1. What is the probability of getting at least one 3 in 2 throws of a dice?
   1. 1/36

   2. 5/36

   3. 11/36

   4. 35/36
   Answer & Explanation
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   Sol : Option 3
   Total possible outcomes = 36.
   Favorable cases = 11 i.e. (3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6). (1, 3) (2, 3) (4, 3) ,(5, 3) ,(6, 3)
   Therefore, Reqd. Probability = 11 / 36
2. The probability of A’s winning a game of chess against B is 2/3. What is the probability that A will win at least 1 of a total of two games?
   1. 1/27

   2. 19/27

   3. 2/27

   4. 8/9
   Answer & Explanation
   Sol : Option 4
   Reqd. Probability = 1 – (A not winning even one Game out of 3) =1 - (1/3)² = 1 - (1/9) = (8/9)
3. What is the probability that a non-leap year will have 53 Mondays?
   1. 52/53

   2. 51/52

   3. 1

   4. None of these
   Answer & Explanation
   Sol : Option 4
   Non Leap Year = 52 Weeks and one Day.
   Prob. of one days Day being Monday = 1 / 7
   Therefore, Reqd. Prob. = 1 / 7

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4. What is the probability that both A and B will pass the examination?
   1. 1/6

   2. 1

   3. 2/3

   4. 1/3
   Answer & Explanation
   Sol : Option 1
   Probability that A will pass in exam = 1/2
   So, Probability that A will fail in exam = 1/2.
   Probability that B will pass in exam = 1/3
   So, Probability that B will fail in exam = 2/3.
   Probability that both will pass in the exam = Probability that A will pass and probability that B will pass
   = 1/2 x 1/3 = 1/6
5. What is the probability that only 1 person [either A or B] will pass the examination?
   1. 1

   2. 1/2

   3. 1/3

   4. 2/3
   Answer & Explanation
   Sol : Option 2
   Probability that only one person will pass is when A passes and B fails or A fails and B passes.
   = (1/2) x (2/3) + (1/2) x (1/3) = 1/2

6. What is the probability that at least one person will pass the examination?
   1. 1

   2. 1/2

   3. 1/3

   4. 2/3
   Answer & Explanation
   Sol : Option 4
   Probability that at least one person, will pass, so the possibilities
   can be (A pass and B fails), or (A fails and B pass) or (Both A and B pass)
   (1/2) x (2/3) + (1/2) x (1/3) + (1/2) x (1/3) = 2/3
7. Three cards are drawn together from a pack of 52 cards at random. What is the probability that all the cards are Diamonds?
   1. ⁴C₃ / ²C₃

   2. ¹³C₃ / ⁵²C₃

   3. ²⁶C₃ / ⁵²C₃

   4. ⁸C₃ / ⁵²C₃
   Answer & Explanation
   Sol : Option 2
   There are 13 diamonds. Three diamonds out of 13 diamonds can be taken out in ¹³C₃ ways.
   Total number of sample spaces = ⁵²C₃
   Required probability = ¹³C₃ / ⁵²C₃
8. A bag contains 8 blue balls and 6 black balls. Three balls are drawn one by one with replacement. What is the probability that all the 3 balls are black?
   1. 27 / 343

   2. 1 / 343

   3. 18 / 343

   4. None of these
   Answer & Explanation
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   Sol : Option 1
   P (of black ball in first attempt) = 6 / 14 = 3 / 7
   Here probability will remain same for the next two attempt.
   Therefore, Required Probability = (3/7) x (3/7) x (3/7) = 27 / 343
9. One bag contains 8 blue balls and 6 Green balls; another bag contains 7 blue balls and 5 green balls. If one ball is drawn from each bag, determine the probability that both are blue?
   1. 1/2

   2. 1/3

   3. 1/4

   4. 1/5
   Answer & Explanation
   Sol : Option 2

   Bag I	Bag II
   8 blue	7 blue
   6 green	5 green

   Probability of getting blue ball from bag I = 8 / 14
   Probability of getting blue ball from bag II = 7 / 12
   Hence reqd. Prob. = (8/14) x (7/12) = (1/3)
10. 1 ball is drawn at random from a box containing 4 red balls, 5 white balls and 6 blue balls, what is the probability that the ball is a red ball?
    1. 1/7

    2. 2/15

    3. 4/15

    4. 1/15
    Answer & Explanation
    Sol : Option 3
    Total possible outcomes = 15. (i.e. One out of 15 balls).
    Favorable outcomes (one out of 4 red balls) = 4.
    Reqd. Probability = 4 / 15