Mathematics Grade 9 15 min

Identify independent and dependent events

Identify independent and dependent events

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Introduction & Learning Objectives

Learning Objectives Define independent and dependent events in their own words. Differentiate between independent and dependent events by analyzing if one event's outcome affects the other's. Identify independent events in mathematical problems and real-world scenarios. Identify dependent events in mathematical problems and real-world scenarios. Explain the key difference between scenarios 'with replacement' and 'without replacement'. Determine how the sample space is affected in a sequence of dependent events. If you pull a red sock from your drawer, does that change your chance of pulling out another red sock next? 🤔 Let's find out! This tutorial will teach you how to tell the difference between two key types of events in probability: i...
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Key Concepts & Vocabulary

TermDefinitionExample EventA specific outcome or a set of outcomes from an experiment. It's the 'thing' we are measuring the probability of.In a coin flip, 'getting heads' is an event. When rolling a die, 'rolling an even number' is an event. Independent EventsTwo or more events where the outcome of one event has absolutely no effect on the probability of the other event(s) occurring.Flipping a coin and getting heads, and then rolling a die and getting a 6. The coin flip does not change the probability of what you'll roll on the die. Dependent EventsTwo or more events where the outcome of the first event changes the probability of the second event occurring.Drawing a card from a deck and not replacing it, then drawing a second card. The first draw c...
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Core Formulas

The Conceptual Test Question Ask: 'Does the outcome of the first event change the probability of the second event?' This is the most important question to ask when identifying event types. If the answer is 'No', the events are independent. If the answer is 'Yes', the events are dependent. The Mathematical Test for Independence Two events, A and B, are independent if and only if P(A \text{ and } B) = P(A) \times P(B). This formula is the formal mathematical way to prove independence. If the probability of both events happening together equals the product of their individual probabilities, they are independent. Conditional Probability and Dependence For dependent events, the probability of B occurring, given that A has already occurred, is w...

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Sample Practice Questions

Challenging
The probability that a student studies for a test (Event S) is P(S) = 0.8. The probability that the student passes the test (Event P) is P(P) = 0.7. The probability that the student studies AND passes is P(S and P) = 0.6. Are the events 'studying' and 'passing' independent or dependent?
A.Independent, because 0.8 × 0.7 is close to 0.6.
B.Dependent, because P(S) × P(P) ≠ P(S and P).
C.Independent, because studying usually leads to passing.
D.Dependent, because P(S) + P(P) > 1.
Challenging
A student claims: 'If I draw a card from a deck and it's a heart, and then I draw a second card without replacement, the events are dependent. But if the first card was a spade, the events would be independent.' What is the flaw in this reasoning?
A.The reasoning is correct; the suit of the first card determines the dependency.
B.The dependency arises from not replacing the card, regardless of its suit. The sample space is always altered.
C.The events are always independent as long as the deck is shuffled.
D.The flaw is that drawing a heart first makes the second draw more likely to be a heart too.
Challenging
There are three traffic lights on a route. The probability of each light being green is 0.5. Let A, B, and C be the events that the first, second, and third lights are green, respectively. Assuming the lights operate independently, which statement is true?
A.If the first two lights are red, the third is more likely to be green.
B.The events are dependent because you encounter the lights in a sequence.
C.The probability of the third light being green, given the first two were green, is still 0.5.
D.The probability of all three being green is the sum of their individual probabilities.

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