Three family-related probability questions that will make you scratch your head
If you have ever applied for, or are considering applying for, a role in a quantitative field like quantitative finance, research engineering, or data science, you have likely encountered probability questions in your preparation. Questions on basic combinatorics, probability, and statistics are favorites in quant interviews. This is because it is relatively easy to create challenging and sometimes counterintuitive problems using simple scenarios. While the math behind basic probability is straightforward, solving these problems can be difficult. It requires careful and systematic analysis, the ability to translate problem statements into mathematical expressions, and a solid understanding of quantitative methods. These are skills that are needed in a quantitative jobs, and can be probed by asking probability questions.
Let’s dive into three common questions asked during such interviews (with solutions!). They all revolve around calculating the probability of having children with a certain set of characteristics.
Question 1 — A family has 6 children, 3 girls and 3 boys. What is the probability that the 3 eldest children are all boys? You can assume that all birth orders are equally likely.
Solution — This is first and foremost a combinatorics question, rather than a probability question. We can get the solution by using the naive definition of probability:
P(X) = N_X/N_S
where N_X is the number of possibilities for event X, and N_S is the total number of possibilities in the total event space. In this case, X is the event that all the three eldest children are boys, and encompasses multiple possibilities. Thus, N_X is the number of birth orders compatible with X, whereas N_S is the total number of all possible birth orders, i.e. all possible permutations of six elements.
This definition is naive in the sense that it does not encompass all the axioms of probability. For instance, it only applies to finite event spaces and assumes that each outcome for X is equally likely. But for problems that involve counting, such as enumerating sequences like in our case, it works just fine.
For concreteness, let’s give names to the 6 children to identify them uniquely: Amy, Bob, Claire, David, Elle, Fred. For short, we will identify them by their initials: A, B, C, D, E, F. Thus, to solve this problem, we need to enumerate all the birth orders that end with any permutation of B, D, F (it is implied that the enumeration is given from youngest to eldest, i.e. in this case Bob is born before David who is born before Fred). These are some realizations of the event X:
- A, C, E, B, D, F.
- C, A, E, B, F, D.
- A, E, C, D, F, B.
- …
These are birth orders that are not compatible with event X:
- A, D, E, B, A, F.
- B, F, D, C, A, E.
- B, E, C, D, F, A.
- …
How many combinations are compatible with event X? Well, we know that B, D, F have to be at the end of the sequence, but we are free to permute them amongst themselves because these children are all boys. The number of permutations of n elements is given by the factorial n!. Thus, we have 3! such permutations. Likewise, the remaining children are all girls and can also be permuted amongst themselves to occur in any arbitrary order. So, for each ordering of the boys, we have 3! possible ordering for the girls. In total, the number of birth order permutations compatible with X is therefore N_X = (3!)². On the other hand, the total number of permutations of all the children is 6! as we have 6 elements, which means N_S = 6!. Putting, these two results together, we arrive at the solution:
P(X) = (3!)²/6! = 0.05
The probability that the 3 eldest children are all boys is 5%.
Question 2— Mary is expecting twin boys. Twins can be either identical or fraternal. Identical twins are of the same sex, while fraternal twins might or might not be. We can assume that identical twins are equally likely to be both boys or both girls, while for fraternal twins all the possible combinations are equally likely. Furthermore, we know that 1/3 of all twins are identical. What is the probability that Mary’s twins are identical?
Solution — Whenever we are given additional pieces of information about some event — in this case that Mary is pregnant and that the children are twin boys — we should instinctively think about conditional probability and its cardinal theorem: Bayes’ rule. Bayes’ rule can be state in many ways. One of these is that the probability of an event A occurring, given that another event B has occurred (i.e. conditional on B, written A|B) can be related to the probability of B occurring, given that A has occurred (written B|A). To be more precise:
P(A|B) = P(B|A)*P(A)/P(B)
This is extremely useful, because sometimes we want to obtain information about the event A|B, but we only have direct knowledge of B|A instead. For this problem, let us identify the events of interest:
- A: Mary is expecting identical twins.
- B: Mary is expecting twin boys.
We want to determine P(A|B), the probability that Mary is expecting identical twins given the information that she is pregnant with twin boys. However, we have access to the inverse event P(B|A), the probability that Mary is pregnant with twin boys given that they are identical. This is P(B|A)=1/2, since identical twins can only be both girls or both boys. Moreover, we know P(A)=1/3 from the problem statement, and we can calculate P(B) by decomposing it with the law of total probability (I wrote about this in previous posts):
P(B) = P(B|A)*P(A) + P(B|A^c)*P(A^c)=1/2*1/3 + 1/4*2/3=1/3
where P(B|A^c) is the probability that Mary is expecting twin boys given that they are fraternal (there are four total possibilities for fraternal twins) and P(A^c)=1-P(A)=2/3. We thus have all the information to plug into Bayes’ rule to get the solution:
P(A|B) = (1/2*1/3)/(1/3)=1/2.
The probability that Mary is expecting identical twins is 50%.
Question 3— A family has two children. What is the probability that both children are boys if we know that the eldest is a boy? What is the probability that both are boys if at least one is a boy? What is the probability that both are boys if we know that at least one is a boy born on a Tuesday? You can assume that being born of a given gender and on a given day of the week are independent of each other.
Solution — This multiple question is an old-time favorite for several reasons. The first is the distinction between knowing something about a specific child vs. knowing some fuzzier information about the children in general (the at least one boy part). The second is that at first sight it might seem irrelevant whether we know on what day of the week the children are born. And yet, the result does change if we consider a boy born on a Tuesday, because it provides more information to condition on, and it thus restricts the overall possibilities. This is a very nice exercise in understanding the consequence of conditioning on more and more information.
Let’s start with the simplest case. What is the probability that both children are boys if we know that the eldest is a boy? Since we know that the eldest is a boy already, we only have two possibilities for the youngest child, so the probability is 50%. Let’s see how this reasoning translates into Bayes’ rule. We first identify the events we need to codify:
- A: Both children are boys.
- B: The eldest child is a boy.
We can calculate P(A) and P(B) directly. There are four total possibilities for the gender of the children (we are assuming gender is binary for simplicity), so P(A)=1/4. If the eldest child is a boy, that reduces the possibilities for the youngest child to two (boy or girl), so P(B)=1/2. What we are required to find is P(A|B), but it’s easier to determine that from P(B|A) with Bayes’ rule, since B|A is the event that at least one child is a boy given that both are boys. Since A implies B, P(B|A)=1. Therefore, we get:
P(A|B) = P(B|A)*P(A)/P(B) = (1*1/4)/(1/2) = 1/2.
Exactly 50% as we anticipated.
Now let’s tackle the next question: what is the probability that both children are boys if we know that at least one is a boy? Let’s first modify the event B:
- B: At least one child is a boy.
To calculate the probability for this modified event B, there is a trick that is often used in probability when we encounter the phrase “at least one”. “At least one X” means the same as “all except zero X”. For example if I have “at least one car” I can have all numbers of cars except zero. In probability language, P(at least one X)=1-P(no X). Thus, the probability that at least one child is a boy is one minus the probability that none are boys (i.e. both girls), which is P(B)=1–1/4=3/4. Plugging this updated result in Bayes’ rule we obtain (note that P(B|A) is still 1):
P(A|B) = P(B|A)*P(A)/P(B) = (1*1/4)/(3/4) = 1/3.
The probability has decreased to 33%! At first, it might seem counterintuitive that this result is different than the previous one. However, if we think carefully about the problem statement, saying that the eldest child is a boy identifies a specific child and thus excludes two out of the four possibilities (boy/boy, boy/girl, girl/boy, girl/girl), whereas saying that at least one child is a boy gives less information as it only excludes the girl/girl case.
Now, we are in good shape to tackle the more complicated question: what is the probability that both children are boys if we know that at least one is a boy born on a Tuesday? Let’s first redefine our event B again:
- B: At least one child is a boy born on a Tuesday.
The probability of B has to be modified accordingly. Since there are seven days of the week and two children in total, and that the probability of being born on a specific day of the week is independent of gender, the probability that a specific child (boy) is born on a Tuesday is 1/2*1/7 =1/14. Using the same “at least” trick from above, we can calculate the probability of B as
P(B) = 1-P(B^c)=1-P(no boy is born on a Tuesday)=1–(13/14)²=27/196
Now, we cannot calculate the probability P(B|A) in an easy way anymore, because knowing that the family has two boys does not immediately give information about the day of the week on which one of them is born. Instead, we can use the definition of conditional probability directly:
P(A|B)=P(A∩B)/P(B),
where the event A∩B means the intersection of A and B, i.e. A and B happening simultaneously. Now, the event “both children are boys and at least one boy is born on a Tuesday” is the same as saying “both children are boys and at least one child is born on a Tuesday”. Let’s call “at least one child is born on a Tuesday” the event B’. Since we assume that gender and day of the week are independent variables we write — reusing the “at least” trick:
P(A∩B)=P(A∩B’)=P(A)P(B’)=1/4*(1-P(B’^c))=1/4*(1–(6/7)²)=13/196
Putting this together with P(B), we finally obtain:
P(A|B)=(13/196)/(27/196)=13/27.
The probability that both are boys given that at least one is a boy born on a Tuesday is around 48%.
It seems absurd that knowing the birth day of (at least) one of the children changes the result so drastically! What is happening? The point is that the knowing the birth day gives more information than just knowing that “at least one is a boy” and pushes this statement closer to fully specifying which child is a boy (like in the eldest case), and thus brings the probability closer to the 50% case if we had full information on one child.
I hope you enjoyed these brain teasers about probability! Stay tune for more probability and statistics to come!
