A jar contains 4 marbles. 3 Red & 1 white. Two marbles are drawn with replacement after each draw. What is the probability that the same color marble is drawn twice?
A) 1/2
B) 1/3
C) 5/8
D) 1/8
Ans: (C)
If the marbles are of the same color then it will be 3/4 * 3/4 + 1/4 * 1/4 = 5/8.
Which of the following events is most likely?
A) At least one 6, when 6 dice are rolled
B) At least 2 sixes when 12 dice are rolled
C) At least 3 sixes when 18 dice are rolled
D) All the above have same probability
Ans: (A)
Probability of ‘6’ turning up in a roll of dice is P(6) = (1/6) & P(6’) = (5/6). Thus, probability of
∞ Case 1: (1/6) * (5/6)5 = 0.06698
∞ Case 2: (1/6)2 * (5/6)10 = 0.00448
∞ Case 3: (1/6)3 * (5/6)15 = 0.0003
Thus, the highest probability is Case 1
Suppose you were interviewed for a technical role. 50% of the people who sat for the first interview received the call for second interview. 95% of the people who got a call for second interview felt good about their first interview. 75% of people who did not receive a second call, also felt good about their first interview. If you felt good after your first interview, what is the probability that you will receive a second interview call?
A) 66%
B) 56%
C) 75%
D) 85%
Ans: (B)
Let’s assume there are 100 people that gave the first round of interview. The 50 people got the interview call for the second round. Out of this 95 % felt good about their interview, which is 47.5. 50 people did not get a call for the interview; out of which 75% felt good about, which is 37.5. Thus, the total number of people that felt good after giving their interview is (37.5 + 47.5) 85. Thus, out of 85 people who felt good, only 47.5 got the call for next round. Hence, the probability of success is (47.5/85) = 0.558.
Another more accepted way to solve this problem is the Baye’s theorem. I leave it to you to check for yourself.
You can quote ISLR’s authors Hastie, Tibshirani who asserted that, in presence of few variables with medium / large sized effect, use lasso regression. In presence of many variables with small / medium sized effect, use ridge regression.
Conceptually, we can say, lasso regression (L1) does both variable selection and parameter shrinkage, whereas Ridge regression only does parameter shrinkage and end up including all the coefficients in the model. In presence of correlated variables, ridge regression might be the preferred choice. Also, ridge regression works best in situations where the least square estimates have higher variance. Therefore, it depends on our model objective.
Rise in global average temperature led to decrease in number of pirates around the world. Does that mean that decrease in number of pirates caused the climate change?
After reading this question, you should have understood that this is a classic case of “causation and correlation”. No, we can’t conclude that decrease in number of pirates caused the climate change because there might be other factors (lurking or confounding variables) influencing this phenomenon.
Therefore, there might be a correlation between global average temperature and number of pirates, but based on this information we can’t say that pirated died because of rise in global average temperature.
While working on a data set, how do you select important variables? Explain your methods.
Following are the methods of variable selection you can use:
Remove the correlated variables prior to selecting important variables
Use linear regression and select variables based on p values
Use Forward Selection, Backward Selection, Stepwise Selection
Use Random Forest, Xgboost and plot variable importance chart
Use Lasso Regression
Measure information gain for the available set of features and select top n features accordingly.
What is the difference between covariance and correlation?
Correlation is the standardized form of covariance.
Covariances are difficult to compare. For example: if we calculate the covariances of salary ($) and age (years), we’ll get different covariances which can’t be compared because of having unequal scales. To combat such situation, we calculate correlation to get a value between -1 and 1, irrespective of their respective scale.
Is it possible capture the correlation between continuous and categorical variable? If yes, how?
Yes, we can use ANCOVA (analysis of covariance) technique to capture association between continuous and categorical variables.
Both being tree based algorithm, how is random forest different from Gradient boosting algorithm (GBM)?
The fundamental difference is, random forest uses bagging technique to make predictions. GBM uses boosting techniques to make predictions.
In bagging technique, a data set is divided into n samples using randomized sampling. Then, using a single learning algorithm a model is build on all samples. Later, the resultant predictions are combined using voting or averaging. Bagging is done is parallel. In boosting, after the first round of predictions, the algorithm weighs misclassified predictions higher, such that they can be corrected in the succeeding round. This sequential process of giving higher weights to misclassified predictions continue until a stopping criterion is reached.
Random forest improves model accuracy by reducing variance (mainly). The trees grown are uncorrelated to maximize the decrease in variance. On the other hand, GBM improves accuracy my reducing both bias and variance in a model.
Running a binary classification tree algorithm is the easy part. Do you know how does a tree splitting takes place i.e. how does the tree decide which variable to split at the root node and succeeding nodes?
A classification trees makes decision based on Gini Index and Node Entropy. In simple words, the tree algorithm find the best possible feature which can divide the data set into purest possible children nodes.
Gini index says, if we select two items from a population at random then they must be of same class and probability for this is 1 if population is pure. We can calculate Gini as following:
Calculate Gini for sub-nodes, using formula sum of square of probability for success and failure (p^2+q^2).
Calculate Gini for split using weighted Gini score of each node of that split
Entropy is the measure of impurity as given by (for binary class):
Entropy, Decision Tree
Here p and q is probability of success and failure respectively in that node. Entropy is zero when a node is homogeneous. It is maximum when a both the classes are present in a node at 50% – 50%. Lower entropy is desirable.
You’ve built a random forest model with 10000 trees. You got delighted after getting training error as 0.00. But, the validation error is 34.23. What is going on? Haven’t you trained your model perfectly?
The model has overfitted. Training error 0.00 means the classifier has mimiced the training data patterns to an extent, that they are not available in the unseen data. Hence, when this classifier was run on unseen sample, it couldn’t find those patterns and returned prediction with higher error. In random forest, it happens when we use larger number of trees than necessary. Hence, to avoid these situation, we should tune number of trees using cross validation.
You’ve got a data set to work having p (no. of variable) > n (no. of observation). Why is OLS as bad option to work with? Which techniques would be best to use? Why?
In such high dimensional data sets, we can’t use classical regression techniques, since their assumptions tend to fail. When p > n, we can no longer calculate a unique least square coefficient estimate, the variances become infinite, so OLS cannot be used at all.
To combat this situation, we can use penalized regression methods like lasso, LARS, ridge which can shrink the coefficients to reduce variance. Precisely, ridge regression works best in situations where the least square estimates have higher variance.
Among other methods include subset regression, forward stepwise regression.
What is convex hull ? (Hint: Think SVM)
In case of linearly separable data, convex hull represents the outer boundaries of the two group of data points. Once convex hull is created, we get maximum margin hyperplane (MMH) as a perpendicular bisector between two convex hulls. MMH is the line which attempts to create greatest separation between two groups.
We know that one hot encoding increasing the dimensionality of a data set. But, label encoding doesn’t. How ?
Don’t get baffled at this question. It’s a simple question asking the difference between the two.
Using one hot encoding, the dimensionality (a.k.a features) in a data set get increased because it creates a new variable for each level present in categorical variables. For example: let’s say we have a variable ‘color’. The variable has 3 levels namely Red, Blue and Green. One hot encoding ‘color’ variable will generate three new variables as Color.Red, Color.Blue and Color.Green containing 0 and 1 value.
In label encoding, the levels of a categorical variables gets encoded as 0 and 1, so no new variable is created. Label encoding is majorly used for binary variables.
What cross validation technique would you use on time series data set? Is it k-fold or LOOCV?
Neither.
In time series problem, k fold can be troublesome because there might be some pattern in year 4 or 5 which is not in year 3. Resampling the data set will separate these trends, and we might end up validation on past years, which is incorrect. Instead, we can use forward chaining strategy with 5 fold as shown below:
fold 1 : training [1], test [2]
fold 2 : training [1 2], test [3]
fold 3 : training [1 2 3], test [4]
fold 4 : training [1 2 3 4], test [5]
fold 5 : training [1 2 3 4 5], test [6]
where 1,2,3,4,5,6 represents “year”.
Explain machine learning to me like a 5 year old.
It’s simple. It’s just like how babies learn to walk. Every time they fall down, they learn (unconsciously) & realize that their legs should be straight and not in a bend position. The next time they fall down, they feel pain. They cry. But, they learn ‘not to stand like that again’. In order to avoid that pain, they try harder. To succeed, they even seek support from the door or wall or anything near them, which helps them stand firm.
This is how a machine works & develops intuition from its environment.
I know that a linear regression model is generally evaluated using Adjusted R² or F value. How would you evaluate a logistic regression model?
We can use the following methods:
Since logistic regression is used to predict probabilities, we can use AUC-ROC curve along with confusion matrix to determine its performance.
Also, the analogous metric of adjusted R² in logistic regression is AIC. AIC is the measure of fit which penalizes model for the number of model coefficients. Therefore, we always prefer model with minimum AIC value.
Null Deviance indicates the response predicted by a model with nothing but an intercept. Lower the value, better the model. Residual deviance indicates the response predicted by a model on adding independent variables. Lower the value, better the model.
Considering the long list of machine learning algorithm, given a data set, how do you decide which one to use?
You should say, the choice of machine learning algorithm solely depends of the type of data. If you are given a data set which is exhibits linearity, then linear regression would be the best algorithm to use. If you given to work on images, audios, then neural network would help you to build a robust model.
If the data comprises of non linear interactions, then a boosting or bagging algorithm should be the choice. If the business requirement is to build a model which can be deployed, then we’ll use regression or a decision tree model (easy to interpret and explain) instead of black box algorithms like SVM, GBM etc.
In short, there is no one master algorithm for all situations. We must be scrupulous enough to understand which algorithm to use.
What is power analysis?
An experimental design technique for determining the effect of a given sample size.
What is K-means? How can you select K for K-means?
What is Collaborative filtering?
The process of filtering used by most of the recommender systems to find patterns or information by collaborating viewpoints, various data sources and multiple agents.
What is the difference between Cluster and Systematic Sampling?
Cluster sampling is a technique used when it becomes difficult to study the target population spread across a wide area and simple random sampling cannot be applied. Cluster Sample is a probability sample where each sampling unit is a collection, or cluster of elements. Systematic sampling is a statistical technique where elements are selected from an ordered sampling frame. In systematic sampling, the list is progressed in a circular manner so once you reach the end of the list,it is progressed from the top again. The best example for systematic sampling is equal probability method.
Are expected value and mean value different?
They are not different but the terms are used in different contexts. Mean is generally referred when talking about a probability distribution or sample population whereas expected value is generally referred in a random variable context.
For Sampling Data
Mean value is the only value that comes from the sampling data.
Expected Value is the mean of all the means i.e. the value that is built from multiple samples. Expected value is the population mean.
For Distributions
Mean value and Expected value are same irrespective of the distribution, under the condition that the distribution is in the same population.
What does P-value signify about the statistical data?
P-value is used to determine the significance of results after a hypothesis test in statistics. P-value helps the readers to draw conclusions and is always between 0 and 1.
• P- Value > 0.05 denotes weak evidence against the null hypothesis which means the null hypothesis cannot be rejected.
• P-value <= 0.05 denotes strong evidence against the null hypothesis which means the null hypothesis can be rejected.
• P-value=0.05is the marginal value indicating it is possible to go either way.
If you dealt 13 cards, what is the probability that the 13th card is a King?
A) 1/52
B) 1/13
C) 1/26
D) 1/12
Ans: (B)
Since we are not told anything about the first 12 cards that are dealt, the probability that the 13th card dealt is a King, is the same as the probability that the first card dealt, or in fact any particular card dealt is a King, and this equals: 4/52.
A) 1/2
B) 1/3
C) 5/8
D) 1/8
Ans: (C)
If the marbles are of the same color then it will be 3/4 * 3/4 + 1/4 * 1/4 = 5/8.
Which of the following events is most likely?
A) At least one 6, when 6 dice are rolled
B) At least 2 sixes when 12 dice are rolled
C) At least 3 sixes when 18 dice are rolled
D) All the above have same probability
Ans: (A)
Probability of ‘6’ turning up in a roll of dice is P(6) = (1/6) & P(6’) = (5/6). Thus, probability of
∞ Case 1: (1/6) * (5/6)5 = 0.06698
∞ Case 2: (1/6)2 * (5/6)10 = 0.00448
∞ Case 3: (1/6)3 * (5/6)15 = 0.0003
Thus, the highest probability is Case 1
 |
| Facebook Data Science Recently Asked Interview Questions Answers |
Suppose you were interviewed for a technical role. 50% of the people who sat for the first interview received the call for second interview. 95% of the people who got a call for second interview felt good about their first interview. 75% of people who did not receive a second call, also felt good about their first interview. If you felt good after your first interview, what is the probability that you will receive a second interview call?
A) 66%
B) 56%
C) 75%
D) 85%
Ans: (B)
Let’s assume there are 100 people that gave the first round of interview. The 50 people got the interview call for the second round. Out of this 95 % felt good about their interview, which is 47.5. 50 people did not get a call for the interview; out of which 75% felt good about, which is 37.5. Thus, the total number of people that felt good after giving their interview is (37.5 + 47.5) 85. Thus, out of 85 people who felt good, only 47.5 got the call for next round. Hence, the probability of success is (47.5/85) = 0.558.
Another more accepted way to solve this problem is the Baye’s theorem. I leave it to you to check for yourself.
When is Ridge regression favorable over Lasso regression?
You can quote ISLR’s authors Hastie, Tibshirani who asserted that, in presence of few variables with medium / large sized effect, use lasso regression. In presence of many variables with small / medium sized effect, use ridge regression.
Conceptually, we can say, lasso regression (L1) does both variable selection and parameter shrinkage, whereas Ridge regression only does parameter shrinkage and end up including all the coefficients in the model. In presence of correlated variables, ridge regression might be the preferred choice. Also, ridge regression works best in situations where the least square estimates have higher variance. Therefore, it depends on our model objective.
Rise in global average temperature led to decrease in number of pirates around the world. Does that mean that decrease in number of pirates caused the climate change?
After reading this question, you should have understood that this is a classic case of “causation and correlation”. No, we can’t conclude that decrease in number of pirates caused the climate change because there might be other factors (lurking or confounding variables) influencing this phenomenon.
Therefore, there might be a correlation between global average temperature and number of pirates, but based on this information we can’t say that pirated died because of rise in global average temperature.
While working on a data set, how do you select important variables? Explain your methods.
Following are the methods of variable selection you can use:
Remove the correlated variables prior to selecting important variables
Use linear regression and select variables based on p values
Use Forward Selection, Backward Selection, Stepwise Selection
Use Random Forest, Xgboost and plot variable importance chart
Use Lasso Regression
Measure information gain for the available set of features and select top n features accordingly.
What is the difference between covariance and correlation?
Correlation is the standardized form of covariance.
Covariances are difficult to compare. For example: if we calculate the covariances of salary ($) and age (years), we’ll get different covariances which can’t be compared because of having unequal scales. To combat such situation, we calculate correlation to get a value between -1 and 1, irrespective of their respective scale.
Is it possible capture the correlation between continuous and categorical variable? If yes, how?
Yes, we can use ANCOVA (analysis of covariance) technique to capture association between continuous and categorical variables.
Both being tree based algorithm, how is random forest different from Gradient boosting algorithm (GBM)?
The fundamental difference is, random forest uses bagging technique to make predictions. GBM uses boosting techniques to make predictions.
In bagging technique, a data set is divided into n samples using randomized sampling. Then, using a single learning algorithm a model is build on all samples. Later, the resultant predictions are combined using voting or averaging. Bagging is done is parallel. In boosting, after the first round of predictions, the algorithm weighs misclassified predictions higher, such that they can be corrected in the succeeding round. This sequential process of giving higher weights to misclassified predictions continue until a stopping criterion is reached.
Random forest improves model accuracy by reducing variance (mainly). The trees grown are uncorrelated to maximize the decrease in variance. On the other hand, GBM improves accuracy my reducing both bias and variance in a model.
Running a binary classification tree algorithm is the easy part. Do you know how does a tree splitting takes place i.e. how does the tree decide which variable to split at the root node and succeeding nodes?
A classification trees makes decision based on Gini Index and Node Entropy. In simple words, the tree algorithm find the best possible feature which can divide the data set into purest possible children nodes.
Gini index says, if we select two items from a population at random then they must be of same class and probability for this is 1 if population is pure. We can calculate Gini as following:
Calculate Gini for sub-nodes, using formula sum of square of probability for success and failure (p^2+q^2).
Calculate Gini for split using weighted Gini score of each node of that split
Entropy is the measure of impurity as given by (for binary class):
Entropy, Decision Tree
Here p and q is probability of success and failure respectively in that node. Entropy is zero when a node is homogeneous. It is maximum when a both the classes are present in a node at 50% – 50%. Lower entropy is desirable.
You’ve built a random forest model with 10000 trees. You got delighted after getting training error as 0.00. But, the validation error is 34.23. What is going on? Haven’t you trained your model perfectly?
The model has overfitted. Training error 0.00 means the classifier has mimiced the training data patterns to an extent, that they are not available in the unseen data. Hence, when this classifier was run on unseen sample, it couldn’t find those patterns and returned prediction with higher error. In random forest, it happens when we use larger number of trees than necessary. Hence, to avoid these situation, we should tune number of trees using cross validation.
You’ve got a data set to work having p (no. of variable) > n (no. of observation). Why is OLS as bad option to work with? Which techniques would be best to use? Why?
In such high dimensional data sets, we can’t use classical regression techniques, since their assumptions tend to fail. When p > n, we can no longer calculate a unique least square coefficient estimate, the variances become infinite, so OLS cannot be used at all.
To combat this situation, we can use penalized regression methods like lasso, LARS, ridge which can shrink the coefficients to reduce variance. Precisely, ridge regression works best in situations where the least square estimates have higher variance.
Among other methods include subset regression, forward stepwise regression.
What is convex hull ? (Hint: Think SVM)
In case of linearly separable data, convex hull represents the outer boundaries of the two group of data points. Once convex hull is created, we get maximum margin hyperplane (MMH) as a perpendicular bisector between two convex hulls. MMH is the line which attempts to create greatest separation between two groups.
We know that one hot encoding increasing the dimensionality of a data set. But, label encoding doesn’t. How ?
Don’t get baffled at this question. It’s a simple question asking the difference between the two.
Using one hot encoding, the dimensionality (a.k.a features) in a data set get increased because it creates a new variable for each level present in categorical variables. For example: let’s say we have a variable ‘color’. The variable has 3 levels namely Red, Blue and Green. One hot encoding ‘color’ variable will generate three new variables as Color.Red, Color.Blue and Color.Green containing 0 and 1 value.
In label encoding, the levels of a categorical variables gets encoded as 0 and 1, so no new variable is created. Label encoding is majorly used for binary variables.
What cross validation technique would you use on time series data set? Is it k-fold or LOOCV?
Neither.
In time series problem, k fold can be troublesome because there might be some pattern in year 4 or 5 which is not in year 3. Resampling the data set will separate these trends, and we might end up validation on past years, which is incorrect. Instead, we can use forward chaining strategy with 5 fold as shown below:
fold 1 : training [1], test [2]
fold 2 : training [1 2], test [3]
fold 3 : training [1 2 3], test [4]
fold 4 : training [1 2 3 4], test [5]
fold 5 : training [1 2 3 4 5], test [6]
where 1,2,3,4,5,6 represents “year”.
Explain machine learning to me like a 5 year old.
It’s simple. It’s just like how babies learn to walk. Every time they fall down, they learn (unconsciously) & realize that their legs should be straight and not in a bend position. The next time they fall down, they feel pain. They cry. But, they learn ‘not to stand like that again’. In order to avoid that pain, they try harder. To succeed, they even seek support from the door or wall or anything near them, which helps them stand firm.
This is how a machine works & develops intuition from its environment.
I know that a linear regression model is generally evaluated using Adjusted R² or F value. How would you evaluate a logistic regression model?
We can use the following methods:
Since logistic regression is used to predict probabilities, we can use AUC-ROC curve along with confusion matrix to determine its performance.
Also, the analogous metric of adjusted R² in logistic regression is AIC. AIC is the measure of fit which penalizes model for the number of model coefficients. Therefore, we always prefer model with minimum AIC value.
Null Deviance indicates the response predicted by a model with nothing but an intercept. Lower the value, better the model. Residual deviance indicates the response predicted by a model on adding independent variables. Lower the value, better the model.
Considering the long list of machine learning algorithm, given a data set, how do you decide which one to use?
You should say, the choice of machine learning algorithm solely depends of the type of data. If you are given a data set which is exhibits linearity, then linear regression would be the best algorithm to use. If you given to work on images, audios, then neural network would help you to build a robust model.
If the data comprises of non linear interactions, then a boosting or bagging algorithm should be the choice. If the business requirement is to build a model which can be deployed, then we’ll use regression or a decision tree model (easy to interpret and explain) instead of black box algorithms like SVM, GBM etc.
In short, there is no one master algorithm for all situations. We must be scrupulous enough to understand which algorithm to use.
What is power analysis?
An experimental design technique for determining the effect of a given sample size.
What is K-means? How can you select K for K-means?
What is Collaborative filtering?
The process of filtering used by most of the recommender systems to find patterns or information by collaborating viewpoints, various data sources and multiple agents.
What is the difference between Cluster and Systematic Sampling?
Cluster sampling is a technique used when it becomes difficult to study the target population spread across a wide area and simple random sampling cannot be applied. Cluster Sample is a probability sample where each sampling unit is a collection, or cluster of elements. Systematic sampling is a statistical technique where elements are selected from an ordered sampling frame. In systematic sampling, the list is progressed in a circular manner so once you reach the end of the list,it is progressed from the top again. The best example for systematic sampling is equal probability method.
Are expected value and mean value different?
They are not different but the terms are used in different contexts. Mean is generally referred when talking about a probability distribution or sample population whereas expected value is generally referred in a random variable context.
For Sampling Data
Mean value is the only value that comes from the sampling data.
Expected Value is the mean of all the means i.e. the value that is built from multiple samples. Expected value is the population mean.
For Distributions
Mean value and Expected value are same irrespective of the distribution, under the condition that the distribution is in the same population.
What does P-value signify about the statistical data?
P-value is used to determine the significance of results after a hypothesis test in statistics. P-value helps the readers to draw conclusions and is always between 0 and 1.
• P- Value > 0.05 denotes weak evidence against the null hypothesis which means the null hypothesis cannot be rejected.
• P-value <= 0.05 denotes strong evidence against the null hypothesis which means the null hypothesis can be rejected.
• P-value=0.05is the marginal value indicating it is possible to go either way.
If you dealt 13 cards, what is the probability that the 13th card is a King?
A) 1/52
B) 1/13
C) 1/26
D) 1/12
Ans: (B)
Since we are not told anything about the first 12 cards that are dealt, the probability that the 13th card dealt is a King, is the same as the probability that the first card dealt, or in fact any particular card dealt is a King, and this equals: 4/52.
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