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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).
Wish you all the best ! 🙂
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Question 1 of 5
1. Question
If three balls are picked at random, what 1s the probability that atleast one is red?
Correct
Answer : c
Total number of outcomes= 10C3 = 120
Number of outcomes not containing red balls= 5C3 = 10
∴Probability that at least one is red = 1 – (10/120) = (11/12)
Incorrect
Answer : c
Total number of outcomes= 10C3 = 120
Number of outcomes not containing red balls= 5C3 = 10
∴Probability that at least one is red = 1 – (10/120) = (11/12)
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Question 2 of 5
2. Question
If four balls are picked at random, what is the probability that two are green and two are blue?
Correct
Answer : b
Total number of outcomes= 10C4 = 210
Favourable number of outcomes = 3C2 x 2C2 = 3 x1 = 3
∴Required probability= (3/210) = (1/70)
Incorrect
Answer : b
Total number of outcomes= 10C4 = 210
Favourable number of outcomes = 3C2 x 2C2 = 3 x1 = 3
∴Required probability= (3/210) = (1/70)
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Question 3 of 5
3. Question
Three mangoes and three apples are kept in a box. if two fruits are chosen at random, find the probability that one is a mango and the other 1s an apple.
Correct
Answer : b
Total number of ways = n(S) = 6C2 = 15
Favourable number of ways = n(E) = 3C1x3C1 = 9
∴Required probability = (9/15) = (3/5)
Incorrect
Answer : b
Total number of ways = n(S) = 6C2 = 15
Favourable number of ways = n(E) = 3C1x3C1 = 9
∴Required probability = (9/15) = (3/5)
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Question 4 of 5
4. Question
Find the number of combinations that can be formed with 5 oranges, 4 mangoes and 3 bananas, when one fruit of each kind is taken.
Correct
Answer : a
The required number of combinations when one fruit of each kind is taken
= 5C1 x 4C1 x 3C1 = 5x4x 3 = 60
Incorrect
Answer : a
The required number of combinations when one fruit of each kind is taken
= 5C1 x 4C1 x 3C1 = 5x4x 3 = 60
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Question 5 of 5
5. Question
In how many ways, 12 balls can be divided between 2 boys, one receiving 5 and the other 7 balls?
Correct
Answer : b
Here, order is important then the number of ways in which 12 different then can be divided between two boys who receives 5 and 7 balls respectively, is
= (12!)/(5!7!) x2! = 1584
Incorrect
Answer : b
Here, order is important then the number of ways in which 12 different then can be divided between two boys who receives 5 and 7 balls respectively, is
= (12!)/(5!7!) x2! = 1584
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