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Statistics and Probability - Conditional probability

Grade 10IB

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

🔑Concepts

Conditional Probability Definition: This refers to the probability of an event AA occurring given that event BB has already occurred, denoted as P(AB)P(A|B). Visually, this corresponds to 'shrinking' the sample space; instead of looking at the entire set of possible outcomes, we only consider the outcomes that fall within the circle representing event BB.

Venn Diagram Interpretation: When looking at a Venn diagram of two events AA and BB, P(AB)P(A|B) is calculated by taking the area of the intersection (ABA \cap B) and dividing it by the entire area of circle BB. It ignores any parts of AA that are outside of BB and ignores the universal set outside of BB.

Tree Diagrams and Branching: In a tree diagram, conditional probabilities are represented by the second set of branches. If the first branch represents event AA, the branch following it represents P(BA)P(B|A). To find the probability of both events happening (ABA \cap B), you multiply the probabilities along the path.

Independent Events: Two events are independent if the occurrence of one does not change the probability of the other. Visually, if AA and BB are independent, the proportion of AA within the BB circle is the same as the proportion of AA in the whole rectangular sample space. This is expressed as P(AB)=P(A)P(A|B) = P(A).

Dependent Events and Sampling: Events are dependent if the outcome of the first event affects the second. This is common in 'without replacement' scenarios. Visually, this is shown on a tree diagram where the denominator of the fraction on the second branch decreases by one because one item has been removed from the total.

Two-Way Tables (Contingency Tables): These tables organize data into rows and columns based on two variables. To find a conditional probability like P(Category XCategory Y)P(\text{Category X} | \text{Category Y}), you restrict your focus to the single row or column representing 'Category Y' and find the proportion of 'Category X' within that specific row or column total.

📐Formulae

P(AB)=P(AB)P(B), provided P(B)>0P(A|B) = \frac{P(A \cap B)}{P(B)}, \text{ provided } P(B) > 0

P(AB)=P(B)×P(AB)P(A \cap B) = P(B) \times P(A|B)

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

P(AB)=P(A) (Condition for independence)P(A|B) = P(A) \text{ (Condition for independence)}

💡Examples

Problem 1:

In a group of 100 students, 40 play football, 30 play cricket, and 10 play both. If a student is selected at random and it is known they play football, what is the probability that they also play cricket?

Solution:

  1. Identify the given information: P(Football)=40100=0.4P(\text{Football}) = \frac{40}{100} = 0.4 and P(FootballCricket)=10100=0.1P(\text{Football} \cap \text{Cricket}) = \frac{10}{100} = 0.1. \ 2. Use the conditional probability formula: P(CricketFootball)=P(CricketFootball)P(Football)P(\text{Cricket} | \text{Football}) = \frac{P(\text{Cricket} \cap \text{Football})}{P(\text{Football})}. \ 3. Substitute the values: P(CricketFootball)=0.10.4=14=0.25P(\text{Cricket} | \text{Football}) = \frac{0.1}{0.4} = \frac{1}{4} = 0.25.

Explanation:

Since we are given that the student plays football, our sample space is reduced from 100 students to just the 40 students who play football. Within those 40, only 10 play cricket. Thus, the probability is 10 out of 40.

Problem 2:

A bag contains 5 red balls and 3 green balls. Two balls are drawn one after the other without replacement. Find the probability that the second ball is green, given that the first ball drawn was red.

Solution:

  1. Determine the initial state: Total balls = 5+3=85 + 3 = 8. \ 2. Adjust for the condition: The first ball drawn was red. This means there are now 51=45 - 1 = 4 red balls left and 3 green balls still in the bag. \ 3. Calculate the new total: 4+3=74 + 3 = 7 balls remaining. \ 4. Find the probability of drawing a green ball from the remaining set: P(Green2Red1)=37P(\text{Green}_2 | \text{Red}_1) = \frac{3}{7}.

Explanation:

In 'without replacement' problems, the denominator of the probability fraction decreases for the second event because the total number of items has changed. Since the first ball was red, the number of green balls remained at 3, but the total dropped to 7.