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Statistics and Probability - Probability

Grade 11IB_AI

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

🔑Concepts

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The theoretical probability of an event AA is given by P(A)=n(A)n(U)P(A) = \frac{n(A)}{n(U)}, where n(A)n(A) is the number of favorable outcomes and n(U)n(U) is the total number of possible outcomes in the sample space.

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Complementary events: The probability of an event not occurring is P(A′)=1−P(A)P(A') = 1 - P(A). The sum of the probabilities of all possible outcomes in a sample space is always 11.

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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 given by 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: These are events that cannot happen at the same time. For such events, P(A∩B)=0P(A \cap B) = 0 and therefore 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 written as P(A∣B)P(A|B).

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Tree Diagrams and Venn Diagrams: These are visual tools used to represent sample spaces and calculate probabilities for compound events, especially when dealing with sampling with or without replacement.

📐Formulae

P(A)=n(A)n(U)P(A) = \frac{n(A)}{n(U)}

P(A′)=1−P(A)P(A') = 1 - P(A)

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

P(A∩B)=P(A)×P(B∣A)P(A \cap B) = P(A) \times P(B|A)

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

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

💡Examples

Problem 1:

In a group of 30 students, 18 study Biology, 15 study Chemistry, and 8 study both. If a student is chosen at random, find the probability that they study Biology or Chemistry.

Solution:

Let BB be the event of studying Biology and CC be the event of studying Chemistry. P(B)=1830P(B) = \frac{18}{30}, P(C)=1530P(C) = \frac{15}{30}, and P(B∩C)=830P(B \cap C) = \frac{8}{30}. Using the addition rule: P(B∪C)=P(B)+P(C)−P(B∩C)P(B \cup C) = P(B) + P(C) - P(B \cap C) P(B∪C)=1830+1530−830P(B \cup C) = \frac{18}{30} + \frac{15}{30} - \frac{8}{30} P(B∪C)=2530=56P(B \cup C) = \frac{25}{30} = \frac{5}{6}

Explanation:

We apply the addition rule for inclusive events. Since 8 students overlap in both categories, we subtract the intersection to avoid double-counting.

Problem 2:

Two events XX and YY are such that P(X)=0.6P(X) = 0.6 and P(Y)=0.3P(Y) = 0.3. If XX and YY are independent, find P(X∩Y)P(X \cap Y) and P(X∪Y)P(X \cup Y).

Solution:

Since the events are independent: P(X∩Y)=P(X)×P(Y)P(X \cap Y) = P(X) \times P(Y) P(X∩Y)=0.6×0.3=0.18P(X \cap Y) = 0.6 \times 0.3 = 0.18

Now, find the union: P(X∪Y)=P(X)+P(Y)−P(X∩Y)P(X \cup Y) = P(X) + P(Y) - P(X \cap Y) P(X∪Y)=0.6+0.3−0.18=0.72P(X \cup Y) = 0.6 + 0.3 - 0.18 = 0.72

Explanation:

Independence allows us to multiply probabilities to find the intersection. We then use that intersection in the general addition rule.

Problem 3:

A bag contains 5 red balls and 3 blue balls. Two balls are drawn one after another without replacement. Find the probability that both balls are red.

Solution:

Let R1R_1 be the event the first ball is red and R2R_2 be the event the second ball is red. P(R1)=58P(R_1) = \frac{5}{8} After drawing one red ball, 4 red balls and 7 total balls remain. P(R2∣R1)=47P(R_2 | R_1) = \frac{4}{7} P(R1∩R2)=P(R1)×P(R2∣R1)P(R_1 \cap R_2) = P(R_1) \times P(R_2 | R_1) P(R1∩R2)=58×47=2056=514P(R_1 \cap R_2) = \frac{5}{8} \times \frac{4}{7} = \frac{20}{56} = \frac{5}{14}

Explanation:

Because the drawing is without replacement, the events are dependent. We use conditional probability to find the probability of the second event.