Mastering Combinatorics Interview Questions: The Ultimate Guide for Acing Tech Interviews

We cant lie -Â Data Science Interviews are TOUGH. Top tech companies ask very hard questions about probability and statistics.

That’s why we put together 40 real chances The answers to all 40 questions are in our book, Ace The Data Science Interview. It also has answers to 161 other questions about SQL, Machine Learning, and Product/Business Sense. You can also practice some of these same exact questions on DataLemurs statistics interview questions section.

Solutions To Probability Interview Questions

Problem #1 Solution:

We can use Bayes Theorem here. Let’s call the situation where we flip an unfair coin U and the situation where we flip a fair coin F. Since the coin is chosen randomly, we know that P(U) = P(F) = 0. 5. Let 5T denote the event where we flip 5 heads in a row. Then we are interested in solving for P(U|5T), i. e. , the chance that we are flipping an unfair coin, since we have seen five tails in a row

We know P(5T|U) = 1 since by definition the unfair coin will always result in tails. Additionally, we know that P(5T|F) = 1/2^5 = 1/32 by definition of a fair coin. By Bayes Theorem we have:

[P(U|5T) = frac{P(5T|U) * P(U)}{P(5T|U) * P(U) + P(5T|F) * P(F)} = frac{0. 5}{0. 5 + 0. 5 * 1/32} = 0. 97].

Therefore the probability we picked the unfair coin is about 97%.

Problem #5 Solution:

By definition, a chord is a line segment whereby the two endpoints lie on the circle. Therefore, two arbitrary chords can always be represented by any four points chosen on the circle. If you choose to represent the first chord by two of the four points then you have:

[{4choose2} = 6 ]

pick the two points that will show chord 1 (and the other two points that will show chord 2). However, keep in mind that we are counting each chord twice because a chord with endpoints p1 and p2 is the same as a chord with endpoints p2 and p1. Therefore the proper number of valid chords is:

Among these three configurations, only exactly one of the chords will intersect, hence the desired probability is:

Problem #13 Solution:

Let X be the number of coin flips needed until two heads. Then we want to solve for E[X]. Let H denote a flip that resulted in heads, and T denote a flip that resulted in tails. Note that E[X] can be written in terms of E[X|H] and E[X|T], i. e. the expected number of flips needed, conditioned on a flip being either heads or tails respectively.

Conditioning on the first flip, we have:

[E[X] = frac{1}{2}(1+E[X|H]) + frac{1}{2}(1+E[X|T])]

Keep in mind that E[X|T] = E[X] because we have to start over to get two heads in a row if a tail is flipped.

To find E[X|H], we can make it depend on the next result, which could be heads (HH) or tails (HT).

Therefore, we have:

[E[X|H] = frac{1}{2}(1+E[X|HH]) + frac{1}{2}(1+E[X|HT])]

If the outcome is HH, then E[X|HH] = 0 because the goal was met. If the outcome is HT, then E[X|HT] = E[X] because a tail was flipped, we need to start over.

[E[X|H] = frac{1}{2}(1+0) + frac{1}{2}(1+E[X]) = 1 + frac{1}{2}E[X]]

Plugging this into the original equation yields E[X] = 6 coin flips

Problem #15 Solution:

Consider the first n coins that A flips, versus the n coins that B flips.

There are three possible scenarios:

• A has more heads than B
• A and B have an equal amount of heads
• A has less heads than B

In case 1, A will always win (no matter what coin comes up), and in case 3, A will always lose (no matter what coin comes up). By symmetry, these two scenarios have an equal probability of occurring.

Denote the probability of either scenario as x, and the probability of scenario 2 as y.

We know that 2x + y = 1 since these 3 scenarios are the only possible outcomes. Now let’s consider coin n+1. If the flip results in heads, with probability 0. 5, then A will have won after scenario 2 (which happens with probability y). Therefore, A’s total chances of winning the game are increased by 0. 5y.

Thus, the probability that A will win the game is:

[x + frac{1}{2}y = x + frac{1}{2}(1-2x) = frac{1}{2}]

Problem #18 Solution:

Let B be the event that all n rolls have a value less than or equal to r. Then we have:

since all n rolls must have a value less than or equal to r. Let A be the event that the largest number is r. We have:

[B_r = B_{r-1} cup A_r]

and since the two events on the right hand side are disjoint, we have:

[P(B_r) = P(B_{r-1}) + P(A_r)]

Therefore, the probability of A is given by:

[P(A_r) = P(B_{r}) – P(B_{r-1}) = frac{r^n}{6^n} – frac{(r-1)^n}{6^n}]

20 Probability Interview Problems Asked By Top-Tech Companies & Wall Street

• One side of a fair coin has heads and the other side has tails. There is also an unfair coin with both sides having tails. Suppose you pick one at random and flip it five times. All five times, it comes up tails. How likely is it that you are flipping an unfair coin?
• [Easy Land] You and your friend are having fun. You and your partner will keep flipping a coin until somebody gets HH or TH. If HH shows up first, you win. If TH shows up first, your friend wins. What is the probability of you winning?.
• [Google – Easy] What is the chance that a seven-game series will last seven games?
• Facebook makes it easy to tell if content is spam or not by having a content team that checks it. Ninety percent of them are careful raters and will mark twenty percent of the content as spam and eighty percent as not spam. However, the people who are still there are lazy raters who will label 20% of the content as spam and the other 20% as non-spam. Let’s say that each piece of content is labeled separately, and that each rater If a rater has given four pieces of content the “good” label, how likely is it that they are a careful rater?
• Let’s say you draw a circle and pick two chords at random. What is the probability that those chords will intersect?.
• There is a test that can tell you for sure if you have a certain disease (Amazon Easy). There is a 1% error rate if you don’t have the disease. How likely is it that someone has the disease if they test positive?
• That’s it! There are 50 cards of 5 different colors. Each color has cards numbered between 1 to 10. You pick 2 cards at random. Which of these is more likely: that they are not the same color or number?
• [Tesla – Easy] A fair six-sided die is rolled twice. What is the chance of rolling a 1 on the first roll and not a 6 on the second roll?
• [Facebook – Easy] How many rolls do you think it will take to see all six sides of a fair die?
• Three friends in Seattle all told you it was raining, but there’s a 1/3 chance that each of them was lying. How likely is it that it will rain in Seattle? Let’s say the chance of rain in Seattle on any given day is 0. 25.
• Let’s say you roll three dice one at a time. How likely is it that you’ll get three numbers in a row that rises in value?
• “Bloomberg Medium”: Three ants are sitting on the edges of a triangle with equal sides. No matter what direction it is, each ant starts moving along the triangle’s edge. How likely is it that none of the ants will hit each other? What if there are k ants on each of the k corners of a polygon with equal sides?
• [Two Sigma – Medium] How many flips of the coin do you think it will take to get two heads in a row?
• What number of cards from a standard deck do you think you’ll draw before you see the first ace?
• […] A and B are playing a game where A has n 1 coins and B has n coins. They both flip all of their coins. How likely is it that A will have more heads than B?
• Let’s say you’re given an unfair coin that seems to lean more toward heads or tails. How can you generate fair odds using this coin?.
• [Quora – Medium] Say you have N i. i. d. draws of a normal distribution with parameters μ and σ. How likely is it that k of those draws will be bigger than a certain value Y?
• [Spotify – Hard] A fair die is rolled n times. For each r in 1, what is the chance that the biggest number rolled is r? 6?.
• [Hard] There are n Snapchat users, and they are split into two groups, A and B. Each user in A is friends with users in B, and so on. Each person in A will pick a random person in B as their best friend, and each person in B will pick a random person in A as their best friend. People who have chosen each other are best friends with each other. How likely is it that there won’t be any best friends who are different?
• [Tesla – Hard] Let’s say a new car is about to come out. Initial data suggests that every day there is either a problem with some part of the vehicle or a chance of a crash with probability p, which means that the part needs to be replaced. Also, every vehicle that has been there for n days needs to be replaced. What is the long-term frequency of vehicle replacements?.

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FAQ

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Why is probability and combinatorics important?

It highlights that understanding probability and combinatorics is essential for data scientists and analysts. In your daily life, you’ll encounter situations where a grasp of probability is necessary; otherwise, you might draw incorrect conclusions.

Why should you study combinatorics?

Gain insights into commonly asked questions and learn how to articulate effective responses. Combinatorics, a fascinating branch of mathematics, is primarily concerned with the study of discrete and combinatorial objects.

How do combinatorics and number theory intersect?

Combinatorics and number theory intersect in various ways. One key intersection is the use of combinatorial methods to solve problems in number theory, such as partition function or counting the number of ways an integer can be expressed as a sum of other integers.

What can I learn in combinatorics?

Within combinatorics, you can learn about wide-ranging topics such as the following: Brilliant’s article collection is expanding to cover every topic you might encounter in any pre-collegiate math program.