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Probability - Combined Events and Tree Diagrams

Grade 10IGCSE

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

🔑Concepts

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Combined Events: Calculating the probability of two or more events happening either together or in sequence.

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Independent Events: Two events are independent if the outcome of the first does not affect the outcome of the second (e.g., rolling a die twice).

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Dependent Events (Conditional): The outcome of the first event affects the probability of the second event (e.g., picking a card and not replacing it).

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Tree Diagrams: A visual tool where branches represent possible outcomes. Multiply probabilities along the branches; add probabilities of different paths.

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Mutually Exclusive Events: Events that cannot happen at the same time. The probability of one or the other occurring is the sum of their individual probabilities.

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Replacement vs. Non-replacement: 'With replacement' implies independent events; 'Without replacement' implies dependent events.

📐Formulae

P(A and B)=P(A)×P(B)P(A \text{ and } B) = P(A) \times P(B) (For Independent Events)

P(A or B)=P(A)+P(B)P(A \text{ or } B) = P(A) + P(B) (For Mutually Exclusive Events)

P(Not A)=1−P(A)P(\text{Not } A) = 1 - P(A)

P(At least one)=1−P(None)P(\text{At least one}) = 1 - P(\text{None})

∑P(all possible outcomes)=1\sum P(\text{all possible outcomes}) = 1

💡Examples

Problem 1:

A bag contains 5 red balls and 3 blue balls. Two balls are taken at random without replacement. Find the probability that both balls are red.

Solution:

P(R1)=58P(R1) = \frac{5}{8}, P(R2∣R1)=47P(R2 | R1) = \frac{4}{7}. Therefore, P(R,R)=58×47=2056=514P(R, R) = \frac{5}{8} \times \frac{4}{7} = \frac{20}{56} = \frac{5}{14}.

Explanation:

Since the first ball is not replaced, the total number of balls decreases from 8 to 7, and the number of red balls decreases from 5 to 4 for the second draw. We multiply along the branch.

Problem 2:

The probability that it rains on Monday is 0.6. If it rains on Monday, the probability it rains on Tuesday is 0.8. If it does not rain on Monday, the probability it rains on Tuesday is 0.3. Calculate the probability it rains on at least one of the two days.

Solution:

P(No rain Monday)=0.4P(\text{No rain Monday}) = 0.4. P(No rain Tuesday | No rain Monday)=1−0.3=0.7P(\text{No rain Tuesday | No rain Monday}) = 1 - 0.3 = 0.7. P(No rain both days)=0.4×0.7=0.28P(\text{No rain both days}) = 0.4 \times 0.7 = 0.28. P(At least one rain)=1−0.28=0.72P(\text{At least one rain}) = 1 - 0.28 = 0.72.

Explanation:

Using the '1 - P(none)' rule is faster than adding the probabilities of (Rain, Rain), (Rain, No Rain), and (No Rain, Rain).

Problem 3:

A spinner has three sections: Red (1/2), Blue (1/3), and Green (1/6). If the spinner is spun twice, what is the probability of getting different colors?

Solution:

P(Same)=P(R,R)+P(B,B)+P(G,G)=(1/2×1/2)+(1/3×1/3)+(1/6×1/6)=1/4+1/9+1/36=9/36+4/36+1/36=14/36P(\text{Same}) = P(R,R) + P(B,B) + P(G,G) = (1/2 \times 1/2) + (1/3 \times 1/3) + (1/6 \times 1/6) = 1/4 + 1/9 + 1/36 = 9/36 + 4/36 + 1/36 = 14/36. P(Different)=1−14/36=22/36=11/18P(\text{Different}) = 1 - 14/36 = 22/36 = 11/18.

Explanation:

It is easier to calculate the probability of the outcomes being the same and subtracting from 1, rather than listing all different color combinations (RB, RG, BR, BG, GR, GB).