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
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
Sol : Option 4
Reqd. Probability = 1 – (A not winning even one Game out of 3) =1 - (1/3)2 = 1 - (1/9) = (8/9)
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
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
Directions for questions 4 to 6: Read the following information and answer the questions that follow. The probability that A will pass the examination is 1/2 and the probability that B will pass the examination is 1/3.
What is the probability that both A and B will pass the examination?
1. 1/6
2. 1
3. 2/3
4. 1/3
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
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
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
What is the probability that at least one person will pass the examination?
1. 1
2. 1/2
3. 1/3
4. 2/3
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
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
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
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?
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
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
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)
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
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
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