Paper ZeroSign in

Are the events independent?

Test, then conclude with a reason

  • In 6 of 20 recent exams
  • Paper 1 or 2
  • 1 to 3 marks
How it is set in the exam

Paper 1 (no calculator) and Paper 2 (calculator) equally (3 exam versions each), 1–3 marks, the last part of a Venn, table or tree question, after the numbers it needs have been found. Common (6 of 20 exam versions). Three times the conclusion was given ("show that … are independent", or "… are not independent"), so the marks were all for the numbers.

Spot it in the exam
Exam phrases to spot (4)
  • “determine whether AA and BB are independent”

  • “show that AA and BB are independent”

  • “show that the events are not independent”

  • “justify your answer”

In disguise (3)
  • "does the chance of AA depend on whether BB happens?"
  • "is AA more likely for one group than for the whole population?", with a table of two groups
  • a two-stage experiment where the question is whether the first stage (which bag) and the result (which colour) are independent
Variants (2)
  • source: a Venn diagram (probabilities or counts), a two-way table, a two-stage tree, or the formulas
  • asked: test and conclude, show independent, or show not independent
Looks like it, but isn't (3)
  • independence is given and there is an unknown to find: Probabilities in terms of an unknown
  • independence is given, and only a product or a conditional is asked: Independent events, one line
  • "mutually exclusive": the test is P(A∩B)=0\mathrm{P}(A \cap B) = 0 (Venn diagrams)
Worked example

Show independence after the addition rule gives P(B)

Paper 1 (no calculator)
Question

Events AA and BB are such that P(A)=13\mathrm{P}(A) = \frac{1}{3} and P(A∩B)=15\mathrm{P}(A \cap B) = \frac{1}{5}, and the probability that neither AA nor BB occurs is 415\frac{4}{15}.
(a) Find P(B)\mathrm{P}(B). [3]
(b) Show that AA and BB are independent. [2]

Try it yourself first, then compare.

Now try a question

Questions of this type are on their way. Try another one meanwhile: write your working on paper or on screen and see where each mark went. Free, with no account.

A handwritten page as Paper Zero marks it: ticks at the two lines that earned a mark, a cross on line 3 where +7 was copied as −7, and a short note beside that line.123456f′(x) = 4x − 124x − 12 = 0, so x = 3f(3) = 2(3)² − 12(3) − 7= 18 − 36 − 7 = −25minimum at (3, −25)M1 ✓A1 ✓A1 ✗11The +7 became −7: 18 − 36 + 7 = −11.
Which method?
You already haveCompare
P(A)\mathrm{P}(A), P(B)\mathrm{P}(B) and P(A∩B)\mathrm{P}(A \cap B) (a Venn diagram, or the formulas)P(A) P(B)\mathrm{P}(A)\,\mathrm{P}(B) with P(A∩B)\mathrm{P}(A \cap B)
a row or column of a table, or a branch of a tree, which is P(A∣B)\mathrm{P}(A \mid B)P(A∣B)\mathrm{P}(A \mid B) with P(A)\mathrm{P}(A)

Either comparison scores. Pick the one whose numbers are already on the page.

Steps
  1. Find the numbers

    the comparison needs, usually from the parts before.

  2. Write the comparison with the numbers and a sign:

    P(A) P(B)=0.1875≠0.225=P(A∩B)\mathrm{P}(A)\,\mathrm{P}(B) = 0.1875 \ne 0.225 = \mathrm{P}(A \cap B). Don't round before comparing.

  3. Conclude in words,

    with the reason attached: "so AA and BB are not independent". A conclusion scores only with correct values (markschemes don't allow or ).

  4. Two-stage setting:

    a verbal reason is also accepted: the containers have different proportions, so the first stage changes the chance of the result.

Common mistakes
  • Testing P(A)+P(B)=P(A∪B)\mathrm{P}(A) + \mathrm{P}(B) = \mathrm{P}(A \cup B)

    that is the test for mutually exclusive, not independent.

  • A conclusion with no numbers,

    or numbers with no conclusion: both are needed, and the reason scores only when the values are right.

  • "Independent" and "mutually exclusive" confused.

    Mutually exclusive events (with non-zero probabilities) are never independent.

More mistakes (3)
  • Comparing counts instead of proportions

    the same number of green counters in two bags of different sizes is not the same probability.

  • Rounding before comparing

    0.18750.1875 and 0.2250.225 are different, but close values can round to the same thing.

  • Assuming independence to find P(A∩B)\mathrm{P}(A \cap B), then "testing" it

    the test needs P(A∩B)\mathrm{P}(A \cap B) from somewhere else.

Next typeRegression on the GDC

More in Probability: Venn diagrams, trees and conditional probability