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?
To solve these probability problems, we can use a method called conditional probability.
Let's define some variables:
- F = event that a pupil learns French
- G = event that a pupil learns German
(c) We need to calculate the probability that a randomly chosen pupil in this school who learns German also learns French. This is equivalent to finding P(F|G), the probability that a pupil learns French given that they learn German.
We know that:
P(F) = 55% = 0.55 (given)
P(G|F) = 40% = 0.4 (given)
We're looking for P(F|G), which is equal to P(F ∩ G) / P(G).
Using Bayes' theorem, we can rewrite this as:
P(F|G) = P(G|F) * P(F) / P(G)
P(G) is given as 33% = 0.33.
Substituting the values into the formula, we get:
P(F|G) = 0.4 * 0.55 / 0.33
P(F|G) ≈ 0.67
Therefore, the probability that a randomly chosen pupil who learns German also learns French is approximately 0.67.
(d) We need to calculate the percentage of pupils in this school who study either French or German, or both. This is equivalent to finding P(F ∪ G), the probability that a pupil learns French or German.
We know that:
P(F) = 55% = 0.55 (given)
P(G) = 33% = 0.33 (given)
P(F ∩ G) = P(G|F) * P(F) = 0.4 * 0.55 (given)
To find P(F ∪ G), we can use the formula:
P(F ∪ G) = P(F) + P(G) - P(F ∩ G)
Substituting the values into the formula, we get:
P(F ∪ G) = 0.55 + 0.33 - (0.4 * 0.55)
P(F ∪ G) ≈ 0.68
Therefore, approximately 68% of pupils in this school study either French or German, or both.
To solve these probability problems, we'll use basic probability concepts and information. Let's go step by step:
(c) To calculate the probability that a randomly chosen pupil in this school who learns German also learns French, we need to use conditional probability. We can use the formula for conditional probability:
P(A|B) = P(A ∩ B) / P(B)
Where P(A|B) is the probability of event A given that event B has occurred, P(A ∩ B) is the probability that both events A and B occur at the same time, and P(B) is the probability of event B occurring.
In this case, event A is learning French and event B is learning German. We are given that 40% of pupils who learn French also learn German, so P(A ∩ B) = 0.40.
We are also given that 33% of pupils in this school learn German, so P(B) = 0.33.
Now, we can calculate P(A|B):
P(A|B) = P(A ∩ B) / P(B)
P(A|B) = 0.40 / 0.33
P(A|B) ≈ 0.1212 or 12.12%
Therefore, the probability that a randomly chosen pupil in this school who learns German also learns French is approximately 12.12%.
(d) To calculate the percentage of pupils in this school who study either French or German, or both, we need to use the concepts of union and intersection.
The probability of A or B (P(A ∪ B)), where A and B are two events, can be calculated as:
P(A ∪ B) = P(A) + P(B) - P(A ∩ B)
In this case, event A is learning French and event B is learning German.
We are given that 55% of pupils in this school choose to learn French, so P(A) = 0.55.
We are given that 33% of pupils in this school learn German, so P(B) = 0.33.
We are also given that 40% of pupils who learn French also learn German, so P(A ∩ B) = 0.40.
Now we can calculate P(A ∪ B):
P(A ∪ B) = P(A) + P(B) - P(A ∩ B)
P(A ∪ B) = 0.55 + 0.33 - 0.40
P(A ∪ B) ≈ 0.48 or 48%
Therefore, approximately 48% of the pupils in this school study either French or German, or both.