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Year 10AC9M10P01

Conditional Probability

Use the language of "if … then", "given", "of", "knowing that" to describe and interpret situations involving conditional probability.

A conditional probability question tells you something has already happened. That information shrinks the group you are choosing from — and the whole method is about finding the new denominator.

Builds onYr 10 · Two-Way Tables

Method

The words that signal it
given that · if · knowing that · of the · among those who. Any of these means the group has shrunk.
The formula
P(A∣B)=P(A and B)P(B)P(A \mid B) = \frac{P(A \text{ and } B)}{P(B)}
Read $P(A \mid B)$ as ‘the probability of A, given B’.
From a table, just count
P(A∣B)=celltotal of BP(A \mid B) = \frac{\text{cell}}{\text{total of B}}
The condition tells you which row or column to stay inside. That total becomes your denominator.
Order matters
P(A∣B)≠P(B∣A)P(A \mid B) \neq P(B \mid A)
The probability of rain given clouds is not the probability of clouds given rain.

Worked example

e.g. Using the table below, find the probability that a student is in Year 10, given that they catch the bus.\text{Using the table below, find the probability that a student is in Year 10, given that they catch the bus.}
  1. Write out the table you are working from.BusWalkTotalYear 10342660Year 9112940Total4555100\begin{array}{l|cc|c} & \text{Bus} & \text{Walk} & \text{Total} \\ \hline \text{Year 10} & 34 & 26 & 60 \\ \text{Year 9} & 11 & 29 & 40 \\ \hline \text{Total} & 45 & 55 & 100 \end{array}
  2. Underline the condition. It is the part after given that.given that they catch the bus\text{given that they catch the bus}
  3. Find the total of that group. This is your new denominator — not 100.Bus total=45\text{Bus total} = 45
  4. Find how many of that group also fit the first condition.Year 10 and bus=34\text{Year 10 and bus} = 34
  5. Write the fraction and simplify.P(Yr10∣bus)=3445≈0,76P(\text{Yr10} \mid \text{bus}) = \frac{34}{45} \approx 0{,}76
  6. Sanity check: a conditional probability is almost always larger than the unconditional one, because the group is smaller.P(Yr10)=60100=0,60vs0,76P(\text{Yr10}) = \frac{60}{100} = 0{,}60 \quad \text{vs} \quad 0{,}76

Practice

Use the same table. The condition tells you which total to divide by.

1
P(walks∣Year 10)P(\text{walks} \mid \text{Year 10})
Answer2660≈0,43\frac{26}{60} \approx 0{,}43
2
P(Year 9∣walks)P(\text{Year 9} \mid \text{walks})
Answer2955≈0,53\frac{29}{55} \approx 0{,}53
3
Explain why P(Yr10∣bus)≠P(bus∣Yr10)\text{Explain why } P(\text{Yr10} \mid \text{bus}) \neq P(\text{bus} \mid \text{Yr10})
Answer3445≈0,76but3460≈0,57. Same cell, different denominator.\frac{34}{45} \approx 0{,}76 \quad \text{but} \quad \frac{34}{60} \approx 0{,}57. \text{ Same cell, different denominator.}
4
A test is 95% accurate. 1% of people have the condition. Of 10 000 people, how many test positive in total?\text{A test is 95\% accurate. 1\% of people have the condition. Of 10\,000 people, how many test positive in total?}
Answer95 true positives+495 false positives=59095 \text{ true positives} + 495 \text{ false positives} = 590
5
From that same group, find P(has it∣tested positive)\text{From that same group, find } P(\text{has it} \mid \text{tested positive})
Answer95590≈0,16 — far lower than most people expect\frac{95}{590} \approx 0{,}16 \text{ — far lower than most people expect}
Next step
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