Insta–DART (Daily Aptitude and Reasoning Test) 2020 - 21
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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
One card is drawn from a well shufflled one pack of cards. What is the probability of it being a diamond king ?
Correct
Ans-b
The total number of all possible events = 52C1 = 52
There will be 4 Kings in a pack. But only one Diamond King.
The number of favourable events = 1C1 = 1
Probability = (1C1)/(52C1) = (1/52)
∴Ans = (1/52)
Incorrect
Ans-b
The total number of all possible events = 52C1 = 52
There will be 4 Kings in a pack. But only one Diamond King.
The number of favourable events = 1C1 = 1
Probability = (1C1)/(52C1) = (1/52)
∴Ans = (1/52)
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Question 2 of 5
2. Question
2 cards are drawn from well shufflled one pack of cards. What is the probability of both being aces?
Correct
Ans- D , 1/221
The total number of all possible events =
52C2 = [(52!)]/[(2!(52-2)!]
=(52!)/(2!50!)
=[52* 51* 50 *49* ……..*3* 2 *1]/[2 *1 *50 *49* ………. *3* 2* 1]
=26 * 51
=1326
=The number of favourable events = 4C2
4C2 = (4!/(2!(4-2)! = 4!/(2!2!)=(4 *3* 2* 1)/(2* 1* 2* 1)=6
∴Probability = (4C2)/(52C2) =(6/1326)=(1/221)
∴Ans. (1/221)
Incorrect
Ans- D , 1/221
The total number of all possible events =
52C2 = [(52!)]/[(2!(52-2)!]
=(52!)/(2!50!)
=[52* 51* 50 *49* ……..*3* 2 *1]/[2 *1 *50 *49* ………. *3* 2* 1]
=26 * 51
=1326
=The number of favourable events = 4C2
4C2 = (4!/(2!(4-2)! = 4!/(2!2!)=(4 *3* 2* 1)/(2* 1* 2* 1)=6
∴Probability = (4C2)/(52C2) =(6/1326)=(1/221)
∴Ans. (1/221)
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Question 3 of 5
3. Question
There are 6 yellow and 8 black coloured balls in a bag. 2 balls are drawn from the bag. What is the probability of both being different coloured balles?
Correct
Ans-C
The total number of balls = 6 yellow +8 black
= 14
The total number of all possible events =14C2
14C2 = (14!)/[2!(14-2)!]
= [(14 *13* 12 *11* ….. *3 *2*1)/(2 *1* 12* 11* ……. 3 *2*1)
= 7* 13
=91
Two balls can be drawn from the bag in all in 91 different ways.
Number of favourable events = 6C1 and 8C1
=6* 8
=48
Probability = (6C1 * 8C1)/(14C1) = (6 * 8)/91
= 48/91
∴Ans = (48/91)
Incorrect
Ans-C
The total number of balls = 6 yellow +8 black
= 14
The total number of all possible events =14C2
14C2 = (14!)/[2!(14-2)!]
= [(14 *13* 12 *11* ….. *3 *2*1)/(2 *1* 12* 11* ……. 3 *2*1)
= 7* 13
=91
Two balls can be drawn from the bag in all in 91 different ways.
Number of favourable events = 6C1 and 8C1
=6* 8
=48
Probability = (6C1 * 8C1)/(14C1) = (6 * 8)/91
= 48/91
∴Ans = (48/91)
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Question 4 of 5
4. Question
There are 10 white, 6 red and 4 blue coloured balls in a bag. 3 balls are drawn at random. What is the probability of getting all the three different coloured balls?
Correct
Ans-C
Total number of balls = 10 white + 6 red +4 blue =20 balls
We are drawing 3 balls from the bag.
That can be done in all in 20C3 ways.
The total number of all possible events =20C3
20C3 = (20!)/(3!(20-3)!
=20 *19 *18 *17 ……..*3 *2* 1)/(3* 2 *1* 17* ….. *3* 2* 1)
20C3 = (10 *19* 6)
The number of favourable events = 10C1* 6C1 *4C1
=10* 6 *4
Probability = (10C1 * 6C1 *4C1)/(20C3)
= (10* 6*4)/(10* 19* 6)
Answer : (4/19)
Incorrect
Ans-C
Total number of balls = 10 white + 6 red +4 blue =20 balls
We are drawing 3 balls from the bag.
That can be done in all in 20C3 ways.
The total number of all possible events =20C3
20C3 = (20!)/(3!(20-3)!
=20 *19 *18 *17 ……..*3 *2* 1)/(3* 2 *1* 17* ….. *3* 2* 1)
20C3 = (10 *19* 6)
The number of favourable events = 10C1* 6C1 *4C1
=10* 6 *4
Probability = (10C1 * 6C1 *4C1)/(20C3)
= (10* 6*4)/(10* 19* 6)
Answer : (4/19)
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Question 5 of 5
5. Question
What is the probability of getting a prime number in the first 20 natural numbers?
Correct
Ans- B
Natural Numbers : 1, 2, 3, ……18, 19, 20
∴Total number of all possible events = 20
∴Prime numbers are 2, 3, 5, 7, 11, 13, 17 and 19
∴Number of favourable events = 8
∴Probability = (8/20) = (2/5)
Answer : (2/5)
Incorrect
Ans- B
Natural Numbers : 1, 2, 3, ……18, 19, 20
∴Total number of all possible events = 20
∴Prime numbers are 2, 3, 5, 7, 11, 13, 17 and 19
∴Number of favourable events = 8
∴Probability = (8/20) = (2/5)
Answer : (2/5)
