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Year 9AC9M9P01

Compound Events With and Without Replacement

List all outcomes for compound events both with and without replacement, using lists, tree diagrams, tables or arrays; assign probabilities to outcomes. Calculate relative frequencies from given or collected data to estimate probabilities of events involving and, inclusive or and exclusive or.

Whether the first item goes back changes every probability after it. Deciding that first is the whole method.

Builds onYr 8 · Two Events: Tables, Trees and Venn

Method

With replacement: nothing changes
The second draw faces the same bag as the first.
Without replacement: the total drops
And so does the count of whatever you took.
Along branches multiply, across branches add
Inclusive or means at least one
P(A or B)=P(A)+P(B)−P(A and B)P(A \text{ or } B) = P(A) + P(B) - P(A \text{ and } B)
Subtract the overlap or you count it twice.
Exclusive or means one but not both

Worked example

e.g. A bag has 5 red and 3 blue. Two are drawn without replacement. Find P(both red).\text{A bag has 5 red and 3 blue. Two are drawn without replacement. Find } P(\text{both red}).
  1. Note the condition. Without replacement, so the bag changes.total starts at 8\text{total starts at } 8
  2. First draw.P(red)=58P(\text{red}) = \frac{5}{8}
  3. Now update the bag. One red is gone.4 red left, 7 total4 \text{ red left, } 7 \text{ total}
  4. Second draw.P(red)=47P(\text{red}) = \frac{4}{7}
  5. Multiply along the branch.58×47=2056\frac{5}{8} \times \frac{4}{7} = \frac{20}{56}
  6. Simplify.=514= \frac{5}{14}

Practice

Decide with or without replacement before anything else.

1
Same bag, with replacement. P(both red)?\text{Same bag, with replacement. } P(\text{both red})?
Answer58×58=2564\frac{5}{8} \times \frac{5}{8} = \frac{25}{64}
2
Two coins. P(at least one head)?\text{Two coins. } P(\text{at least one head})?
Answer34\frac{3}{4}
3
Without replacement, P(red then blue)?\text{Without replacement, } P(\text{red then blue})?
Answer58×37=1556\frac{5}{8} \times \frac{3}{7} = \frac{15}{56}
4
P(A)=0,4, P(B)=0,3, P(A and B)=0,1. Find P(A or B).P(A) = 0{,}4,\ P(B) = 0{,}3,\ P(A \text{ and } B) = 0{,}1. \text{ Find } P(A \text{ or } B).
Answer0,60{,}6
5
Why subtract the overlap in inclusive or?\text{Why subtract the overlap in inclusive or?}
AnswerIt would otherwise be counted twice.\text{It would otherwise be counted twice.}
6
In 200 trials an event happens 46 times. Estimate its probability.\text{In 200 trials an event happens 46 times. Estimate its probability.}
Answer0,230{,}23
Next step
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