Top Probabilty Interview Questions [ Updated 2025 ]

This article presents 15 practical and thoughtfully designed case studies that simulate scenarios typically encountered in interviews. Each case is based on real-world challenges across logistics, e-commerce, and digital payment systems. These problems are structured to test your problem-solving skills, logical reasoning, and ability to handle probabilistic models under different conditions.

Q1. Suppose you have ten fair dice. If you randomly throw them simultaneously, what is the probability that the sum of all of the top faces is divisible by six?

Ans. With 10 dices, the possible sums divisible by 6 are 12, 18, 24, 30, 36, 42, 48, 54, and 60.

You don't actually need to calculate the probability of getting each of these numbers as the final sums from 10 dices because no matter what the sum of the first 9 numbers is, you can still choose a number between 1 to 6 on the last die and add to that previous sum to make the final sum divisible by 6. Therefore, we only care about the last die. And the probability to get that number on the last die is 1/6.

So the answer is 1/6

Q.2 If you have three draws from a uniformly distributed random variable between 0 and 2, what is the probability that the median of three numbers is greater than 1.5?

Ans. To get a median greater than 1.5 at least two of the three numbers must be greater than 1.5. The probability of one number being greater than 1.5 in this distribution is 0.25. Then, using the binomial distribution with three trials and a success probability of 0.25 we compute the probability of 2 or more successes to get the probability of the median is more than 1.5, which would be about 15.6%.

Q3. Assume you have a deck of 100 cards with values ranging from 1 to 100 and you draw two cards randomly without replacement, what is the probability that the number of one of them is double the other?

Ans. There are a total of ways to choose two cards at random from the 100 cards and there are only 50 pairs of these 4950 ways that you will get one number and it's double.

Therefore the probability that the number of one of them is double the other is 50/4950.

Q4. Assume two coins, one fair and the other is unfair. You pick one at random, flip it five times, and observe that it comes up as tails all five times. What is the probability that you are fliping the unfair coin? Assume that the unfair coin always results in tails.

Ans. Let's use Baye’s theorem let U denote the case where you are flipping the unfair coin and F denote the case where you are flipping the fair coin. Since the coin is chosen randomly, we know that .

Let 5T denote the event of flipping 5 tails in a row.

Then, we are interested in solving for (the probability that you are flipping the unfair coin given that you obtained 5 tails). Since the unfair coin always results in tails,

therefore and also by the definition of a fair coin.

Lets apply Bayes theorem where

Therefore the probability that you picked the unfair coin is 97%

Q5. Assume you take a stick of length 1 and you break it uniformly at random into three parts. What is the probability that the three pieces can be used to form a triangle?

Ans. Let's say, x and y are the lengths of the two parts, so the length of the third part will be 1-x-y

As per the triangle inequality theorem, the sum of two sides should always be greater than the third side. Therefore, no two lengths can be more than

To achieve this the first breaking point (X) should before the 0.5 mark on the stick and the second breaking point (Y) should be after the 0.5 mark on the stick.

Hence, overall probability =

Q6. Explain the Difference Between Probability and Likelihood?

Ans. Probability and likelihood are two important concepts in statistics and data analysis, but they mean different things and are used in different ways.

Probability measures how likely an event is to happen. It is a value between 0 and 1, where 0 means the event cannot happen, and 1 means it is certain to happen. For example, the probability of getting heads when flipping a fair coin is 0.5.

Likelihood, on the other hand, measures how well a given statistical model or hypothesis explains some observed data. It’s not the same as probability; instead, it shows how plausible the data is based on the model or hypothesis. For instance, if we hypothesise that the average height of a population is 6 feet, and a sample shows an average height of 5 feet, the likelihood of that hypothesis being correct is low.

Q7. Explain the difference between frequentist and Bayesian probability approaches?

The frequentist and Bayesian approaches to probability look at probabilities in different ways:

• Frequentist approach: Probability is seen as how often an event happens in the long run if you repeat an experiment many times. For example, if you flip a fair coin many times, the probability of getting heads is 50%. In this approach, probabilities are fixed and based on the data you collect. Unknown values in a model (like the average height of a population) are treated as fixed numbers that need to be estimated using techniques like maximum likelihood estimation.
  
• Bayesian approach: Probability is seen as a way to express how confident you are about something happening. For example, if you believe there’s a 70% chance it will rain tomorrow, that’s your confidence based on what you know. In this approach, probabilities are personal and can change as you get more information. It starts with a belief (called a prior), and as new data comes in, this belief is updated to form a new belief (called a posterior).

Q8. If there are 30 people in a room, what is the probability that everyone has different birthdays?

Ans. To find the probability that all 30 people in a room have different birthdays:

The first person has 365 choices, the second has 364 (to avoid the first person’s birthday), the third has 363, and so on, down to 336 choices for the 30th person.

• The probability is:

Calculating this gives:

So, there’s about a 29.37% chance that everyone has different birthdays

Q9. A group of 60 students is randomly split into 3 classes of equal size. All partitions are equally likely. Jack and Jill are two students belonging to that group. What is the probability that Jack and Jill will end up in the same class? 

Ans. We have 60 students, each assigned a number from 1 to 60. The numbers are divided into three groups:

• Group 1: Numbers 1 to 20
• Group 2: Numbers 21 to 40
• Group 3: Numbers 41 to 60

Now, we randomly assign numbers to students. To find the probability that Jack and Jill end up in the same group, we follow these steps:

• Assign a random number to Jack.
• Jack is assigned a number between 1 and 60, and we need to find the probability that Jill ends up in the same group.
• If Jack is in Group 1 (numbers 1–20), there are 19 other numbers in Group 1.
• Jill can be assigned one of 59 remaining numbers, and 19 of those will be in the same group as Jack.

Thus, the probability that Jill ends up in the same group as Jack is:

Q10. Suppose a life insurance company sells a $240,000 one year term life insurance policy to a 25-year old female for $210. The probability that the female survives the year is .999592. Find the expected value of this policy for the insurance company.

Ans.

• P(company loses the money ) = 0.99592
• P(company does not lose the money ) = 0.000408

The amount of money company loses if it loses = 240,000 – 210 = 239790

While the money it gains is $210

• Expected money the company will have to give = 239790*0.000408 = 97.8
• Expect money company gets = 210.

Therefore the value = 210 – 98 = $112

Case Studies Related to Probability

Q1. Delivery Partner Availability

Case Study: In a delivery system, there are 10 delivery partners available in a specific zone. Each delivery partner has a 70% chance of being available at any given time. What is the probability that at least one delivery partner is available to take an order?

Ans. The key to solving this is to understand the complement rule. Instead of calculating the probability of "at least one being available" directly, we first calculate the probability of none being available and then subtract that from 1.

• Step 1 - Define unavailability of a single partner: Probability that a single partner is unavailable = 1−0.7=0.3
• Step 2 - Probability all partners are unavailable: If there are 10 independent partners, the probability all are unavailable is: P(all unavailable)=0.310.

Using a calculator:

• Step 3 - At least one available: Using the complement rule: P(at least one available)=1−P(all unavailable).

Substituting values:

The probability that at least one delivery partner is available is approximately 99.94%.

2. Order Delivery Time

Case Study: A food delivery platform guarantees delivery within 30 minutes for most orders. The probability of an order being delivered within 30 minutes is 80%. If 5 orders are placed simultaneously, what is the probability that exactly 3 orders are delivered on time?

Ans: To solve this case study, we can use the binomial distribution formula,

Where:

• n is the number of trials (orders),
• k is the number of successes (orders delivered on time),
• p is the probability of success (order delivered on time),
• (n/k​) is the binomial coefficient, which is calculated as:

In this case:

• n=5 (since 5 orders are placed),
• k=3 (we want exactly 3 orders to be delivered on time),
• p=0.80 (probability of an order being delivered on time).

Step 1: Calculate the binomial coefficient

Step 3: Multiply all the parts together:

P(X=3)=10×0.512×0.04=0.2048

Thus, The probability that exactly 3 orders are delivered on time is 0.2048 or 20.48%.

3. Customer Retention Prediction

Case Study: A service finds that 70% of their customers make repeat purchases. If they randomly select 10 customers who made a purchase in the past week, what is the probability that exactly 7 will make a repeat purchase?

Ans: This case study also follows a binomial distribution because it satisfies the key characteristics of a binomial experiment. The binomial distribution models the number of successes (repeat purchases) in a fixed number of independent trials (10 customers), each with the same probability of success (p=0.7). Since the problem seeks the probability of exactly 7 successes, it aligns perfectly with the binomial probability distribution formula:

Where:

• n=10 (the number of customers selected),
• k=7 (the number of customers making a repeat purchase),
• p=0.7 (the probability of a customer making a repeat purchase).
  

Step 1: Calculate the binomial coefficient

Step 3: Multiply all the parts together:

P(X=7)=120×0.0823543×0.027=0.267

Thus, The probability that exactly 7 out of 10 customers will make a repeat purchase is approximately 0.267 or 26.7%.

4. Stock Demand Forecasting

Case Study: A warehouse stocks a product, and the demand for this product follows a Poisson distribution with a mean of 25 units per day. What is the probability that they will receive exactly 20 orders for the product tomorrow?

Ans: To solve this case study, we will use the Poisson distribution. We use a Poisson experiment because it models the probability of a certain number of events (orders) happening in a fixed time period (one day) when the events occur independently and at a constant average rate (mean demand of 25 units per day). This makes it ideal for predicting discrete counts like product orders.

The formula for the Poisson distribution is:

Where:

• P(X=k) is the probability of receiving exactly k orders,
• λ is the mean number of orders (25 units per day),
• k is the number of orders we are interested in (20 orders),
• e is Euler's number (approximately 2.71828).

Step 1: Set up the given values

• λ=25 (mean number of orders per day),
• k=20k = 20k=20 (we are looking for the probability of exactly 20 orders).

Step 2: Apply the Poisson formula

Step 3: Compute the value

Result - approximately 0.0519 or 5.19%

5. Optimal Delivery Resource Allocation

Case Study: A delivery service wants to ensure that enough delivery partners are available during peak hours. If the probability of a delivery partner being available is 0.75, and there are 8 delivery partners, what is the probability that at least 6 partners will be available during peak hours?

Ans: To solve this problem, we can use the binomial distribution because it models the probability of a certain number of successes (partners being available) in a fixed number of independent trials (8 partners), where each trial has the same probability of success (0.75). This makes it ideal for scenarios with a fixed number of repeated events and a constant success probability.

Where:

• n=8 (the number of delivery partners),
• k is the number of partners we want to be available,
• p=0.75 (the probability that a partner is available),
• (n/k) is the binomial coefficient.

The problem asks for the probability that at least 6 partners will be available, meaning we need to calculate the probability of having 6, 7, or 8 partners available. To do this, we'll sum the probabilities for k=6, k=7, and k=8.

Calculate probabilities for k=6, k=7, and k=8

We will calculate the individual probabilities using the binomial formula and then sum them to get the total probability.

• Calculate the probability for exactly 6 partners available:

• Calculate the probability for exactly 7 partners available:

• Calculate the probability for exactly 8 partners available:

Finally, sum the probabilities

=0.6785 = 67.85%

After performing the calculations, the total probability is approximately 0.6785 or 67.85%

6. Customer Rating Distribution

Case Study: A restaurant or service provider has customer ratings that follow a discrete uniform distribution. If 5 customers rate a service, what is the probability that exactly 3 will give a rating of 4 stars or higher?

Ans. To solve this case study, we need to calculate the probability that exactly 3 out of 5 customers give a rating of 4 stars or higher. We use a binomial distribution because it models the probability of a certain number of successes (customers giving a rating of 4 stars or higher) in a fixed number of independent trials (5 customers), where each trial has the same probability of success. This makes it suitable for calculating the likelihood of specific outcomes in repeated events like customer ratings.

Step 1: Understand the distribution

The ratings follow a discrete uniform distribution, meaning each rating (e.g., 1, 2, 3, 4, 5 stars) has an equal probability of being chosen. For a 5-star rating system:

Thus, the probability of a rating being 4 stars or higher is 2/5​, and the probability of a rating being lower than 4 stars is 3/5​.

Step 2: Use the binomial probability formula

The problem involves a binomial distribution where:

• n=5: The number of customers
• k=3: The number of customers giving 4 stars or higher
• p=2/5​: Probability of success (rating 4 or higher)

The binomial probability formula is:

Step 3: Calculate the probability

Substitute the values:

4. Combining the terms

The probability that exactly 3 out of 5 customers will give a rating of 4 stars or higher is approximately 0.2304 or 23.04%.

7. Food Quality Control

Case Study: A service finds that 90% of their orders pass the quality control checks. If 12 orders are checked, what is the probability that exactly 10 orders pass the quality check?

Ans. This is a binomial probability problem because each order either passes or does not pass the quality check We use a binomial distribution because it models the probability of a certain number of successes (orders passing the quality check) in a fixed number of independent trials (12 orders), where each trial has the same probability of success (90%). This makes it ideal for quality control scenarios..

Given:

• Probability of passing p=0.9
• Probability of failing q=1−p=0.1
• Number of trials n=12
• Number of successes k=10

The binomial probability formula is:

Combine the values:

The probability that exactly 10 out of 12 orders pass the quality check is approximately 23.01%.

8. Delivery Time Prediction

• Case Study: The average delivery time in a city is 25 minutes with a standard deviation of 5 minutes. Assuming the delivery times follow a normal distribution, what is the probability that the delivery time will be between 20 and 30 minutes?

This problem involves finding the probability of a value falling within a range in a normal distribution. We use a normal distribution because it models continuous data like delivery times, which are symmetrically distributed around the mean (25 minutes) with a known standard deviation (5 minutes). This makes it ideal for calculating the probability of delivery times falling within a specific range.

Given:

• Mean (μ) = 25 minutes
• Standard deviation (σ) = 5 minutes
• Range = 20 to 30 minutes
• Normal distribution assumption.

We calculate the z-scores for 20 and 30 minutes and use the standard normal distribution table (z-table) to find probabilities.

• Step 1: Calculate z-scores

The z-score formula is:

1. For x=20

2. For x=30

• Step 2: Find probabilities from the z-table

Using the z-table:

The probability of the delivery time being between 20 and 30 minutes is:

The probability that the delivery time is between 20 and 30 minutes is approximately 68.26%.

9. Customer Churn Rate

Case Study: The probability that a customer stops using a service after 1 year is 0.3. What is the probability that exactly 2 customers will churn in a group of 5 customers?

Ans. This is a binomial probability problem. We use a binomial distribution because it models the probability of a certain number of successes (customers churning) in a fixed number of independent trials (5 customers), where each trial has the same probability of success (0.3). This makes it suitable for predicting customer churn.

The formula is:

Substituting:

2. Calculate the probabilities:

3. Combine

The probability that exactly 2 customers will churn is approximately 30.87%.

10. Order Volume Prediction

Case Study: A service estimates that 40% of its customers will place an order during a special sale event. If 12 customers are selected, what is the probability that exactly 5 will place an order?

Ans. This is also a binomial probability problem, We use a binomial distribution because it models the probability of a certain number of successes (customers placing an order) in a fixed number of independent trials (12 customers), where each trial has the same probability of success (40%). This makes it suitable for predicting order volumes

Substituting:

Calculate probabilities:

Combine:

The probability that exactly 5 customers will place an order is approximately 22.6%.