Probability distribution tables
Find the unknown, then E(X) and probabilities
- In 14 of 20 recent exams
- 2% of the marks
- Paper 1 or 2
- 1 to 5 marks
How it is set in the exam
Paper 1 (no calculator) and Paper 2 (calculator) (6 and 8 of its 14 exam versions), 1–5 marks per part, usually 2. Half the time it is in a Section A question of 5–6 marks, usually most or all of it, and half the time part of a Section B question, usually the opening parts. Very common (14 of 20 exam versions), and it is the step the other two types in this guide end with.
Exam phrases to spot (6)
“find the value of k”
“find E(X)”
“find P(X>c)”
“given that E(X)=…, find p and q”
“find the possible values of p”
“show that …k2…=0”
In disguise (4)
- a frequency table of data, and an "expected" value per person: the relative frequencies are the probabilities
- P(X=x) given as a formula in x: write out the table first
- the probabilities follow a pattern (an arithmetic or geometric sequence): they still add up to 1 (Where it hides)
- one value of X is unknown and E(X) is given: the E(X) equation is linear in it
Variants (3)
- the table: multiples of k; a quadratic in k; powers of k (, GDC); a formula in x; a frequency table; a table built in an earlier part from an experiment
- given as well: E(X) (two unknowns, or an unknown value of X)
- asked: k, E(X), the probability of a range, the mode, the possible values of a parameter, the range of E(X)
Looks like it, but isn't (4)
- k is in Venn regions, tree branches or a two-way table of events, with a fact such as independence: Probabilities in terms of an unknown
- the entries must first be built by listing outcomes: Build the distribution from the experiment
- unknown frequencies of a data set, with its mean: Work backwards from the mean or median (the same two equations, with n in place of 1
- "given that", with both events ranges of X: Given that X is in a range
Linear in k, a cut-off between two values, then E(X)
The random variable X has this probability distribution.
| x | P(X=x) |
|---|---|
| 0 | 0.05 |
| 1 | 3k |
| 2 | 0.35 |
| 3 | 2k |
| 4 | k |
| x | 0 | 1 | 2 | 3 | 4 |
|---|---|---|---|---|---|
| P(X=x) | 0.05 | 3k | 0.35 | 2k | k |
(a) Find the value of k. [2]
(b) Find P(X>2.5). [2]
(c) Find E(X). [2]
Try it yourself first, then compare.
Now try one like it
Write your working on paper or on screen. Paper Zero marks it step by step, like the worked example, and shows where each mark went. Free, with no account.
The equation from "total =1" (and from E(X), if it is given) decides the route.
| The equation | Solve by |
|---|---|
| linear in k | rearranging |
| quadratic in k | factorising on ; on , solve a polynomial equation. Then test each root in every entry |
| powers of k, with a second unknown | eliminate the second unknown, then solve the polynomial equation; or write the second unknown from each equation and intersect the two graphs |
| two unknowns, both linear (total 1 and E(X)) | simultaneous equations, by hand or on the GDC |
| no equation: the total is 1 whatever the parameter is | 0≤ each entry ≤1: one interval per entry, then their overlap |
- 1
Total =1:
add every entry, including the one at x=0, and set the sum equal to 1. If E(X) is given, write ∑xP(X=x)=E(X) as a second equation.
- 2
Solve
(the table above). Then test every root in every entry: a root is valid only if all the entries lie between 0 and 1. Reject the others, naming the entry that fails, and only after all the roots have been found.
- 3
A probability of a range:
translate the words into values of X (Inequality words), then add those entries, or use the complement. The mode is the value with the largest entry.
- 4
E(X):
write the sum with every term, then its value. Check that it lies between the smallest and the largest value of X.
- 5
The possible values of a parameter:
write 0≤entry≤1 for every entry containing it, and take the overlap. When every entry is linear in the parameter, so is E(X), and its least and greatest values come from the ends of that interval.
Checking a given value instead of solving.
In a "show that", substituting the given k (or p and q) to check the total earns the method mark at most: set up the equations and solve them forwards.
Rejecting only negative roots.
A positive root can still make an entry negative or bigger than 1 (k=0.5 in Example 1.2). Test every entry.
Leaving the rejected root in
(), or rejecting it with no reason (). When the root is itself an entry, "a probability can't be negative" is enough. When it isn't, name the entry that fails: P(X=2)=0.4−0.5<0. The reason () only counts after both roots have been found.
More mistakes (8)
E(X) for both roots
. Work out E(X) with the valid value only.
Leaving out P(X=0)
from the total: it adds nothing to E(X), but it is part of the 1.
Dividing E(X) by the number of values,
as if it were the mean of the x values: the probabilities already do the averaging.
Translating the words wrongly
"fewer than 3" is X≤2, and "more than 2.5" is X≥3 (Inequality words).
Giving the probability as the mode.
The mode is a value of X.
The ends of the range of E(X) the wrong way round
when E(X) decreases as the parameter increases: 1≤E(X)≤2.2, not 2.2≤E(X)≤1.
Rounding k on
before using it again: keep the stored value, and give answers to 3 s.f.
Carrying on with an invalid k
later parts earn follow-through marks only when your k makes every entry a valid probability (one markscheme needed 0<k<1, and a probability from the table below 1). If an entry comes out negative or above 1, find the slip before going on.