At a large school, students are required to learn at least one language, Spanish or French. It is known that \(75\% \) of the students learn Spanish, and \(40\% \) learn French.

Find the percentage of students who learn both Spanish and French.

At a large school, students are required to learn at least one language, Spanish or French. It is known that \(75\% \) of the students learn Spanish, and \(40\% \) learn French.

Find the percentage of students who learn Spanish, but not French.

At a large school, students are required to learn at least one language, Spanish or French. It is known that \(75\% \) of the students learn Spanish, and \(40\% \) learn French.

At this school, \(52\% \) of the students are girls, and \(85\% \) of the girls learn Spanish.

A student is chosen at random. Let G be the event that the student is a girl, and let S be the event that the student learns Spanish.

(i)     Find \({\rm{P}}(G \cap S)\) .

(ii)    Show that G and S are not independent.

At a large school, students are required to learn at least one language, Spanish or French. It is known that \(75\% \) of the students learn Spanish, and \(40\% \) learn French.

At this school, \(52\% \) of the students are girls, and \(85\% \) of the girls learn Spanish.

A boy is chosen at random. Find the probability that he learns Spanish.

valid approach     (M1)

e.g. Venn diagram with intersection, union formula,

\({\rm{P}}(S \cap F) = 0.75 + 0.40 - 1\)

\(15\) (accept \(15\% \))     A1     N2

[2 marks]

valid approach involving subtraction     (M1)

e.g. Venn diagram, \(75 - 15\)

60 (accept \(60\% \))     A1     N2

[2 marks]

(i) valid approach     (M1)

e.g. tree diagram, multiplying probabilities, \({\rm{P}}(S|G) \times {\rm{P(}}G{\rm{)}}\)

correct calculation     (A1)

e.g. \(0.52 \times 0.85\)

\({\rm{P}}(G \cap S) = 0.442\) (exact)     A1     N3

(ii) valid reasoning, with words, symbols or numbers (seen anywhere)     R1

e.g. \({\rm{P(}}G{\rm{)}} \times {\rm{P}}(S) \ne {\rm{P}}(G \cap S)\) , \({\rm{P}}(S|G) \ne {\rm{P(}}S{\rm{)}}\) , not equal,

one correct value     A1

e.g. \({\rm{P}}(G) \times {\rm{P}}(S) = 0.39\) , \({\rm{P}}(S|G) = 0.85\) , \(0.39 \ne 0.442\)

G and S are not independent     AG     N0

[5 marks]

METHOD 1

\(48\% \) are boys (seen anywhere)     A1

e.g. \({\rm{P}}(B) = 0.48\)

appropriate approach     (M1)

e.g. \({\text{P(girl and Spanish)}} + {\text{P(boy and Spanish)}} = {\text{P(Spanish)}}\)

correct approach to find P(boy and Spanish)     (A1)

e.g. \({\rm{P(}}B \cap S{\rm{) = P(}}S{\rm{)}} - {\rm{P}}(G \cap S)\) , \({\rm{P(}}B \cap S{\rm{) = P(}}S|B) \times {\rm{P}}(B)\) , 0.308

correct substitution     (A1)

e.g. \(0.442 + 0.48x = 0.75\) , \(0.48x = 0.308\)

correct manipulation     (A1)

e.g. \({\rm{P}}(S|B) = \frac{{0.308}}{{0.48}}\)

\({\rm{P}}({\rm{Spanish}}|{\rm{boy}}) = 0.641666 \ldots \) , \(0.641\bar 6\)

\({\rm{P}}({\rm{Spanish}}|{\rm{boy}}) = 0.642\) \([0.641{\text{, }}0.642]\)     A1     N3

[6 marks]

METHOD 2

\(48\% \) are boys (seen anywhere)     A1

e.g. 0.48 used in tree diagram

appropriate approach     (M1)

e.g. tree diagram

correctly labelled branches on tree diagram     (A1)

e.g. first branches are boy/girl, second branches are Spanish/not Spanish

correct substitution     (A1)

e.g. \(0.442 + 0.48x = 0.75\)

correct manipulation     (A1)

e.g. \(0.48x = 0.308\) , \({\rm{P}}(S|B) = \frac{{0.308}}{{0.48}}\)

\({\rm{P}}({\rm{Spanish}}|{\rm{boy}}) = 0.641666 \ldots \) , \(0.641\bar 6\)

\({\rm{P}}({\rm{Spanish}}|{\rm{boy}}) = 0.642\) \([0.641{\text{, }}0.642]\)

[6 marks]

Parts (a) and (b) were generally done well although some candidates left answers as decimals rather than the required percentages.

Parts (a) and (b) were generally done well although some candidates left answers as decimals rather than the required percentages.

In part (c) (i), candidates failed to find the intersection of the events as, in general, they multiplied probabilities, assuming the events were independent or they incorrectly attempted to use the union formula. Independence in (c) (ii) caused difficulty with some candidates attempting to use the conditions for mutually exclusive events while others assumed the events were independent in part (i) and then found \({\rm{P}}(G \cap S)\) by multiplying \({\rm{P}}(S|G) \times {\rm{P(}}G{\rm{)}}\) . 

Part (d) proved quite challenging as a great majority could only find the probability of being a boy. Those who did attempt it, and successfully connected the problem with conditional probability, often had difficulties in reaching the correct final answer.