CAIE A-Level Further Math A2 4.5 Probability Generating Functions QuestionsPractise constructing PGFs, extracting probabilities, differentiating for moments and combining independent variables.Syllabus2028–2030CourseFurther Mathematics 9231LevelA2
Exam pointsconstruct G_X(t)=ΣP(X=x)t^x from a probability table or discrete modelread probabilities from coefficients and use derivatives at t=1 for momentsmultiply PGFs only when the summed random variables are independent
Question 1[Maximum number: 10]Keira has two unbiased coins. She tosses both coins. The number of heads obtained by Keira is denoted by X.Question (a)(a)Find the probability generating function GX(t)\mathrm{G}_{X}(t)GX(t) of X.Hassan has three coins, two of which are biased so that the probability of obtaining a head when the coin is tossed is 13\frac{1}{3}31. The corresponding probability for the third coin is 14\frac{1}{4}41. The number of heads obtained by Hassan when he tosses these three coins is denoted by Y.[ 1 ]Show AnswerGX(t)=14+12t+14t2\mathrm{G}_{X}(t)=\frac{1}{4}+\frac{1}{2} t+\frac{1}{4} t^{2}GX(t)=41+21t+41t2Accept (0.5+0.5t) 2{ }^{2}21Question (b)(b)Find the probability generating function GY(t)\mathrm{G}_{Y}(t)GY(t) of Y.The random variable Z is the total number of heads obtained by Keira and Hassan.[ 3 ]Show AnswerP(0H)=1236P(1H)=1636P(2H)=736P(3H)=136\mathrm{P}(0 \mathrm{H})=\frac{12}{36} \quad \mathrm{P}(1 \mathrm{H})=\frac{16}{36} \quad \mathrm{P}(2 \mathrm{H})=\frac{7}{36} \quad \mathrm{P}(3 \mathrm{H})=\frac{1}{36}P(0H)=3612P(1H)=3616P(2H)=367P(3H)=361M1 A1Attempt at probs, at least 2 correctAll correctGY(t)=1236+1636t+736t2+136t3\mathrm{G}_{Y}(t)=\frac{12}{36}+\frac{16}{36} t+\frac{7}{36} t^{2}+\frac{1}{36} t^{3}GY(t)=3612+3616t+367t2+361t3B1 FTFT their probabilities, must be cubic with 4 non-zero terms3Question (c)(c)Find the probability generating function of Z, expressing your answer as a polynomial.[ 3 ]Show AnswerGZ(t)=(14+12t+14t2)(1236+1636t+736t2+136t3)\mathrm{G}_{Z}(t)=\left(\frac{1}{4}+\frac{1}{2} t+\frac{1}{4} t^{2}\right)\left(\frac{12}{36}+\frac{16}{36} t+\frac{7}{36} t^{2}+\frac{1}{36} t^{3}\right)GZ(t)=(41+21t+41t2)(3612+3616t+367t2+361t3)Attempt to multiply their two PGF=1144(12+40t+51t2+31t3+9t4+t5)=\frac{1}{144}\left(12+40 t+51 t^{2}+31 t^{3}+9 t^{4}+t^{5}\right)=1441(12+40t+51t2+31t3+9t4+t5)Obtain quintic expression and collect terms3Question (d)(d)Use the probability generating function of Z to find E(Z).[ 2 ]Show AnswerGZ′(t)=1144(40+102t+93t2+36t3+5t4)\mathrm{G}_{Z}^{\prime}(t)=\frac{1}{144}\left(40+102 t+93 t^{2}+36 t^{3}+5 t^{4}\right)GZ′(t)=1441(40+102t+93t2+36t3+5t4)DifferentiateE(Z)=GZ′(1)=2312=(=1.92)\mathrm{E}(Z)=\mathrm{G}_{Z}^{\prime}(1)=\frac{23}{12}=(=1.92)E(Z)=GZ′(1)=1223=(=1.92)2Question (e)(e)Use the probability generating function of Z to find the most probable value of Z.[ 1 ]Show Answer2B1 FTFT power of term with largest coefficient in their GZ(t)\mathrm{G}_{Z}(t)GZ(t)1Add to Test
Question (a)(a)Find the probability generating function GX(t)\mathrm{G}_{X}(t)GX(t) of X.Hassan has three coins, two of which are biased so that the probability of obtaining a head when the coin is tossed is 13\frac{1}{3}31. The corresponding probability for the third coin is 14\frac{1}{4}41. The number of heads obtained by Hassan when he tosses these three coins is denoted by Y.[ 1 ]Show AnswerGX(t)=14+12t+14t2\mathrm{G}_{X}(t)=\frac{1}{4}+\frac{1}{2} t+\frac{1}{4} t^{2}GX(t)=41+21t+41t2Accept (0.5+0.5t) 2{ }^{2}21
Question (b)(b)Find the probability generating function GY(t)\mathrm{G}_{Y}(t)GY(t) of Y.The random variable Z is the total number of heads obtained by Keira and Hassan.[ 3 ]Show AnswerP(0H)=1236P(1H)=1636P(2H)=736P(3H)=136\mathrm{P}(0 \mathrm{H})=\frac{12}{36} \quad \mathrm{P}(1 \mathrm{H})=\frac{16}{36} \quad \mathrm{P}(2 \mathrm{H})=\frac{7}{36} \quad \mathrm{P}(3 \mathrm{H})=\frac{1}{36}P(0H)=3612P(1H)=3616P(2H)=367P(3H)=361M1 A1Attempt at probs, at least 2 correctAll correctGY(t)=1236+1636t+736t2+136t3\mathrm{G}_{Y}(t)=\frac{12}{36}+\frac{16}{36} t+\frac{7}{36} t^{2}+\frac{1}{36} t^{3}GY(t)=3612+3616t+367t2+361t3B1 FTFT their probabilities, must be cubic with 4 non-zero terms3
Question (c)(c)Find the probability generating function of Z, expressing your answer as a polynomial.[ 3 ]Show AnswerGZ(t)=(14+12t+14t2)(1236+1636t+736t2+136t3)\mathrm{G}_{Z}(t)=\left(\frac{1}{4}+\frac{1}{2} t+\frac{1}{4} t^{2}\right)\left(\frac{12}{36}+\frac{16}{36} t+\frac{7}{36} t^{2}+\frac{1}{36} t^{3}\right)GZ(t)=(41+21t+41t2)(3612+3616t+367t2+361t3)Attempt to multiply their two PGF=1144(12+40t+51t2+31t3+9t4+t5)=\frac{1}{144}\left(12+40 t+51 t^{2}+31 t^{3}+9 t^{4}+t^{5}\right)=1441(12+40t+51t2+31t3+9t4+t5)Obtain quintic expression and collect terms3
Question (d)(d)Use the probability generating function of Z to find E(Z).[ 2 ]Show AnswerGZ′(t)=1144(40+102t+93t2+36t3+5t4)\mathrm{G}_{Z}^{\prime}(t)=\frac{1}{144}\left(40+102 t+93 t^{2}+36 t^{3}+5 t^{4}\right)GZ′(t)=1441(40+102t+93t2+36t3+5t4)DifferentiateE(Z)=GZ′(1)=2312=(=1.92)\mathrm{E}(Z)=\mathrm{G}_{Z}^{\prime}(1)=\frac{23}{12}=(=1.92)E(Z)=GZ′(1)=1223=(=1.92)2
Question (e)(e)Use the probability generating function of Z to find the most probable value of Z.[ 1 ]Show Answer2B1 FTFT power of term with largest coefficient in their GZ(t)\mathrm{G}_{Z}(t)GZ(t)1