Edexcel A-level Further Maths coverage

Probability generating functions

Section FS1-7
3 spec leafs

Notes and three levels of exam-style practice for each registered specification leaf in this section.

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FS1-7.1

Definitions, derivations and applications. Use of the probability generating function for the negative binomial, geometric, binomial and Poisson distributions.

  • For a non-negative integer-valued random variable, the probability generating function is GX(t)=E(tX)=xP(X=x)txG_X(t)=E(t^X)=\sum_xP(X=x)t^x.
  • The standard PGFs are (1p+pt)n(1-p+pt)^n for binomial, eλ(t1)e^{\lambda(t-1)} for Poisson, pt1(1p)t\dfrac{pt}{1-(1-p)t} for geometric, and (pt1(1p)t)r\left(\dfrac{pt}{1-(1-p)t}\right)^r for negative binomial.
  • The coefficient of txt^x in GX(t)G_X(t) equals P(X=x)P(X=x), so a PGF encodes the whole distribution.
  • A common error is to omit the factor tt in the geometric PGF; the Edexcel convention starts at trial 11, not at zero failures.

Tier 1 · Easy

3 marks
ORIGINAL

Given XBin(4,0.3)X\sim\operatorname{Bin}(4,0.3), write down GX(t)G_X(t) and use its coefficient of t2t^2 to find P(X=2)P(X=2).

Tier 2 · Standard

5 marks
ORIGINAL

A geometric random variable XX has parameter p=0.4p=0.4 and counts the trial of the first success. Derive its PGF and hence find P(X=5)P(X=5).

Tier 3 · Hard

7 marks
ORIGINAL

The random variable XX counts the trial on which the third success occurs in independent trials with success probability 0.250.25. Derive GX(t)G_X(t) and use it to find P(X=6)P(X=6).

FS1-7.2

Use to find the mean and variance.

  • For a PGF GXG_X, GX(1)=E(X)G'_X(1)=E(X) and GX(1)=E[X(X1)]G''_X(1)=E[X(X-1)].
  • Therefore Var(X)=GX(1)+GX(1)[GX(1)]2\operatorname{Var}(X)=G''_X(1)+G'_X(1)-[G'_X(1)]^2.
  • The check GX(1)=1G_X(1)=1 confirms that the encoded probabilities sum to one before derivatives are used.
  • A common error is to treat GX(1)G''_X(1) as E(X2)E(X^2); the missing GX(1)G'_X(1) must be added.

Tier 1 · Easy

4 marks
ORIGINAL

The PGF of XX is GX(t)=(0.8+0.2t)6G_X(t)=(0.8+0.2t)^6. Use derivatives of the PGF to find E(X)E(X) and Var(X)\operatorname{Var}(X).

Tier 2 · Standard

5 marks
ORIGINAL

A random variable has PGF GX(t)=2t3tG_X(t)=\dfrac{2t}{3-t}. Use the PGF to find its mean and variance.

Tier 3 · Hard

6 marks
ORIGINAL

A random variable has PGF GX(t)=k(1+2t+3t2+4t3)G_X(t)=k(1+2t+3t^2+4t^3). Find kk, then use derivatives to calculate E(X)E(X) and Var(X)\operatorname{Var}(X).

FS1-7.3

Probability generating function of the sum of independent random variables.

  • If XX and YY are independent, then GX+Y(t)=GX(t)GY(t)G_{X+Y}(t)=G_X(t)G_Y(t).
  • Multiply and simplify the PGFs before identifying a familiar distribution or extracting coefficients.
  • For example, multiplying eλ(t1)e^{\lambda(t-1)} and eμ(t1)e^{\mu(t-1)} gives e(λ+μ)(t1)e^{(\lambda+\mu)(t-1)}, the PGF of Po(λ+μ)\operatorname{Po}(\lambda+\mu).
  • A common error is to multiply PGFs without establishing independence; dependence prevents the expectation from factorising.

Tier 1 · Easy

4 marks
ORIGINAL

Independent variables have distributions XPo(1.4)X\sim\operatorname{Po}(1.4) and YPo(2.6)Y\sim\operatorname{Po}(2.6). Use PGFs to identify the distribution of X+YX+Y and find P(X+Y=0)P(X+Y=0).

Tier 2 · Standard

5 marks
ORIGINAL

Independent variables satisfy XBin(5,0.4)X\sim\operatorname{Bin}(5,0.4) and YBin(7,0.4)Y\sim\operatorname{Bin}(7,0.4). Use PGFs to identify S=X+YS=X+Y and calculate P(S=3)P(S=3).

Tier 3 · Hard

7 marks
ORIGINAL

Independent geometric variables XX and YY each have parameter 0.40.4 and count trials to first success. Use PGFs to identify the distribution of S=X+YS=X+Y and find P(S>5)P(S>5).