Worksheet: Probability

A sample space is the set of all possible outcomes of a random experiment. An event is a subset of the sample space. For example, consider the experiment of tossing a coin once. The sample space S = {H, T}, where H represents heads and T represents tails. An event could be E: 'the coin shows heads', which corresponds to the subset {H}. Thus, any occurrence within the sample space that satisfies this condition is part of the event.

A simple event consists of a single outcome from the sample space, while a compound event includes two or more outcomes. For instance, in the experiment of tossing a die, rolling a 3 is a simple event (E = {3}), whereas rolling an even number (E = {2, 4, 6}) is a compound event as it includes multiple outcomes. Therefore, the key difference is in the number of outcomes they represent.

Mutually exclusive events refer to two or more events that cannot occur simultaneously. For instance, considering the experiment of rolling a die, let event A be 'rolling an even number' (A = {2, 4, 6}) and event B be 'rolling an odd number' (B = {1, 3, 5}). Here, A and B cannot happen at the same time; hence, they are mutually exclusive. The intersection of A and B is the empty set, A ∩ B = φ.

The arithmetic of probability explains how to calculate the probability of the occurrence of events A or B or both. The formula is P(A ∪ B) = P(A) + P(B) - P(A ∩ B). For example, if P(A) = 0.5 and P(B) = 0.3, and the probability of both A and B occurring is P(A ∩ B) = 0.1, then: P(A ∪ B) = 0.5 + 0.3 - 0.1 = 0.7. This demonstrates combining probabilities while accounting for any overlap.

To find P(not A), we use the formula: P(not A) = 1 - P(A). Given P(A) = 0.4, we have P(not A) = 1 - 0.4 = 0.6. This means 60% of the time, event A does not occur. The approach is straightforward; it stems from the fundamental principle that the total probability of all possible outcomes equals 1.

Exhaustive events cover all possible outcomes of a sample space so that at least one of them will occur. For instance, consider the experiment of rolling a die. Let event A be 'rolling a number less than 4' (A = {1, 2, 3}), event B be 'rolling a 4' (B = {4}), and event C be 'rolling a number greater than 4' (C = {5, 6}). The union of A, B, and C encompasses the whole sample space S = {1, 2, 3, 4, 5, 6}, demonstrating that these events are exhaustive.

The complement of an event A, denoted A', includes all outcomes in the sample space S that are not part of A. For example, if the sample space for the roll of a die is S = {1, 2, 3, 4, 5, 6} and event A is defined as 'rolling a number greater than 3' (A = {4, 5, 6}), then the complement A' would be 'rolling a number not greater than 3', represented as A' = {1, 2, 3}. Hence, P(A') can be calculated knowing P(A).

When outcomes are equally likely, the probability of an event E is calculated as P(E) = n(E)/n(S), where n(E) is the number of favorable outcomes and n(S) the total outcomes. For instance, in tossing a fair coin, the sample space S = {H, T} has 2 outcomes. If event E is 'getting heads', then n(E) = 1. Thus, P(E) = 1/2 = 0.5. This shows how each outcome has the same chance of occurring.

Conditional probability measures the probability of an event A occurring given that another event B has already occurred, denoted as P(A | B). For instance, if the probability of drawing a red card from a deck (event A) is 26/52, and event B is knowing a card drawn is a heart, then P(A | B) = P(A ∩ B)/P(B). If there are 13 hearts, P(B) = 13/52. If we only consider red hearts, P(A ∩ B)=13/52. Hence, P(A | B) = (13/52)/(13/52) = 1.

Mutually exclusive events are those that cannot occur at the same time. For example, if A = {1, 3, 5} and B = {2, 4, 6} when rolling a die, A and B are mutually exclusive because you cannot roll a single die and get both an odd and even number simultaneously. The probability of both A and B occurring is zero (P(A ∩ B) = 0). Thus, P(A or B) = P(A) + P(B).

The sample space S consists of 52 cards. The event A (drawing a heart) has 13 outcomes, and event B (drawing a queen) has 4 outcomes. However, one outcome (the queen of hearts) falls in both A and B. The probability of A or B is calculated as: P(A ∪ B) = P(A) + P(B) - P(A ∩ B) = (13/52) + (4/52) - (1/52) = 16/52 = 4/13.

The total sample space for two dice is 36. To find the probability that the sum is greater than 8, identify the successful outcomes: (3, 6), (4, 5), (5, 4), (6, 3), (4, 6), (5, 5), (6, 4), (5, 6), (6, 5), (6, 6). There are 10 combinations where the sum is greater than 8. Thus, P(sum > 8) = 10/36 = 5/18.

Complementary events are two mutually exclusive events whose probabilities add up to 1. For example, if A is the event 'rolling a 1' when a die is thrown, its complement A' (not rolling a 1) would include outcomes {2, 3, 4, 5, 6}. Thus, P(A) = 1/6 and P(A') = 5/6. Their relationship is given by: P(A) + P(A') = 1.

The probability of drawing an ace first is P(A) = 4/52. After drawing one ace, the probability of drawing a second ace is P(A|A1) = 3/51. The joint probability of both events happening without replacement is: P(A and A1) = P(A) * P(A|A1) = (4/52) * (3/51) = 12/2652 = 1/221.

Let A = {students studying Mathematics}, B = {students studying Physics}. Using the principle of inclusion-exclusion: P(A ∪ B) = P(A) + P(B) - P(A ∩ B). Here, P(A) = 30/60, P(B) = 40/60, P(A ∩ B) = 15/60. Thus, P(A ∪ B) = 30/60 + 40/60 - 15/60 = 55/60 = 11/12.

The word 'PROBABILITY' contains 11 letters, with vowels being O, A, I, I (4 vowels total). The probability is thus calculated as: P(vowel) = 4/11.

The Law of Total Probability states that if events B1, B2,..., Bn form a partition of the sample space, then the probability of any event A can be expressed as: P(A) = Σ P(A|Bi)P(Bi). As an example, consider a factory producing defective and non-defective parts; knowing the probabilities of defects based on machine type allows for better predictions of overall defect rates.

The Axiomatic approach defines probability through axioms rather than relying solely on empirical observations. This provides a solid foundation for probability theory. A significant implication is that it allows the calculation of probabilities even in situations where empirical data may not exist, e.g., in random processes like gambling or insurance, ensuring consistency and coherence in applied probability.

Delve into how the law of large numbers asserts that as the number of trials increases, the sample mean will converge to the expected value. Discuss real-life applications such as gambling and insurance, and evaluate potential pitfalls in interpretations.

Use the example of drawing cards from a deck. Event A could be drawing a heart, while Event B could involve drawing a face card. Explain the probability calculations, emphasizing how independence affects their probability.

Develop a case study based on diagnostic tests and discuss sensitivity, specificity, and prevalence. Critically evaluate how these factors influence overall testing outcomes.

Explore an example with weather predictions (e.g. rain, snow, or neither). Discuss how considering complements simplifies probability computation.

Investigate the assumption of equal likelihood of outcomes, using examples from loaded dice. Examine consequences for fairness and decision-making.

Construct an experiment around a simple dice roll, exploring different event definitions (e.g., rolling an odd number, rolling a number less than four). Analyze how these definitions interact.

Consider a supply chain scenario where the availability of a product affects purchasing decisions. Discuss how conditional probabilities can optimize stock levels.

Run a comparative analysis on trials with fair vs unfair coins. Discuss how bias distorts long-term predictions in probability.

Detail a scenario with teams matched in tournaments, exploring how teams are paired using combinations versus their order using permutations.

Analyze the Monty Hall Problem in-depth, explaining the counterintuitive outcome when switching vs. staying and reinforcing key probability principles.