frac{15}{50} + \frac{8}{50} = \frac{23}{50}$.
7. From a group of 5 boys and 4 girls, a committee of three is chosen at random. What is the probability that the committee consists of exactly 2 boys and 1 girl?
   Answer: 10/21 First, we find the total number of ways to choose a committee of 3 from the 9 people. This is the size of our sample space. $n(S) = \binom{9}{3} = \frac{9!}{3!(9-3)!} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 3 \times 4 \times 7 = 84$. Next, we find the number of ways to form the desired committee. We need to choose 2 boys from 5, and 1 girl from 4. The number of ways to do this is the product of the combinations: Number of ways to choose 2 boys from 5 is $\binom{5}{2} = \frac{5!}{2!3!} = \frac{5 \times 4}{2} = 10$. Number of ways to choose 1 girl from 4 is $\binom{4}{1} = 4$. The total number of favourable outcomes is $10 \times 4 = 40$. The probability is the ratio of favourable outcomes to the total number of outcomes: $P(\text{2 boys, 1 girl}) = \frac{40}{84}$. Simplifying this fraction by dividing the numerator and denominator by 4 gives $\frac{10}{21}$.
8. Let A and B be two events. Given that $P(A \cup B) = p$, $P(A) = q$, and $P(B) = r$. Which expression represents the probability that exactly one of the events A or B occurs?
   Answer: 2p - q - r The event that 'exactly one of A or B occurs' can be represented as $(A \cap B') \cup (A' \cap B)$. The probability of this event is $P(A \cup B) - P(A \cap B)$. First, we need to find $P(A \cap B)$ using the general addition rule: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$. Rearranging this gives $P(A \cap B) = P(A) + P(B) - P(A \cup B)$. Substituting the given variables: $P(A \cap B) = q + r - p$. Now, we can find the probability of exactly one event occurring: $P(\text{exactly one}) = P(A \cup B) - P(A \cap B) = p - (q + r - p) = p - q - r + p = 2p - q - r$.
9. Two events A and B are such that $P(A) = 3/5$ and $P(A \cup B) = 4/5$. Let $P(B) = p$. If A and B are mutually exclusive, what is the value of $p$?
   Answer: 1/5 If two events A and B are mutually exclusive, it means they cannot happen at the same time, so $P(A \cap B) = 0$. The addition rule for probability is $P(A \cup B) = P(A) + P(B) - P(A \cap B)$. Since A and B are mutually exclusive, this simplifies to $P(A \cup B) = P(A) + P(B)$. We are given $P(A \cup B) = 4/5$, $P(A) = 3/5$, and $P(B) = p$. Substituting these values into the simplified formula gives $4/5 = 3/5 + p$. Solving for $p$ gives $p = 4/5 - 3/5 = 1/5$.
10. Events A, B, and C are defined on the same sample space. It is given that B is a subset of A ($B \subseteq A$) and that A and C are mutually exclusive. Which of the following statements must be true?
    Answer: $P(B \cup C) = P(B) + P(C)$ Let's analyze the given conditions. $B \subseteq A$ means that any outcome in event B is also in event A. A and C are mutually exclusive means $A \cap C = \emptyset$, so $P(A \cap C) = 0$. Since every outcome of B is in A, and A has no outcomes in common with C, it must be that B also has no outcomes in common with C. Therefore, B and C are mutually exclusive, and $P(B \cap C) = 0$. The addition rule for probability is $P(B \cup C) = P(B) + P(C) - P(B \cap C)$. Since $P(B \cap C) = 0$, this simplifies to $P(B \cup C) = P(B) + P(C)$. Thus, statement A must be true. B is false because if $B \subseteq A$, then $A \cap B = B$, so $P(A \cap B) = P(B)$. C is incorrect because the addition rule is $P(A \cup C) = P(A) + P(C) - P(A \cap C)$. D is false because if $B \subseteq A$, then $P(B) \le P(A)$.
11. Let A and B be events with $P(A \cup B) = 0.8$, $P(A' \cap B) = 0.3$ and $P(A \cap B') = 0.4$. Find $P(A \cap B)$.
    Answer: 0.1 The event $A \cup B$ is the union of three disjoint events: $A \cap B'$ (A only), $A' \cap B$ (B only), and $A \cap B$ (both A and B). Therefore, the probability of the union is the sum of the probabilities of these three events: $P(A \cup B) = P(A \cap B') + P(A' \cap B) + P(A \cap B)$. We are given the values for $P(A \cup B)$, $P(A' \cap B)$, and $P(A \cap B')$. We can substitute these into the equation to find $P(A \cap B)$. $0.8 = 0.4 + 0.3 + P(A \cap B)$. $0.8 = 0.7 + P(A \cap B)$. Solving for $P(A \cap B)$ gives $P(A \cap B) = 0.8 - 0.7 = 0.1$.
12. Find the number of students who study exactly two of these subjects.
    Answer: 22 We need to find the number of students in the regions corresponding to exactly two subjects. These are the intersection regions minus the central region where all three subjects overlap. - Number studying Physics and Chemistry only: $n(P \cap C) - n(P \cap C \cap B) = 15 - 5 = 10$. - Number studying Physics and Biology only: $n(P \cap B) - n(P \cap C \cap B) = 12 - 5 = 7$. - Number studying Chemistry and Biology only: $n(C \cap B) - n(P \cap C \cap B) = 10 - 5 = 5$. The total number of students studying exactly two subjects is the sum of these values: $10 + 7 + 5 = 22$.
13. In a class, the probability that a student takes Chemistry (C) is 0.7 and the probability that a student takes Physics (P) is 0.6. The probability that a student takes neither subject is 0.1. What is the probability that a student takes Chemistry but not Physics?
    Answer: 0.3 Given: $P(C) = 0.7$, $P(P) = 0.6$, and $P(C' \cap P') = 0.1$. By De Morgan's law, $P(C' \cap P') = P((C \cup P)') = 0.1$. Therefore, $P(C \cup P) = 1 - 0.1 = 0.9$. Using the addition rule: $P(C \cup P) = P(C) + P(P) - P(C \cap P)$. Substituting: $0.9 = 0.7 + 0.6 - P(C \cap P)$, so $P(C \cap P) = 1.3 - 0.9 = 0.4$. The probability of taking Chemistry but not Physics is $P(C \cap P') = P(C) - P(C \cap P) = 0.7 - 0.4 = 0.3$.
14. Events A, B, and C are mutually exclusive and exhaustive. Given $P(A) = 2x$, $P(B) = 3x$, and $P(C) = 4x$, find $P(A \cup B)$.
    Answer: 5/9 Since the events A, B, and C are mutually exclusive and exhaustive, the sum of their probabilities must be 1. $P(A) + P(B) + P(C) = 1$ $2x + 3x + 4x = 1 \implies 9x = 1 \implies x = 1/9$. Now we can find the probabilities of A and B: $P(A) = 2x = 2/9$ $P(B) = 3x = 3/9 = 1/3$ The question asks for $P(A \cup B)$. Since A and B are mutually exclusive, $P(A \cap B) = 0$, so we can use the simplified addition rule: $P(A \cup B) = P(A) + P(B) = 2/9 + 3/9 = 5/9$. Alternatively, since the events are exhaustive, $P(A \cup B) = P(C') = 1 - P(C) = 1 - 4x = 1 - 4/9 = 5/9$.

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