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“I have applied for various data science roles but I always get rejected because of a lack of experience.”

This is easily the most common issue I’ve heard from aspiring data scientists. They put in the hard work to learn the theoretical aspect of data science but when it comes to applying it in the real world, not many organizations are willing to take them on.

No matter how well you do in the interview round – the hiring manager always finds the lack of data science experience as the main sticking point.

So what can you do about this? It is a seemingly unassailable obstacle in your quest to become a rockstar data scientist.

We at Analytics Vidhya understand this challenge and are thrilled to launch the Data Science Immersive Bootcamp to help you overcome it!This is an unmissable opportunity where you will get to learn on the job from data science experts with decades of industry experience.

“In the Data Science Immersive Bootcamp, we are not only focusing on classroom training – we provide hands-on internship to enrich you with practical experience.”

So you get the best of both worlds – you learn data science AND get to work on real-world projects.

Let’s Gauge the Benefits of the Data Science Immersive BootcampThis Bootcamp has been created by keeping Data Science professionals at heart and industry requirements in mind. Let’s dive in to understand the benefits of Data Science Immersive Bootcamp:

Learn on the job from Day 1: This is a golden opportunity where you can learn data science and apply your learnings in various projects manned by you at Analytics Vidhya during the course of this Bootcamp

Work with experienced data scientists and experts: The best experts from different verticles will come together to teach and mentor you at the Bootcamp – it is bound to boost your experience and knowledge exponentially!

Work on real-world projects: Apply all that you learn on the go! Real challenges are faced when you dive in to solve a practical problem and cruising through that successfully will hone and fine-tune your blossoming data science portfolio

Peer Groups and Collaborative Learning: Best solutions are derived when you learn with the community! And this internship gives you an opportunity to be part of several focused teams working on different data projects

Build your data science profile: You will get to present your work in front of Analytics Vidhya’s thriving and burgeoning community with over 400,000 data science enthusiasts and practitioners. You are bound to shine like a star after getting such an exhaustive learning and hands-on experience

Mock interviews: Get the best hack to crack data science interviews

Unique Features of the Data Science Immersive BootcampThere are so many unique features that come with this Bootcamp. Here’s a quick summary of the highlights:

Curriculum of Data Science Immersive BootcampData Science Immersive Bootcamp is one of the most holistic and intensive programs in the data science space. Here’s a high-level overview of what we will cover during this Bootcamp:

Python for Data Science

Linear Algebra

SQL and other Databases

Statistics – Descriptive and Inferential

Data Visualization

Structured Thinking & Communications

Basics of Machine Learning

Advanced Machine Learning Algorithms

Deep Learning Basics

Recurrent Neural Networks (RNN)

Natural Language Processing (NLP)

Convolutional Neural Networks (CNN)

Computer Vision

Building Data Pipelines

Big Data Engineering

Big Data Machine Learning

Wholesome Data Science is what we call it – everything you need to learn is presented in a single platter!

How to apply for the Data Science Immersive Bootcamp?Here are the steps for the admission process to the Data Science Immersive Bootcamp:

Online Application – Apply with a simple form

Take the Fit Test – No knowledge of data science expected

Interaction with Analytics Vidhya team (Gurgaon) – Interview round to screen the best candidates

Offer Rollouts – The chosen candidates will be sent the official offer to be a part of the Bootcamp!

Offer Acceptance

Welcome to AV’s Data Science Immersive Bootcamp!

Fee Structure and Duration for the Data Science Immersive BootcampAdmission Fees: INR 25,000/-

Note: It is non-refundable and will get adjusted with the entire upfront payment (INR 3,50,000) or with 1st Installment (in case of Installment Payment Plan).

Option 1: (If paying all upfront)

INR 3,51,000/-

Option 2: (Installment Payment Plan)

1st Installment – INR 99,000/- (30 days before the start date)

2nd Installment – INR 1,25,000/- (within 60 days of the start date)

3rd Installment – INR 1,25,000/- (within 120 days of the start date)

Here are the details of the program:Duration of Program: 9 months / 40 weeks

Internship Stipend (from month 1): Rs. 15,000/- per month

Number of Projects: 10+ real-world projects

No. of Seats in the Bootcamp: Maximum 30

And The Most Awaited Aspect – You Get A Job Guarantee!As mentioned above, this Bootcamp will enrich you with knowledge and industry experience thus making you the perfect fit for any role in Data Science. Bridging the gap between education and what employers want – the ultimate jackpot Analytics Vidhya is providing!

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And do not forget to register with us TODAY! Only 30 candidates will get a chance to unravel the best of Data Science in this specialized Bootcamp.

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## Introduction To Git For Data Science

The data science and engineering fields are interacting more and more because data scientists are working on production systems and joining R&D teams. We want to make it simpler for data scientists without prior engineering experience to understand the core engineering best practices.

We are building a manual on engineering subjects like Git, Docker, cloud infrastructure, and model serving that we hear data science practitioners think about.

Introduction to GitA version control system called Git is made to keep track of changes made to a source code over time.

Typically, each user will clone a single central repository to their local system (referred to as “origin” or “remote”) which the individual users will clone to their local machine (called “local” or “clone”). Users “push” and “merge” their completed work back into the central repository once they have stored relevant work (referred to as “commits”) on their computers.

Difference between Git and GitHubGit serves as both the foundational technology, for tracking and merging changes in a source code, and its command-line client (CLI).

An online platform called GitHub was created on top of git technology to make it simpler. Additionally, it provides capabilities like automation, pulls requests, and user management. GitLab and Sourcetree are two additional options.

Git for Data ScienceIn data science we are going to analyze the data using some models and algorithms, a model might be created by more than one person which makes it hard to handle and makes updates at the same time, but Git makes this all easy by storing the previous versions and allowing many peoples to work on the same project at a single time.

Let’s look into some terms of Git which are very common among developers

Terms

Repository − “Database” containing all of a project’s branches and commits

Branch − A repository’s alternative state or route of development.

Merge − Merging two (or more) branches into one branch, one truth is the definition of the merge.

Clone − The process of locally copying a remote repository.

Origin − The local clone was made from a remote repository, which is referred to as the origin.

Main/Master − Common names for the root branch, which is the main repository of truth, include “main” and “master.”

Stage − Choosing which files to include in the new commit at this stage

Commit − A stored snapshot of the staged modifications made to the file(s) in the repository is known as a “commit.”

HEAD − Abbreviation for the current commit in your local repository.

Push − Sending changes to a remote repository for public viewing is known as pushing.

Pull − Pulling is the process of adding other people’s updates to your personal repository.

Pull Request − Before merging your modifications to main/master, use the pull request mechanism to examine and approve them.

As we have discussed above do for that we need some commands that are generally used, lets discussed them below −

git init − Create a new repository on your local computer.

git clone − begin editing an already-existing remote repository.

git add − Select the file or files to save (staging).

Show the files you have modified with git status.

git commit − Store a copy of the selected file(s) as a snapshot (commit).

Send your saved snapshots (commits) into the distant repository using the git push command.

Pull current commits made by others into your own computer using the git pull command.

Create or remove branches with the git branch.

git checkout − Change branches or reverse local file(s) modifications.

git merge − merges branches with git to create a single branch or a single truth.

Rules for Handling Git Process SmoothThere are some rules for handling the smooth process of uploading a project over GitHub

Don’t push datasetsGit is used to tracking, manage, and store the codes but it is not a good practice to put the datasets over it. Keep track of the data there are many good data trackers available.

Don’t push secrets Don’t use the –force−force method is used in various situations but it is not recommended to use it mostly because while pushing the code to git if there is an error, it will be displayed by the compiler or the CLI to use the force method to put the data on the server but it is not a good approach.

Do small commits with clear descriptionsBeginners developers may not be as good with the small commits but it is recommended to do the small commits as they make the view of the development process much clear and helps out in future updates. Also writing a good and clear description makes the same process much easier.

ConclusionA version control system called Git is made to keep track of changes made to a source code over time. Without a version control system, a collaboration between multiple people working on the same project is complete confusion. Git serves as both the foundational technology, for tracking and merging changes in a source code, and its command-line client (CLI). An online platform called GitHub was created on top of git technology to make it simpler. Additionally, it provides capabilities like automation, pulls requests, and user management.

## 40 Questions On Probability For Data Science

Introduction

Probability forms the backbone of many important data science concepts from inferential statistics to Bayesian networks. It would not be wrong to say that the journey of mastering statistics begins with probability. This skilltest was conducted to help you identify your skill level in probability.

A total of 1249 people registered for this skill test. The test was designed to test the conceptual knowledge of probability. If you are one of those who missed out on this skill test, here are the questions and solutions. You missed on the real time test, but can read this article to find out how you could have answered correctly.

Here are the leaderboard ranking for all the participants.

Are you preparing for your next data science interview? Then look no further! Check out the comprehensive ‘Ace Data Science Interviews‘ course which encompasses hundreds of questions like these along with plenty of videos, support and resources. And if you’re looking to brush up your probability sills even more, we have covered it comprehensively in the ‘Introduction to Data Science‘ course!

Overall ScoresBelow are the distribution scores, they will help you evaluate your performance.

You can access the final scores here. More than 300 people participated in the skill test and the highest score obtained was 38. Here are a few statistics about the distribution.

Mean Score: 19.56

Median Score: 20

Mode Score: 15

This was also the first test where some one scored as high as 38! The community is getting serious about DataFest

Useful ResourcesBasics of Probability for Data Science explained with examples

Introduction to Conditional Probability and Bayes theorem for data science professionals

1) Let A and B be events on the same sample space, with P (A) = 0.6 and P (B) = 0.7. Can these two events be disjoint?

A) Yes

B) No

Solution: (B)

P(AꓴB) = P(A)+P(B)-P(AꓵB).

An event is disjoint if P(AꓵB) = 0. If A and B are disjoint P(AꓴB) = 0.6+0.7 = 1.3

And Since probability cannot be greater than 1, these two mentioned events cannot be disjoint.

2) Alice has 2 kids and one of them is a girl. What is the probability that the other child is also a girl?

You can assume that there are an equal number of males and females in the world.

A) 0.5

B) 0.25

C) 0.333

D) 0.75

Solution: (C)

The outcomes for two kids can be {BB, BG, GB, GG}

Since it is mentioned that one of them is a girl, we can remove the BB option from the sample space. Therefore the sample space has 3 options while only one fits the second condition. Therefore the probability the second child will be a girl too is 1/3.

3) A fair six-sided die is rolled twice. What is the probability of getting 2 on the first roll and not getting 4 on the second roll?

A) 1/36

B) 1/18

C) 5/36

D) 1/6

E) 1/3

Solution: (C)

The two events mentioned are independent. The first roll of the die is independent of the second roll. Therefore the probabilities can be directly multiplied.

P(getting first 2) = 1/6

P(no second 4) = 5/6

Therefore P(getting first 2 and no second 4) = 1/6* 5/6 = 5/36

4)

A) TrueB) False

Solution: (A)

P(AꓵCc) will be only P(A). P(only A)+P(C) will make it P(AꓴC). P(BꓵAcꓵCc) is P(only B) Therefore P(AꓴC) and P(only B) will make P(AꓴBꓴC)

5) Consider a tetrahedral die and roll it twice. What is the probability that the number on the first roll is strictly higher than the number on the second roll?

Note: A tetrahedral die has only four sides (1, 2, 3 and 4).

A) 1/2

B) 3/8

C) 7/16

D) 9/16

Solution: (B)

(1,1)

(2,1)

(3,1)

(4,1)

(1,2)

(2,2)

(3,2)

(4,2)

(1,3)

(2,3)

(3,3)

(4,3)

(1,4)

(2,4)

(3,4)

(4,4)

There are 6 out of 16 possibilities where the first roll is strictly higher than the second roll.

6) Which of the following options cannot be the probability of any event?

C) 1.001

A) Only A

B) Only B

C) Only C

D) A and B

E) B and C

F) A and C

A) -0.00001B) 0.5C) 1.001

Solution: (F)

Probability always lie within 0 to 1.

7) Anita randomly picks 4 cards from a deck of 52-cards and places them back into the deck ( Any set of 4 cards is equally likely ). Then, Babita randomly chooses 8 cards out of the same deck ( Any set of 8 cards is equally likely). Assume that the choice of 4 cards by Anita and the choice of 8 cards by Babita are independent. What is the probability that all 4 cards chosen by Anita are in the set of 8 cards chosen by Babita?

A)48C4 x 52C4

B)48C4 x 52C8

C)48C8 x 52C8

D) None of the above

Solution: (A)

The total number of possible combination would be 52C4 (For selecting 4 cards by Anita) * 52C8 (For selecting 8 cards by Babita).

Since, the 4 cards that Anita chooses is among the 8 cards which Babita has chosen, thus the number of combinations possible is 52C4 (For selecting the 4 cards selected by Anita) * 48C4 (For selecting any other 4 cards by Babita, since the 4 cards selected by Anita are common)

Question Context 8:

A player is randomly dealt a sequence of 13 cards from a deck of 52-cards. All sequences of 13 cards are equally likely. In an equivalent model, the cards are chosen and dealt one at a time. When choosing a card, the dealer is equally likely to pick any of the cards that remain in the deck.

8) 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

Solution: (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

9) A fair six-sided die is rolled 6 times. What is the probability of getting all outcomes as unique?

A) 0.01543

B) 0.01993

C) 0.23148

D) 0.03333

Solution: (A)

For all the outcomes to be unique, we have 6 choices for the first turn, 5 for the second turn, 4 for the third turn and so on

Therefore the probability if getting all unique outcomes will be equal to 0.01543

10) 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?

A) 1/3

B) 19/59

C) 18/58

D) 1/2

Solution: (B)

Assign a different number to each student from 1 to 60. Numbers 1 to 20 go in group 1, 21 to 40 go to group 2, 41 to 60 go to group 3.

All possible partitions are obtained with equal probability by a random assignment if these numbers, it doesn’t matter with which students we start, so we are free to start by assigning a random number to Jack and then we assign a random number to Jill. After Jack has been assigned a random number there are 59 random numbers available for Jill and 19 of these will put her in the same group as Jack. Therefore the probability is 19/59

A) 2.75

B) 3.35

C) 4.13

D) 5.33

Solution: (A)

Tosses = 2 * (1/4)[probability of selecting coin A] + 3*(3/4)[probability of selecting coin B]

= 2.75

12) 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.

A) $131

B) $140

C) $112

D) $125

Solution: (C)

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

13)

A) TrueB) False

Solution: (A)

The above statement is true. You would need to know that

P(A/B) = P(AꓵB)/P(B)

Multiplying the three we would get – P(AꓵBꓵCc), hence the equations holds true

14) When an event A independent of itself?

A) Always

B) If and only if P(A)=0

C) If and only if P(A)=1

D) If and only if P(A)=0 or 1

Solution: (D)

The event can only be independent of itself when either there is no chance of it happening or when it is certain to happen. Event A and B is independent when P(AꓵB) = P(A)*P(B). Now if B=A, P(AꓵA) = P(A) when P(A) = 0 or 1.

15) Suppose you’re in the final round of “Let’s make a deal” game show and you are supposed to choose from three doors – 1, 2 & 3. One of the three doors has a car behind it and other two doors have goats. Let’s say you choose Door 1 and the host opens Door 3 which has a goat behind it. To assure the probability of your win, which of the following options would you choose.

A) Switch your choice

B) Retain your choice

C) It doesn’t matter probability of winning or losing is the same with or without revealing one door

Solution: (A)

I would recommend reading this article for a detailed discussion of the Monty Hall’s Problem.

16) Cross-fertilizing a red and a white flower produces red flowers 25% of the time. Now we cross-fertilize five pairs of red and white flowers and produce five offspring. What is the probability that there are no red flower plants in the five offspring?

A) 23.7%

B) 37.2%

C) 22.5%

D) 27.3%

Solution: (A)

The probability of offspring being Red is 0.25, thus the probability of the offspring not being red is 0.75. Since all the pairs are independent of each other, the probability that all the offsprings are not red would be (0.75)5 = 0.237. You can think of this as a binomial with all failures.

17) A roulette wheel has 38 slots – 18 red, 18 black, and 2 green. You play five games and always bet on red slots. How many games can you expect to win?

A) 1.1165

B) 2.3684C) 2.6316

C) 2.6316

D) 4.7368

Solution: (B)

The probability that it would be Red in any spin is 18/38. Now, you are playing the game 5 times and all the games are independent of each other. Thus, the number of games that you can win would be 5*(18/38) = 2.3684

18) A roulette wheel has 38 slots, 18 are red, 18 are black, and 2 are green. You play five games and always bet on red. What is the probability that you win all the 5 games?

A) 0.0368

B) 0.0238

C) 0.0526

D) 0.0473

Solution: (B)

The probability that it would be Red in any spin is 18/38. Now, you are playing for game 5 times and all the games are independent of each other. Thus, the probability that you win all the games is (18/38)5 = 0.0238

19) Some test scores follow a normal distribution with a mean of 18 and a standard deviation of 6. What proportion of test takers have scored between 18 and 24?

A) 20%

B) 22%

C) 34%

D) None of the above

Solution: (C)

So here we would need to calculate the Z scores for value being 18 and 24. We can easily doing that by putting sample mean as 18 and population mean as 18 with σ = 6 and calculating Z. Similarly we can calculate Z for sample mean as 24.

Z= (X-μ)/σ

Therefore for 26 as X,

Z = (18-18)/6 = 0 , looking at the Z table we find 50% people have scores below 18.

For 24 as X

Z = (24-18)/6 = 1, looking at the Z table we find 84% people have scores below 24.

Therefore around 34% people have scores between 18 and 24.

20) 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

Solution: (C)

If the marbles are of the same color then it will be 3/4 * 3/4 + 1/4 * 1/4 = 5/8.

21) Which of the following events is most likely?

C) At least 3 sixes when 18 dice are rolled

D) All the above have same probability

A) At least one 6, when 6 dice are rolledB) At least 2 sixes when 12 dice are rolled

Solution: (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

22) 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%

Solution: (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.

23) A coin of diameter 1-inches is thrown on a table covered with a grid of lines each two inches apart. What is the probability that the coin lands inside a square without touching any of the lines of the grid? You can assume that the person throwing has no skill in throwing the coin and is throwing it randomly.

You can assume that the person throwing has no skill in throwing the coin and is throwing it randomly.

A) 1/2

B) 1/4

C) Π/3

D) 1/3

Solution: (B)

Think about where all the center of the coin can be when it lands on 2 inches grid and it not touching the lines of the grid.

If the yellow region is a 1 inch square and the outside square is of 2 inches. If the center falls in the yellow region, the coin will not touch the grid line. Since the total area is 4 and the area of the yellow region is 1, the probability is ¼ .

24) There are a total of 8 bows of 2 each of green, yellow, orange & red. In how many ways can you select 1 bow?

A) 1

B) 2

C) 4

D) 8

Solution: (C)

You can select one bow out of four different bows, so you can select one bow in four different ways.

25) Consider the following probability density function: What is the probability for X≤6 i.e. P(x≤6)

What is the probability for X≤6 i.e. P(x≤6)

A) 0.3935

B) 0.5276

C) 0.1341

D) 0.4724

Solution: (B)

To calculate the area of a particular region of a probability density function, we need to integrate the function under the bounds of the values for which we need to calculate the probability.

Therefore on integrating the given function from 0 to 6, we get 0.5276

26) In a class of 30 students, approximately what is the probability that two of the students have their birthday on the same day (defined by same day and month) (assuming it’s not a leap year)?

For example – Students with birthday 3rd Jan 1993 and 3rd Jan 1994 would be a favorable event.

A) 49%

B) 52%

C) 70%

D) 35%

Solution: (C)

The total number of combinations possible for no two persons to have the same birthday in a class of 30 is 30 * (30-1)/2 = 435.

Now, there are 365 days in a year (assuming it’s not a leap year). Thus, the probability of people having a different birthday would be 364/365. Now there are 870 combinations possible. Thus, the probability that no two people have the same birthday is (364/365)^435 = 0.303.

Thus, the probability that two people would have their birthdays on the same date would be 1 – 0.303 = 0.696

27) Ahmed is playing a lottery game where he must pick 2 numbers from 0 to 9 followed by an English alphabet (from 26-letters). He may choose the same number both times.

If his ticket matches the 2 numbers and 1 letter drawn in order, he wins the grand prize and receives $10405. If just his letter matches but one or both of the numbers do not match, he wins $100. Under any other circumstance, he wins nothing. The game costs him $5 to play. Suppose he has chosen 04R to play.

What is the expected net profit from playing this ticket?

A) $-2.81

B) $2.81C) $-1.82

C) $-1.82

D) $1.82

Solution: (B)

Expected value in this case

E(X) = P(grand prize)*(10405-5)+P(small)(100-5)+P(losing)*(-5)

P(grand prize)= (1/10)*(1/10)*(1/26)

P(small) = 1/26-1/2600, the reason we need to do this is we need to exclude the case where he gets the letter right and also the numbers rights. Hence, we need to remove the scenario of getting the letter right.

P(losing ) = 1-1/26-1/2600

Therefore we can fit in the values to get the expected value as $2.81

28) Assume you sell sandwiches. 70% people choose egg, and the rest choose chicken. What is the probability of selling 2 egg sandwiches to the next 3 customers?

A) 0.343

B) 0.063

C) 0.44

D) 0.027

Solution: (C)

Question context: 29 – 30

HIV is still a very scary disease to even get tested for. The US military tests its recruits for HIV when they are recruited. They are tested on three rounds of Elisa( an HIV test) before they are termed to be positive.

The prior probability of anyone having HIV is 0.00148. The true positive rate for Elisa is 93% and the true negative rate is 99%.

29) What is the probability that a recruit has HIV, given he tested positive on first Elisa test? The prior probability of anyone having HIV is 0.00148. The true positive rate for Elisa is 93% and the true negative rate is 99%.

A) 12%

B) 80%

C) 42%

D) 14%

Solution: (A)

I recommend going through the Bayes updating section of this article for the understanding of the above question.

30) What is the probability of having HIV, given he tested positive on Elisa the second time as well.

The prior probability of anyone having HIV is 0.00148. The true positive rate for Elisa is 93% and the true negative rate is 99%.

A) 20%

B) 42%

C) 93%

D) 88%

Solution: (C)

I recommend going through the Bayes updating section of this article for the understanding of the above question.

C) You have the same probability of winning in guessing either, hence whatever you guess there is just a 50-50 chance of winning or losing

D) None of these

Solution: (C)

32) The inference using the frequentist approach will always yield the same result as the Bayesian approach.

A) TRUE

B) FALSE

Solution: (B)

The frequentist Approach is highly dependent on how we define the hypothesis while Bayesian approach helps us update our prior beliefs. Therefore the frequentist approach might result in an opposite inference if we declare the hypothesis differently. Hence the two approaches might not yield the same results.

33) Hospital records show that 75% of patients suffering from a disease die due to that disease. What is the probability that 4 out of the 6 randomly selected patients recover?

A) 0.17798

B) 0.13184

C) 0.03295

D) 0.35596

Solution: (C)

Think of this as a binomial since there are only 2 outcomes, either the patient dies or he survives.

Here n =6, and x=4. p=0.25(probability if living(success)) q = 0.75(probability of dying(failure))

P(X) = nCx pxqn-x = 6C4 (0.25)4(0.75)2 = 0.03295

34) The students of a particular class were given two tests for evaluation. Twenty-five percent of the class cleared both the tests and forty-five percent of the students were able to clear the first test.

Calculate the percentage of students who passed the second test given that they were also able to pass the first test.

A) 25%

B) 42%

C) 55%

D) 45%

Solution: (C)

This is a simple problem of conditional probability. Let A be the event of passing in first test.

B is the event of passing in the second test.

P(AꓵB) is passing in both the events

P(passing in second given he passed in the first one) = P(AꓵB)/P(A)

= 0.25/0.45 which is around 55%

35) While it is said that the probabilities of having a boy or a girl are the same, let’s assume that the actual probability of having a boy is slightly higher at 0.51. Suppose a couple plans to have 3 children. What is the probability that exactly 2 of them will be boys?

A) 0.38

B) 0.48

C) 0.58

D) 0.68

E) 0.78

Solution: (A)

Think of this as a binomial distribution where getting a success is a boy and failure is a girl. Therefore we need to calculate the probability of getting 2 out of three successes.

P(X) = nCx pxqn-x = 3C2 (0.51)2(0.49)1 = 0.382

36) Heights of 10 year-olds, regardless of gender, closely follow a normal distribution with mean 55 inches and standard deviation 6 inches. Which of the following is true?

A) We would expect more number of 10 year-olds to be shorter than 55 inches than the number of them who are taller than 55 inches

B) Roughly 95% of 10 year-olds are between 37 and 73 inches tall

C) A 10-year-old who is 65 inches tall would be considered more unusual than a 10-year-old who is 45 inches tall

D) None of these

Solution: (D)

None of the above statements are true.

37) About 30% of human twins are identical, and the rest are fraternal. Identical twins are necessarily the same sex, half are males and the other half are females. One-quarter of fraternal twins are both males, one-quarter both female, and one-half are mixed: one male, one female. You have just become a parent of twins and are told they are both girls. Given this information, what is the probability that they are identical?

A) 50%

B) 72%

C) 46%

D) 33%

Solution: (C)

This is a classic problem of Bayes theorem.

P(I) denoted Probability of being identical and P(~I) denotes Probability of not being identical

P(Identical) = 0.3

P(not Identical)= 0.7

38) Rob has fever and the doctor suspects it to be typhoid. To be sure, the doctor wants to conduct the test. The test results positive when the patient actually has typhoid 80% of the time. The test gives positive when the patient does not have typhoid 10% of the time. If 1% of the population has typhoid, what is the probability that Rob has typhoid provided he tested positive?

A) 12%

B) 7%

C) 25%

D) 31.5%

Solution: (B)

We need to find the probability of having typhoid given he tested positive.

=P(testing +ve and having typhoid) / P(testing positive)

39) Jack is having two coins in his hand. Out of the two coins, one is a real coin and the second one is a faulty one with Tails on both sides. He blindfolds himself to choose a random coin and tosses it in the air. The coin falls down with Tails facing upwards. What is the probability that this tail is shown by the faulty coin?

A) 1/3

B) 2/3

C) 1/2

D) 1/4

Solution: (B)

We need to find the probability of the coin being faulty given that it showed tails.

P(Faulty) = 0.5

P(getting tails) = 3/4

P(faulty and tails) =0.5*1 = 0.5

Therefore the probability of coin being faulty given that it showed tails would be 2/3

40) A fly has a life between 4-6 days. What is the probability that the fly will die at exactly 5 days?

A) 1/2

B) 1/4

C) 1/3

D) 0

Solution: (D)

Here since the probabilities are continuous, the probabilities form a mass function. The probability of a certain event is calculated by finding the area under the curve for the given conditions. Here since we’re trying to calculate the probability of the fly dying at exactly 5 days – the area under the curve would be 0. Also to come to think of it, the probability if dying at exactly 5 days is impossible for us to even figure out since we cannot measure with infinite precision if it was exactly 5 days.

End NotesIf you missed out on this competition, make sure you complete in the ones coming up shortly. We are giving cash prizes worth $10,000+ during the month of April 2023.

If you have any questions or doubts feel free to post them below.

Check out all the upcoming skilltests here.## Python Treatment For Outliers In Data Science

What is Feature Engineering?

When we have a LOT OF FEATURES in the given dataset, feature engineering can become quite a challenging and interesting module.

The number of features could significantly impact the model considerably, So that feature engineering is an important task in the Data Science life cycle.

Feature ImprovementsIn the Feature Engineering family, we are having many key factors are there, let’s discuss Outlier here. This is one of the interesting topics and easy to understand in Layman’s terms.

Outlier

An outlier is an observation of a data point that lies an abnormal distance from other values in a given population. (odd man out)

Like in the following data point (Age)

18,22,45,67,89,125,30

An outlier is an object(s) that deviates significantly from the rest of the object collection.

List of Cities

New York, Las Angles, London, France, Delhi, Chennai

It is an abnormal observation during the Data Analysis stage, that data point lies far away from other values.

List of Animals

cat, fox, rabbit, fish

An outlier is an observation that diverges from well-structured data.

The root cause for the Outlier can be an error in measurement or data collection error.

Quick ways to handling Outliers.

Outliers can either be a mistake or just variance. (As mentioned, examples)

If we found this is due to a mistake, then we can ignore them.

If we found this is due to variance, in the data, we can work on this.

In the picture of the Apples, we can find the out man out?? Is it? Hope can Yes!

But the huge list of a given feature/column from the .csv file could be a really challenging one for naked eyes.

First and foremost, the best way to find the Outliers are in the feature is the visualization method.

What are the Possibilities for an Outlier?Of course! It would be below quick reasons.

Missing values in a dataset.

Data did not come from the intended sample.

Errors occur during experiments.

Not an errored, it would be unusual from the original.

Extreme distribution than normal.

That’s fine, but you might have questions about Outlier if you’re a real lover of Data Analytics, Data mining, and Data Science point of view.

Let’s have a quick discussion on those.

Understand more about Outlier

Outliers tell us that the observations of the given data set, how the

data point(s) differ significantly from the overall perspective. Simply saying

odd one/many. this would be an

error during

data collection.

Generally,

Outliers

affect

statistical results while doing the EDA process, we could say a quick example is the MEAN and MODE of a given set of data set, which will be misleading that the

data

values would be higher than they really are.

Positive Relationship

When the correlation coefficient is closer to value 1

Negative Relationship

When the correlation coefficient is closer to value -1

Independent

When X and Y are independent

, then the

correlation coefficient

is close to

zero (0)

We could understand the data collection process from the Outliers and its observations. An analysis of how it occurs and how to minimize and set the process in future data collection guidelines.

Even though the Outliers increase the inconsistent results in your dataset during analysis and the power of statistical impacts significant, there would challenge and roadblocks to remove them in few situations.

DO or DO NOT (Drop Outlier)

Before dropping the Outliers, we must analyze the dataset with and without outliers and understand better the impact of the results.

If you observed that it is obvious due to incorrectly entered or measured, certainly you can drop the outlier. No issues on that case.

If you find that your assumptions are getting affected, you may drop the outlier straight away, provided that no changes in the results.

If the outlier affects your assumptions and results. No questions simply drop the outlier and proceed with your further steps.

Finding OutliersSo far we have discussed what is Outliers, how it affects the given dataset, and Either can we drop them or NOT. Let see now how to find from the given dataset. Are you ready!

We will look at simple methods first, Univariate and Multivariate analysis.

Univariate method: I believe you’re familiar with Univariate analysis, playing around one variable/feature from the given data set. Here to look at the Outlier we’re going to apply the BOX plot to understand the nature of the Outlier and where it is exactly.

Let see some sample code. Just I am taking chúng tôi as a sample for my analysis, here I am considering age for my analysis.

plt.figure(figsize=(5,5)) sns.boxplot(y='age',data=df_titanic)

You can see the outliers on the top portion of the box plot visually in the form of dots.

Multivariate method: Just I am taking titanic.csv as a sample for my analysis, here I am considering age and passenger class for my analysis.

plt.figure(figsize=(8,5)) sns.boxplot(x='pclass',y='age',data=df_titanic)We can very well use Histogram and Scatter Plot visualization technique to identify the outliers.

mathematically to find the Outliers as follows Z-Score and Inter Quartile Range (IQR) Score methods

Z-Score method: In which the distribution of data in the form mean is 0 and the standard deviation (SD) is 1 as Normal Distribution format.

Let’s consider below the age group of kids, which was collected during data science life cycle stage one, and proceed for analysis, before going into further analysis, Data scientist wants to remove outliers. Look at code and output, we could understand the essence of finding outliers using the Z-score method.

import numpy as np kids_age = [1, 2, 4, 8, 3, 8, 11, 15, 12, 6, 6, 3, 6, 7, 12,9,5,5,7,10,10,11,13,14,14] mean = np.mean(voting_age) std = np.std(voting_age) print('Mean of the kid''s age in the given series :', mean) print('STD Deviation of kid''s age in the given series :', std) threshold = 3 outlier = [] for i in voting_age: z = (i-mean)/std outlier.append(i) print('Outlier in the dataset is (Teen agers):', outlier) OutputThe outlier in the dataset is (Teenagers): [15]

(IQR) Score method: In which data has been divided into quartiles (Q1, Q2, and Q3). Please refer to the picture Outliers Scaling above. Ranges as below.

25th percentile of the data – Q1

50th percentile of the data – Q2

75th percentile of the data – Q3

Let’s have the junior boxing weight category series from the given data set and will figure out the outliers.

import numpy as np import seaborn as sns # jr_boxing_weight_categories jr_boxing_weight_categories = [25,30,35,40,45,50,45,35,50,60,120,150] Q1 = np.percentile(jr_boxing_weight_categories, 25, interpolation = 'midpoint') Q2 = np.percentile(jr_boxing_weight_categories, 50, interpolation = 'midpoint') Q3 = np.percentile(jr_boxing_weight_categories, 75, interpolation = 'midpoint') IQR = Q3 - Q1 print('Interquartile range is', IQR) low_lim = Q1 - 1.5 * IQR up_lim = Q3 + 1.5 * IQR print('low_limit is', low_lim) print('up_limit is', up_lim) outlier =[] for x in jr_boxing_weight_categories: outlier.append(x) print(' outlier in the dataset is', outlier) Outputthe outlier in the dataset is [120, 150]

sns.boxplot(jr_boxing_weight_categories)Loot at the boxplot we could understand where the outliers are sitting in the plot.

So far, we have discussed what is Outliers, how it looks like, Outliers are good or bad for data set, how to visualize using matplotlib /seaborn and stats methods.

Now, will conclude correcting or removing the outliers and taking appropriate decision. we can use the same Z- score and (IQR) Score with the condition we can correct or remove the outliers on-demand basis. because as mentioned earlier Outliers are not errors, it would be unusual from the original.

Hope this article helps you to understand the Outliers in the zoomed view in all aspects. let’s come up with another topic shortly. until then bye for now! Thanks for reading! Cheers!!

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## Closing The Data Science Skills Gap In India

The field of Data Science today has evolved much more and become the most demanded domain across industries. This field of study involves scientific methods, algorithms, processes, and systems to pull out meaningful insights from both structured and unstructured data for effective decision making and predictions. As a multi-disciplinary domain, data science has enabled businesses across the world to assess market trends, analyze users’ metrics, envisage potential business risks and make better decisions.

The field of Data Science today has evolved much more and become the most demanded domain across industries. This field of study involves scientific methods, algorithms, processes, and systems to pull out meaningful insights from both structured and unstructured data for effective decision making and predictions. As a multi-disciplinary domain, data science has enabled businesses across the world to assess market trends, analyze users’ metrics, envisage potential business risks and make better decisions. In India, businesses are quickly capitalizing on this newly emerging field to garner more from their data and deliver more value to customers, leading to a rise in demand of data scientists. According to a Great Learning report , the country is expected to see 1.5 lakh new openings in Data Science in 2023 , an increase of nearly 62% compared to the last year. As the competitive business landscape is evolving than ever, understanding the users and their preferences accurately has become critical for companies. This is where the role of data science comes into the scenario, making it possible to create and leverage time-saving automated models to get insights into a user’s purchase history, age, income level, and related demographics. Since the data science field is rising at an unprecedented rate, the huge demand for data science talent with less than five years of work experience is most among BFSI (38%), followed by Energy (13%), Pharma and Healthcare (12%), and E-Commerce (11%), among others. BFSI is the highest average salary offered industry to a Data Scientist stood at INR 13.56 LPA, which is followed by manufacturing and healthcare (INR 11.8 LPA each), and IT (INR 10.06 LPA), the report noted. Meanwhile, as companies across almost every industry are looking to acquire skilled and qualified talent who can navigate through everyday issues with innovative and right solutions, the supply of skilled professionals is far lesser than the demand. According to our estimations, there are major skills gaps in the field of big data professionals (58%) in 2023 at a global level. In order to close these gaps, academia, governments as well as organizations must carry out innovative ways to support future minds and derive value from that to spur economic growth. There is a need to implement re-skilling and up-skilling initiatives for data scientists at both corporate and academic levels. The data science professionals will also be expected to have specialized skills in data across all industries, making up-skilling a mandatory job. While the Great Learning report identified four unique career paths in the data science field including Data Scientist Data Analyst , Data Engineer and Business Intelligence Developer, the need for rapid up-skilling and adequate guidance has become indispensable. Moreover, investing in data-driven decisions and technology can also add value to businesses, easing the specialized talent shortage and enabling companies to drive efficiency.

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