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
Romeo speaks the truth 3 times out of 8 times he speaks. While, Juliet speaks the truth 5 times out of 6 times he speaks. Find the probability of them contradicting each other when asked to answer a Yes/No question.
Correct
Solution: Option (c)
Explanation:
Probability that Romeo speaks the truth = 3/8
Probability that Juliet speaks the truth = 5/6
Romeo and Juliet do not contradict with each other if they both speak the truth or if they both speak lie.
Probability that they both speak the truth = (3/8) × (5/6) = 5/16
Probability that they both speak lie = (5/8) × (1/6) = 5/48
Required probability = 1 – (5/16) – (5/48) = 1 – [15/48 + 5/48] = 1 – (20/48) = 1 – 5/12 = 7/12
Incorrect
Solution: Option (c)
Explanation:
Probability that Romeo speaks the truth = 3/8
Probability that Juliet speaks the truth = 5/6
Romeo and Juliet do not contradict with each other if they both speak the truth or if they both speak lie.
Probability that they both speak the truth = (3/8) × (5/6) = 5/16
Probability that they both speak lie = (5/8) × (1/6) = 5/48
Required probability = 1 – (5/16) – (5/48) = 1 – [15/48 + 5/48] = 1 – (20/48) = 1 – 5/12 = 7/12
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Question 2 of 5
2. Question
A dice is thrown. If an even number comes, then a coin is tossed once, but if a prime number comes, then two coins are tossed. What is the probability of getting exactly one head?
Correct
Solution: Option (a)
Explanation:
Probability of getting even number when dice is thrown = 3/6 = 1/2
Probability of getting prime number = 3/6 = 1/2
Total events when 1 coin is tossed = {H, T}
Probability of getting 1 head when 1 coin is tossed = 1/2
Total events when 2 coins are tossed = {HH, HT, TH, TT}
Probability of getting 1 head when 2 coins are tossed = 3/4
Probability of getting exactly one head = (1/2 × 1/2 + 1/2 × 3/4) = 1/4 + 3/8 = 5/8 Hence, option (a) is the correct answer.
Incorrect
Solution: Option (a)
Explanation:
Probability of getting even number when dice is thrown = 3/6 = 1/2
Probability of getting prime number = 3/6 = 1/2
Total events when 1 coin is tossed = {H, T}
Probability of getting 1 head when 1 coin is tossed = 1/2
Total events when 2 coins are tossed = {HH, HT, TH, TT}
Probability of getting 1 head when 2 coins are tossed = 3/4
Probability of getting exactly one head = (1/2 × 1/2 + 1/2 × 3/4) = 1/4 + 3/8 = 5/8 Hence, option (a) is the correct answer.
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Question 3 of 5
3. Question
A natural number is selected such that it satisfies 1 ≤x ≤ 100. The probability that the number satisfy the inequality 3x + 4 ≥ 229, is
Correct
Solution: Option (c)
Explanation:
13x + 4 ≥ 30, x ≥ 75 means all numbers from 75 to 100 = 25 numbers
Total available numbers from 1 to 100 = 100
Required probability = 25/100 =1/4 Hence option (c) is correct answer.
Incorrect
Solution: Option (c)
Explanation:
13x + 4 ≥ 30, x ≥ 75 means all numbers from 75 to 100 = 25 numbers
Total available numbers from 1 to 100 = 100
Required probability = 25/100 =1/4 Hence option (c) is correct answer.
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Question 4 of 5
4. Question
An anti-aircraft gun can take a maximum of 4 shots at an enemy plane moving away from it. The probabilities of hitting the plane at the first, second, third and fourth shot are 0.4, 0.3, 0.2 and 0.1, respectively. What is the probability that the gun hits the plane?
Correct
Solution: Option (a)
Explanation:
Required probability = P(hitting in 1st shot) + P(hitting in 2nd shot) + P(hitting in
3rd shot) + P(hitting in 4th shot)
= 0.4 + (1-0.4)(0.3) + (1-0.4)(1-0.3)(0.2) + (1-0.4)(1-0.3)(1-0.2)0.1 =0 .6976 Hence option (a) is correct answer.
Incorrect
Solution: Option (a)
Explanation:
Required probability = P(hitting in 1st shot) + P(hitting in 2nd shot) + P(hitting in
3rd shot) + P(hitting in 4th shot)
= 0.4 + (1-0.4)(0.3) + (1-0.4)(1-0.3)(0.2) + (1-0.4)(1-0.3)(1-0.2)0.1 =0 .6976 Hence option (a) is correct answer.
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Question 5 of 5
5. Question
In a musical chair game, the person playing the music has been advised to stop playing the music at any time within 2 minutes after she starts playing. What is the probability that the music will stop within the first half-minute after starting?
Correct
Solution: Option (d)
Explanation:
Here, the possible outcomes are all the numbers between 0 and 2.
Let E be the event that “the music is stopped within the first half-minute”. The outcomes favourable to E are points on the number line from 0 to 1/2
Since all the outcomes are equally likely, we can argue that, of the total distance of 2, the distance favourable to the event E is 1/2.
So, P(E) = Distance favourable to the event E/ Total distance in which outcomes can lie = (1/2)/2 =1/4 Hence, option (d) is the correct answer.
Incorrect
Solution: Option (d)
Explanation:
Here, the possible outcomes are all the numbers between 0 and 2.
Let E be the event that “the music is stopped within the first half-minute”. The outcomes favourable to E are points on the number line from 0 to 1/2
Since all the outcomes are equally likely, we can argue that, of the total distance of 2, the distance favourable to the event E is 1/2.
So, P(E) = Distance favourable to the event E/ Total distance in which outcomes can lie = (1/2)/2 =1/4 Hence, option (d) is the correct answer.