Question

A large school in the UK offers its pupils the opportunity to learn several
non-English languages. It turns out that:
• 55% of pupils in this school choose to learn French;
• of the pupils who learn French, 40% also learn German.

(c) Additional information is now given that 33% of pupils in this school
learn German. Calculate the probability that a randomly chosen pupil
in this school who learns German also learns French.
(d) What percentage of pupils in this school study either French or
German, or both?

Answer 1

From the initial information:

Of the 55% that learn French, we have a crossover
of 40% of that 55% also learn German.

.55(.4) = .220 = 22%

Thus, of the French learning students we
have 33% learn only French, 22% learn French and German.

(c) we know 33% of the students learn German.
We already know that 22% learn French and German.
This means that 11% of the students learn only German.

Thus we have twice as many students learning German plus
French as we have learning only German.

The probability that a randomly chosen student who learns
German also learns French is (G+F):G = 22:11 = 2:1

d) The total percentage of students that learn either
French or German or both is:

F + (F+G) + G = 33% + 22% + 11% = 66%

Answer 2

Draw a Venn diagram.
Pr(F)=.55
Pr(G)=.33
Pr(G|F)=.4*.55

c. Pr(F|G) is what you wish, so
Bayes Theorem says that.

P(F|G)= (P(G|F)P(F))/(Pr(G|F)Pr(F)+Pr(G|notF)(Pr(notF))
= .4*.55 /(.4*.55+.67*.45)
check those numbers