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?
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%
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
To calculate the probability that a randomly chosen pupil in this school who learns German also learns French, we can use conditional probability.
Let's denote the events:
A = student learns French
B = student learns German
Given information:
P(A) = 0.55 (55% of pupils learn French)
P(B|A) = 0.40 (40% of French learners also learn German)
P(B) = 0.33 (33% of pupils learn German)
We want to find P(A|B), the probability that a student who learns German also learns French.
We can use Bayes' theorem to calculate this probability:
P(A|B) = (P(B|A) * P(A)) / P(B)
Substituting the known values into the formula:
P(A|B) = (0.40 * 0.55) / 0.33
Calculating this:
P(A|B) ≈ 0.664 (or 66.4%)
Therefore, there is approximately a 66.4% chance that a randomly chosen pupil who learns German also learns French.
To calculate the percentage of pupils in this school who study either French or German, or both, we can use the concept of union and intersection.
Let's denote the events:
A = student learns French
B = student learns German
We want to find P(A U B), the probability that a student studies either French or German or both.
P(A U B) = P(A) + P(B) - P(A ∩ B) (since A and B might overlap)
From the given information, we know:
P(A) = 0.55 (55% of pupils learn French)
P(B) = 0.33 (33% of pupils learn German)
P(A ∩ B) = P(A) * P(B|A) = 0.55 * 0.40
Substituting these values into the formula:
P(A U B) = 0.55 + 0.33 - (0.55 * 0.40)
Calculating this:
P(A U B) ≈ 0.748 (or 74.8%)
Therefore, approximately 74.8% of pupils in this school study either French or German, or both.