In the Venn diagram, 𝜉 represents 29 students in a class, 𝑆 is students who have at least one sister, and 𝐵 is students who have at least one brother. Part a) One student from the class is picked at random. Work out the probability that the student has a sister.
Let’s look at this Venn diagram more closely. This Greek letter 𝜉 represents the universal set. That’s the set of all elements — in this case students — who are included on the Venn diagram. We know that 𝜉 represents 29 students in the class. So this is the total.
𝑆 and 𝐵 are each subsets of 𝜉, representing those students who have at least one sister and those students who have at least one brother. In the center of the Venn diagram, we have the overlap or intersection of the sets 𝑆 and 𝐵 in which there are four students. This tells us that there are four students in the class who have at least one sister and at least one brother.
On the outside of the Venn diagram, we have this number eight which is included within the universal set but it’s outside of each of the sets 𝑆 and 𝐵. So this eight tells us that there are eight students in the class who have neither sisters nor brothers.
Let’s look at part a more closely. We’re told that one student from the class is picked at random and asked to work out the probability that this student has a sister. As this student is picked at random, this means that every student in the class has an equal chance of being selected. So to work out the probability that this student has a sister, we just need to divide the number of students in the class who have a sister by the total number of students.
The total number of students is easy. We know that this is 29. But what about the number who have a sister? Well, from our Venn diagram, that’s everybody who’s included inside the circle for 𝑆. So that’s 𝑥 plus four for those who only have a sister, but not a brother and also four for those who have both sisters and brothers.
By adding these together, we get an expression for the number of people who have a sister. It’s 𝑥 plus four plus four which simplifies to 𝑥 plus eight. But in order to work out our probability, we’re going to need to know the value of 𝑥.
To work 𝑥 out, we can use the fact that the total number of students in the class is 29. So all of the numbers on the inside of the Venn diagram need to sum to 29. We can, therefore, form an equation: 𝑥 plus four plus four plus 𝑥 plus one plus eight is equal to 29.
We can then simplify this equation by collecting like terms. 𝑥 plus 𝑥 gives two 𝑥. And the sum of four, four, one, and eight is 17. So our equation becomes two 𝑥 plus 17 equals 29. We now need to solve this equation for 𝑥.
First, we subtract 17 from each side giving two 𝑥 is equal to 12. And then, we divide by two giving 𝑥 equals six. We can then return to our Venn diagram and fill in the values of 𝑥 plus four which is 10 and 𝑥 plus one which is seven.
A quick check confirms that the sum of all of the values inside our Venn diagram — so that’s 10, four, seven, and eight — is indeed equal to the total of 29. Returning then to the probability that this student has a sister, the number who had a sister was represented by 𝑥 plus eight which we can now work out to be 14 because this is six plus eight. The denominator of our fraction is 29.
So we found that the probability that this student has a sister is equal to 14 over 29. And we’ll leave our answer as a fraction.
Part b) of the question says, “The student who has been picked at random has a brother. Work out the probability that this student has a sister.”
In this part of the question, we’re actually being asked to find a conditional probability. We know that the student chosen has a brother. And we’re asked to find the probability that they also have a sister. This is a conditional probability: the probability that they have a sister given that they have a brother.
As we know the student has a brother, we’re no longer looking at all 29 students in the full Venn diagram. We’re just looking at the circle of those students who have a brother. That’s the circle that I’ve highlighted in pink. The total number of students in this circle is the sum of the numbers inside it. So that’s four plus seven which is 11. And so, this is the denominator of our fraction.
The numerator for our fraction is the number of students who have at least one sister as well as at least one brother. That’s those students who are in the intersection of the two circles which is four.
Our conditional probability then that this student has a sister given that they also have a brother is four elevenths.