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Statistics and Probability - Probability

Grade 11IB_AA

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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The Sample Space SS is the set of all possible outcomes of an experiment. The probability of an event AA is given by P(A)=n(A)n(S)P(A) = \frac{n(A)}{n(S)}, provided all outcomes are equally likely.

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Complementary Events: The probability of an event not occurring is P(A′)=1−P(A)P(A') = 1 - P(A).

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Combined Events (Addition Rule): For any two events AA and BB, the probability that AA or BB (or both) occurs is P(A∪B)=P(A)+P(B)−P(A∩B)P(A \cup B) = P(A) + P(B) - P(A \cap B).

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Mutually Exclusive Events: Events that cannot happen at the same time. For these events, P(A∩B)=0P(A \cap B) = 0, which simplifies the addition rule to P(A∪B)=P(A)+P(B)P(A \cup B) = P(A) + P(B).

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Independent Events: Two events are independent if the occurrence of one does not affect the probability of the other. Mathematically, P(A∩B)=P(A)×P(B)P(A \cap B) = P(A) \times P(B).

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Conditional Probability: The probability of event AA occurring given that event BB has already occurred is denoted by P(A∣B)=P(A∩B)P(B)P(A|B) = \frac{P(A \cap B)}{P(B)}.

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Discrete Random Variables: A variable XX that takes on a countable number of values. The sum of all probabilities in a probability distribution must equal 11, i.e., ∑P(X=x)=1\sum P(X=x) = 1.

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Expected Value: The mean or average outcome of a discrete random variable, denoted by E(X)=∑x⋅P(X=x)E(X) = \sum x \cdot P(X=x).

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Binomial Distribution: A discrete distribution where there are nn independent trials, each with two possible outcomes (success or failure) and a constant probability of success pp. It is denoted as X∼B(n,p)X \sim B(n, p).

📐Formulae

P(A′)=1−P(A)P(A') = 1 - P(A) icon

P(A∪B)=P(A)+P(B)−P(A∩B)P(A \cup B) = P(A) + P(B) - P(A \cap B) icon

P(A∣B)=P(A∩B)P(B)P(A | B) = \frac{P(A \cap B)}{P(B)} icon

P(A∩B)=P(A)×P(B) (for independent events)P(A \cap B) = P(A) \times P(B) \text{ (for independent events)} icon

E(X)=∑i=1nxiP(X=xi)E(X) = \sum_{i=1}^{n} x_i P(X = x_i) icon

P(X=r)=(nr)pr(1−p)n−rP(X=r) = \binom{n}{r} p^r (1-p)^{n-r} icon

💡Examples

Problem 1:

Given that P(A)=0.6P(A) = 0.6, P(B)=0.4P(B) = 0.4, and P(A∪B)=0.8P(A \cup B) = 0.8, find P(A∣B)P(A | B).

Solution:

  1. Find P(A∩B)P(A \cap B) using the addition rule: P(A∪B)=P(A)+P(B)−P(A∩B)P(A \cup B) = P(A) + P(B) - P(A \cap B) 0.8=0.6+0.4−P(A∩B)0.8 = 0.6 + 0.4 - P(A \cap B) 0.8=1.0−P(A∩B)0.8 = 1.0 - P(A \cap B) P(A∩B)=0.2P(A \cap B) = 0.2

  2. Use the conditional probability formula: P(A∣B)=P(A∩B)P(B)P(A | B) = \frac{P(A \cap B)}{P(B)} P(A∣B)=0.20.4=0.5P(A | B) = \frac{0.2}{0.4} = 0.5

Explanation:

First, we use the general addition rule to solve for the intersection. Then, we apply the definition of conditional probability to find the probability of AA given BB.

Problem 2:

A discrete random variable XX has the following probability distribution:

x123P(X=x)0.2k0.5\begin{array}{|c|c|c|c|} \hline x & 1 & 2 & 3 \\ \hline P(X=x) & 0.2 & k & 0.5 \\ \hline \end{array}

Find the value of kk and calculate E(X)E(X).

Solution:

  1. Since the sum of probabilities must be 11: 0.2+k+0.5=10.2 + k + 0.5 = 1 0.7+k=10.7 + k = 1 k=0.3k = 0.3

  2. Calculate the Expected Value E(X)E(X): E(X)=∑x⋅P(X=x)E(X) = \sum x \cdot P(X=x) E(X)=(1×0.2)+(2×0.3)+(3×0.5)E(X) = (1 \times 0.2) + (2 \times 0.3) + (3 \times 0.5) E(X)=0.2+0.6+1.5=2.3E(X) = 0.2 + 0.6 + 1.5 = 2.3

Explanation:

We use the property that the total probability must equal 11 to find the missing parameter kk. Then, we multiply each value of xx by its corresponding probability and sum them up to find the mean (expected value).

Problem 3:

Calculate the difference between 10001000 and 456456 using vertical subtraction.

Solution:

1000−456544\begin{array}{r} 1000 \\ - 456 \\ \hline 544 \end{array}

Explanation:

Standard vertical subtraction used to demonstrate the formatting requirement for arithmetic.