# Important Probability Questions and Answers For Competitive Exams

Probability is essentially the possibility of happening of an event. Probability is an important topic usually taught in classes 10th and 11th. Increasingly probability is finding a place in competitive exams. Due to the nature of section this is a relatively difficult section for many. Acing probability will be essential to get a good rank in any exam. In day to day life too probability is important, as all the events have certain likelihood of happening. Here we have covered few fundamentals for probability followed by few question for your practice.

The theory of probabilities is simply the Science of logic quantitatively treated. – C.S. PEIRCE

## Important Probability Questions and Answers For Government Job Exams

Definition 1 If E and F are two events associated with the same sample space of a random experiment, the conditional probability of the event E given that F has occurred, i.e.
P (E|F) is given by P(E|F) = P(E F)/ P(F) ∩ provided P(F) ≠ 0
Definition 2
Two events E and F are said to be independent, if
P(F|E) = P (F) provided P (E) ≠ 0
and P (E|F) = P (E) provided P (F) ≠ 0
Thus, in this definition we need to have P (E) ≠ 0 and P(F) ≠ 0
Now, by the multiplication rule of probability, we have
P(E ∩ F) = P(E) . P (F|E) ... (1)
If E and F are independent, then (1) becomes
P(E ∩ F) = P(E) . P(F) ... (2)
Thus, using (2), the independence of two events is also defined as follows:
Definition 3 Let E and F be two events associated with the same random experiment,
then E and F are said to be independent if
P(E ∩ F) = P(E) . P (F)
Impossible event: The probability of an occurrence/event impossible to happen is 0. Such an event is called an impossible event.
Sure event: The probability of an event which is sure to occur is 1. Such an event is known as sure event or a certain event.

Q.1) The letters of the word NATURE are to be arranged so that three vowels should not come together. Find the number of arrangement.
[1]576
[2]225
[3]486
[4]785
[5]512

Q.2) From group of 9 men and 6 women a committee has to be formed of 6 members in which at least 3 women are required. Find the number of ways to form the committee.
[1]2225
[2]2250
[3]4236 v [4]2275
[5]5134

Q.3) Find the probability of forming a seven number Team for a project from a group of 8 workers and 6 managers such that no even numbers of managers are selected in the project?

[1] 215 429
[2] 115 231
[3] 155 231
[4] 217 319
[5] 217 429

Q.4) Find the probability of forming a five number Team for a project from a group of 6 workers and 5 managers such that no odd numbers of managers are selected in the project?
[1] 215 231
[2] 115 231
[3] 155 231
[4] 115 241
[5] 125 231

Q.5) From a group of 9 men and 8 women, 7 persons are to be selected to form a committee so that at least 5 men are there on the committee. In how many ways can this be done?
[1]5634
[2]3563
[3]4236
[4]4754
[5]5134

Q.6) In how many ways can the letter of the word ‘ARRANGEMENT’ be arranged?
[1]6027000
[2]5343232
[3]7590000
[4]6957000
[5]6237000

Q.7) In how many ways the letter of the word ‘MANUAL’ can be arranged so that vowels come together?
[1]72
[2]65
[3]82
[4]94
[5]88

Q.8) A basket contains 8 black, 5 red, 7 pink and 3 purple balls. If 2 balls are picked randomly, then what is the possibility that either both balls are red or pink?
[1] 13 253
[2] 31 245
[3] 51 203
[4] 31 253
[5] 47 211

Q.9)A bag contains 12 brown balls and 9 grey balls. Two balls are drawn randomly. What is the probability that they are of the same colour?
[1] 10 31
[2] 17 35
[3] 16 33
[4] 17 33
[5] 11 35
Q.10)There are 5 red balls, 4 yellow balls and 3 pink balls in a bag. Two balls are drawn randomly. What is the probability that none of the drawn balls is of pink colour?
[1] 3 5
[2] 5 12
[3] 7 13
[4] 6 11
[5] 6 17

Q.11)A bag contains 43 cards (numbered 1, 2, 3 ……., 43). Two cards are picked at random from the bag. What is the probability that the sum of numbers of both the cards that are drawn is even?
[1] 31 53
[2] 21 43
[3] 22 43
[4] 21 41
[5] 11 13

Q.12)A bag contains 50 cards (numbered 1,2, 3, 4, …….,50) Two cards are picked at random from the bag. What is the probability that the first card drawn has a prime number and the second card drawn has a number which is multiple of 15?
[1] 7 400
[2] 9 410
[3] 9 488
[4] 11 490
[5] 9 490

Directions (13- 15): Study the following information carefully and answer the questions:
A bag contains 5 red balls, 8 blue balls and 3 pink balls.

Q.13)Three balls are drawn randomly. What is the probability that exactly one of them is pink ball?
[1] 110 343
[2] 110 291
[3] 117 280
[4] 113 220
[5] 117 200

Q.14) One ball is drawn randomly. What is the probability that it is either red or blue ball?
[1] 13 16
[2] 11 16
[3] 13 17
[4] 9 16
[5] 7 11

Q.15) Three balls are drawn randomly. What is the probability that all of them are red?
[1] 1 61
[2] 1 58
[3] 11 56
[4] 1 56
[5] 5 56

Solution:

Total letters = 6
Number of vowels = 3 (AUE)
Number of words when three vowels come together = AUE NTR
= > 4! × 3!
Required arrangement = 6! – (4! × 3!) = 720 – 144 = 576

Solution:
Possible cases = (3W 3M) (4W 2M) (5W 1M) (6W)
Number of ways = 6C3 9C3 + 6C4 9C2 + 6C5 9C1 + 6C6
= >1680 + 540 + 54 + 1
= > 2275

Possible cases = (6W 1M) (4W 3M) (2W 5M)
= 8C6 × 6C1 + 8C4 × 6C3 + 8C2 × 6C5
=168 + 1400 + 168 = 1736 Required probability = 1736 14𝐶 7 = 1736 3432 = 217 429

Possible cases = (3W 2M) (1W 4M)
= 6C3 × 5C2 + 6C1 × 5C4
=200 + 30 = 230 Required probability = 230 11𝐶5 = 230 462 = 115 231

Number of committee = 9C5 × 8C2 + 9C6 × 8C1 + 9C7
= 126 × 28 + 84 × 8 + 36
= 3528 + 672 + 36
= 4236

Word ‘ARRANGEMENT’ consist of 11 letters in which alphabets A, R, N and E comes twice.
Therefore, number of arrangements’ = 11! 2! ×2! ×2! ×2! = 11 ×10 ×9 ×8 ×7 ×6 ×5 ×4 ×3 ×2 ×1 2 ×2 ×2 ×2
= 6237000

MANUAL = (MNL) (AUA)
Number of arrangement,
= 4! ×3! 2!
= 72

Solution:
Required probability,
= 5C2 +7C2 23C2 = 10+21 253 = 31 253

Solution:
Required probability,
= 12C2 +9C2 21C2 = 66+36 210 = 102 210 = 17 35

Solution:
Required probability, = 9C2 12C2 = 36 66 = 6 11

Solution:
Total even numbers between 1 and 43 = 21
Total odd numbers between 1 and 43 = 22 When both cards are odd = 22 43 × 21 42 = 11 43 When both cards are even = 21 43 × 20 42 = 10 43 Therefore, required probability = 11 43 + 10 43 = 21 43

Solution:
Prime numbers = 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43 and 47
Total = 15
Multiples of 15 = 15, 30 and 45 Therefore, required probability = 15 50 × 3 49 = 9 490

Solution:
Required probability = 3C1 ×13 C2 16C3
= 234 560
= 117 280