UPSC Insta–DART (Daily Aptitude and Reasoning Test) 18 Oct 2024
Kartavya Desk Staff
Considering the alarming importance of CSAT in UPSC CSE Prelims exam and with enormous requests we received recently, InsightsIAS has started Daily CSAT Test to ensure students practice CSAT Questions on a daily basis. Regular Practice would help one overcome the fear of CSAT too.We are naming this initiative as Insta– DART – Daily Aptitude and Reasoning Test. We hope you will be able to use DART to hit bull’s eye in CSAT paper and comfortably score 100+ even in the most difficult question paper that UPSC can give you in CSP-2021. Your peace of mind after every step of this exam is very important for us.
Looking forward to your enthusiastic participation (both in sending us questions and solving them on daily basis on this portal).
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• Question 1 of 5 1. Question An instructor has a question bank consisting of 200 easy True / False questions, 300 difficult True / False questions, 400 easy multiple-choice questions and 500 difficult multiple-choice questions. If a question is selected at random from the question bank, what is the probability that it will be an easy question given that it is a multiple-choice question? a) 4/5 b) 4/9 c) 1/9 d) 1/2 Correct Answer: B Explanation Let us denote E = easy questions, M = multiple choice question, D = difficult questions and T = true/false questions. Total number of questions = 1400 Total number of multiple-choice questions = 900 Probability of selecting easy multiple-choice question is P ( E ∩ M) = 400/1400 = 2/7 Probability of selecting multiple is P ( M) = 900/1400 = 9/14 P(E|M) represents the probability that a randomly selected question will be an easy question, given that is a multiple-choice question. Therefore, P(E|M) = P( E ∩ M)/P(M) = (2/7)/(9/14) = 4/9 Therefore, required probability = 4/9 Incorrect Answer: B Explanation Let us denote E = easy questions, M = multiple choice question, D = difficult questions and T = true/false questions. Total number of questions = 1400 Total number of multiple-choice questions = 900 Probability of selecting easy multiple-choice question is P ( E ∩ M) = 400/1400 = 2/7 Probability of selecting multiple is P ( M) = 900/1400 = 9/14 P(E|M) represents the probability that a randomly selected question will be an easy question, given that is a multiple-choice question. Therefore, P(E|M) = P( E ∩ M)/P(M) = (2/7)/(9/14) = 4/9 Therefore, required probability = 4/9
#### 1. Question
An instructor has a question bank consisting of 200 easy True / False questions, 300 difficult True / False questions, 400 easy multiple-choice questions and 500 difficult multiple-choice questions. If a question is selected at random from the question bank, what is the probability that it will be an easy question given that it is a multiple-choice question?
Explanation
Let us denote E = easy questions,
M = multiple choice question,
D = difficult questions and
T = true/false questions.
Total number of questions = 1400
Total number of multiple-choice questions = 900
Probability of selecting easy multiple-choice question is P ( E ∩ M) = 400/1400 = 2/7
Probability of selecting multiple is P ( M) = 900/1400 = 9/14
P(E|M) represents the probability that a randomly selected question will be an easy question, given that is a multiple-choice question.
Therefore, P(E|M) = P( E ∩ M)/P(M) = (2/7)/(9/14) = 4/9
Therefore, required probability = 4/9
Explanation
Let us denote E = easy questions,
M = multiple choice question,
D = difficult questions and
T = true/false questions.
Total number of questions = 1400
Total number of multiple-choice questions = 900
Probability of selecting easy multiple-choice question is P ( E ∩ M) = 400/1400 = 2/7
Probability of selecting multiple is P ( M) = 900/1400 = 9/14
P(E|M) represents the probability that a randomly selected question will be an easy question, given that is a multiple-choice question.
Therefore, P(E|M) = P( E ∩ M)/P(M) = (2/7)/(9/14) = 4/9
Therefore, required probability = 4/9
• Question 2 of 5 2. Question A volleyball team of 6 players is to be selected from a group of 8 male and 7 female players. In how many ways is the team selected such that at most two female players are there in the team. a) 1470 b) 1598 c) 1762 d) 1890 Correct Answer: D Explanation Case I: Two female players are there in the team Number of ways to select the team = 8C4 × 7C2 = 70 × 21 = 1470 Case II: Only one female player is there in the team Number of ways to select the team = 8C5 × 7C1 = 7 × 56 = 392 Case III: No female players is there in the team Number of ways to select the team = 8C6 = 28 So, total number of ways = (1470 + 392 + 28) = 1890 Incorrect Answer: D Explanation Case I: Two female players are there in the team Number of ways to select the team = 8C4 × 7C2 = 70 × 21 = 1470 Case II: Only one female player is there in the team Number of ways to select the team = 8C5 × 7C1 = 7 × 56 = 392 Case III: No female players is there in the team Number of ways to select the team = 8C6 = 28 So, total number of ways = (1470 + 392 + 28) = 1890
#### 2. Question
A volleyball team of 6 players is to be selected from a group of 8 male and 7 female players. In how many ways is the team selected such that at most two female players are there in the team.
Explanation
Case I: Two female players are there in the team
Number of ways to select the team = 8C4 × 7C2 = 70 × 21 = 1470
Case II: Only one female player is there in the team
Number of ways to select the team = 8C5 × 7C1 = 7 × 56 = 392
Case III: No female players is there in the team
Number of ways to select the team = 8C6 = 28
So, total number of ways = (1470 + 392 + 28) = 1890
Explanation
Case I: Two female players are there in the team
Number of ways to select the team = 8C4 × 7C2 = 70 × 21 = 1470
Case II: Only one female player is there in the team
Number of ways to select the team = 8C5 × 7C1 = 7 × 56 = 392
Case III: No female players is there in the team
Number of ways to select the team = 8C6 = 28
So, total number of ways = (1470 + 392 + 28) = 1890
• Question 3 of 5 3. Question Three houses are available in a locality. Three persons apply for the houses. Each applies for one house without consulting others. The probability that all the three apply for the same house is a) 1/9 b) 2/9 c) 5/9 d) 7/9 Correct Answer: A Explanation One person can select one house out of 3= 3C1 ways =3. Hence, three persons can select one house out of 3 in 3 x 3 x 3 =9 Therefore, probability that all thre apply for the same house is 1/9 Incorrect Answer: A Explanation One person can select one house out of 3= 3C1 ways =3. Hence, three persons can select one house out of 3 in 3 x 3 x 3 =9 Therefore, probability that all thre apply for the same house is 1/9
#### 3. Question
Three houses are available in a locality. Three persons apply for the houses. Each applies for one house without consulting others. The probability that all the three apply for the same house is
Explanation
One person can select one house out of 3= 3C1 ways =3.
Hence, three persons can select one house out of 3 in 3 x 3 x 3 =9
Therefore, probability that all thre apply for the same house is 1/9
Explanation
One person can select one house out of 3= 3C1 ways =3.
Hence, three persons can select one house out of 3 in 3 x 3 x 3 =9
Therefore, probability that all thre apply for the same house is 1/9
• Question 4 of 5 4. Question There are 5 English, 4 Hindi and 3 regional newspaper options available in a library. In how many ways the owner can subscribe to five newspapers such that there are at least two English and two Hindi newspapers? a) 230 b) 240 c) 220 d) 280 Correct Answer: D Explanation There are three possible cases Case 1: 3 English + 2 Hindi – No of ways = 5C3× 4C2 = 60 Case 2: 2 English + 3 Hindi – No of ways = 5C2× 4C3 = 40 Case 3: 2 English + 2 Hindi + 1 regional – No of ways = 5C2× 4C2 × 3C1 = 180 Total number of ways = (60 +40 + 180) = 280 Incorrect Answer: D Explanation There are three possible cases Case 1: 3 English + 2 Hindi – No of ways = 5C3× 4C2 = 60 Case 2: 2 English + 3 Hindi – No of ways = 5C2× 4C3 = 40 Case 3: 2 English + 2 Hindi + 1 regional – No of ways = 5C2× 4C2 × 3C1 = 180 Total number of ways = (60 +40 + 180) = 280
#### 4. Question
There are 5 English, 4 Hindi and 3 regional newspaper options available in a library. In how many ways the owner can subscribe to five newspapers such that there are at least two English and two Hindi newspapers?
Explanation
There are three possible cases
Case 1: 3 English + 2 Hindi – No of ways = 5C3× 4C2 = 60
Case 2: 2 English + 3 Hindi – No of ways = 5C2× 4C3 = 40
Case 3: 2 English + 2 Hindi + 1 regional – No of ways = 5C2× 4C2 × 3C1 = 180
Total number of ways = (60 +40 + 180) = 280
Explanation
There are three possible cases
Case 1: 3 English + 2 Hindi – No of ways = 5C3× 4C2 = 60
Case 2: 2 English + 3 Hindi – No of ways = 5C2× 4C3 = 40
Case 3: 2 English + 2 Hindi + 1 regional – No of ways = 5C2× 4C2 × 3C1 = 180
Total number of ways = (60 +40 + 180) = 280
• Question 5 of 5 5. Question A black and a red dice are rolled. Find the conditional probability of obtaining a sum greater than 9, given that the black die resulted in a 5. a) 1/2 b) 1/4 c) 1/3 d) 1/6 Correct Answer: C Explanation Let the first observation be from the black die and second from the red die. When two dice (one black and another red) are rolled, the sample space S = 6 × 6 = 36 Now, A : obtaining a sum greater than 9 = {(4, 6), (5, 5), (5, 6), (6, 4), (6, 5), (6, 6)} B: Black die results in a 5. = {(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)} Therefore, (A ∩ B) = {(5, 5), (5, 6)} The conditional probability of obtaining a sum greater than 9, given by the black die resulting in a 5, is given by P( A|B). Therefore, P(A|B) = P(A ∩ B)/ P(B) = (2/36)/(6/36) = 2/6 = 1/3 Incorrect Answer: C Explanation Let the first observation be from the black die and second from the red die. When two dice (one black and another red) are rolled, the sample space S = 6 × 6 = 36 Now, A : obtaining a sum greater than 9 = {(4, 6), (5, 5), (5, 6), (6, 4), (6, 5), (6, 6)} B: Black die results in a 5. = {(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)} Therefore, (A ∩ B) = {(5, 5), (5, 6)} The conditional probability of obtaining a sum greater than 9, given by the black die resulting in a 5, is given by P( A|B). Therefore, P(A|B) = P(A ∩ B)/ P(B) = (2/36)/(6/36) = 2/6 = 1/3
#### 5. Question
A black and a red dice are rolled. Find the conditional probability of obtaining a sum greater than 9, given that the black die resulted in a 5.
Explanation
Let the first observation be from the black die and second from the red die.
When two dice (one black and another red) are rolled, the sample space S = 6 × 6 = 36
Now, A : obtaining a sum greater than 9 = {(4, 6), (5, 5), (5, 6), (6, 4), (6, 5), (6, 6)}
B: Black die results in a 5. = {(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)}
Therefore, (A ∩ B) = {(5, 5), (5, 6)}
The conditional probability of obtaining a sum greater than 9, given by the black die resulting in a 5, is given by P( A|B).
Therefore, P(A|B) = P(A ∩ B)/ P(B) = (2/36)/(6/36) = 2/6 = 1/3
Explanation
Let the first observation be from the black die and second from the red die.
When two dice (one black and another red) are rolled, the sample space S = 6 × 6 = 36
Now, A : obtaining a sum greater than 9 = {(4, 6), (5, 5), (5, 6), (6, 4), (6, 5), (6, 6)}
B: Black die results in a 5. = {(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)}
Therefore, (A ∩ B) = {(5, 5), (5, 6)}
The conditional probability of obtaining a sum greater than 9, given by the black die resulting in a 5, is given by P( A|B).
Therefore, P(A|B) = P(A ∩ B)/ P(B) = (2/36)/(6/36) = 2/6 = 1/3
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