# Probability Class 11 Mathematics Exam Questions

Please refer to Probability Class 11 Mathematics Exam Questions provided below. These questions and answers for Class 11 Mathematics have been designed based on the past trend of questions and important topics in your class 11 Mathematics books. You should go through all Class 11 Mathematics Important Questions provided by our teachers which will help you to get more marks in upcoming exams.

## Class 11 Mathematics Exam Questions Probability

Class 11 Mathematics students should read and understand the important questions and answers provided below for Probability which will help them to understand all important and difficult topics.

Question. Suppose that each child born is equally likely to be a boy or a girl. Consider a family with exactly three children.
(i) List the eight elements in the sample space whose outcomes are all possible genders of the three children.
(ii) Write each of the following events as a set and find its probability.
(a) The event that exactly one child is a girl.
(b) The event that atleast two children are girls.
(c) The event that no child is a girl.
Answer. (i) Let B denotes a boy and G denotes a girl. Then, all possible genders are expressed as
S = {BBB, BBG, BGB, GBB, BGG,GBG, GGB, GGG}
(ii) (a) Let E1 denotes the event that exactly one child is a girl.
Then, E1 = BBG, BGB, GBB
⇒P (E1) = 3/8
(b) Let E2 denotes the event that atleast two children are girls.
Then, E2 = BGG, GBG, GGB, GGG}

Question. A fair coin with 1 marked on one face and 6 marked on the other and a fair die are both tossed. Find the probability that the sum of numbers that turn up is
(i) 3 (ii) 12.
Answer. Let S be a sample space associated with the given experiment.
Then, S = {(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),
(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)} ⇒ n(S) = 12
(i) Let A be the event of getting the sum of numbers 3.
Then, A = {(1,2)}

Question. Describe the sample space for the experiment of tossing a coin four times.
Answer. Let us denote head by H and tail by T, then the possibilities are

Thus, the sample space for this experiment is given by
[{HHHH}, {HHHT}, {HHTH}, {HHTT}, {HTHH},
{HTHT}, {HTTH}, {HTTT}, {THHH}, {THHT}, {THTH},
{THTT}, {TTHH}, {TTHT}, {TTTH}, {TTTT}]

Question. Three coins are tossed once. Let A denotes the event ‘‘three heads show’’, B denotes the event
‘‘two heads and one tail show’’, C denotes the event
‘‘three tails show’’ and D denotes the event ‘‘a head shows on the first coin’’. Which pair of events are
(i) mutually exclusive events.
(ii) compound events.
Answer. When three coins are tossed, then the sample space is
S = {HHH,HHT,HTH, THH,HTT,THT, TTH,TTT}
Given, A = event of getting three heads
∴ A = {HHH}
B = event of getting two heads and one tail
∴ B = {HHT, HTH, THH}
C = event of getting three tails
∴ C = {TTT}
and D = event of getting a head on first coin.
∴ D = {HHH, HTT, HHT, HTH}
(i) Clearly, A ∩ B =f, A ∩ C = f,
A ∩ D = {HHH}, B ∩ C = f,
B ∩ D = {HHT, HTH} and C ∩ D = f
∴We can say A and B, A and C, B and C, C and D are mutually exclusive events.
(ii) B and D are compound event, since they have more than one sample point.

Question. A die is thrown. Find
(i) P (a prime number) (ii) P(a number ≥ 3)
(iii) P(a number ≤1)
Answer. In throwing a die, the sample space will be
S = {1, 2, 3, 4, 5, 6}
∴ n(S) = 6
(i) Let E1 be the event of getting a prime number.
Then, E1 = {2, 3, 5}
∴ n(E1) = 3
Now, probability of getting a prime number,

Question. Three of six vertices of a regular hexagon are chosen at random, what is the probability that the triangle with these vertices is equilateral?
Answer. Let ABCDEF be a regular hexagon.

Clearly, out of 6 vertices, 3 vertices can be chosen in 6C3 ways = 20ways.
Thus, total number of triangles so formed = 20
[no three points are collinear]
Note that, out of these 20 triangles only
ΔACE and ΔBDF are equilateral triangles.
∴ Required probability = 2/20 = 1/10

Question. In a relay race, there are five teams A, B, C, D and E.
(i) What is the probability that A, B and C finish first, second and third, respectively.
(ii) What is the probability that A, B and C are first three to finish (in any order), assume that all finishing orders are equally likely.
Answer. The sample space consisting of all finishing orders in the first three places, will have

(i) If A, B and C finish first, second and third, respectively, then there is only one finishing order for this i.e. ABC.
= 1/60
(ii) If A, B and C are the first three finishers, then there will be 3! arrangements for A, B and C. Thus, the sample points corresponding to this event will be 3! in number.
Hence, P(A, B and C are first three to finish)
= 2/60 = 1/10

Question. Determine the probability p, for each of the following events.
(i) An odd number appears in a single toss of a fair die.
(ii) Atleast one head appears in two tosses of a fair coin.
(iii) 4 king, 2 of hearts or 3 of spades appears in drawing a single card from a well shuffled ordinary deck of 52 cards.
(iv) The sum of 6 appears in a single toss of a pair of fair dice.
Answer. 6. (i) When a die is thrown the possible outcomes are
S = {1, 2, 3, 4, 5, 6} out of which 1, 3, 5 are odd.
∴ Required probability = 3/6 = 1/2
(ii) When a fair coin is tossed two times, the sample space is
S = {HH,HT, TH, TT}
In atleast one head favourable outcomes are
HH,HT, TH.
∴ Required probability = 3/4
(iii) Total cards = 52
Favourable = 4 king + 2 of heart + 3 of spade
= 4 + 1 + 1 = 6
∴ Required probability = 6/52 = 2/26
(iv) When a pair of dice is rolled total sample parts are 36.
Out of which (1, 5), (5, 1), (2, 4), (4, 2) and (3, 3) represent the sum as 6.
∴ Required probability = 5/36

Question. A die is loaded in such a way that each odd number is twice as likely to occur as each even number.
Find P(G), where G is the event that a number greater than 3 occurs on a single roll of the die.
Answer. Let E1 denote the event ‘getting number i on the upper face of the die’, i = 1, 2, 3, 4, 5, 6.
Clearly, E1, i = 1, 2, 3, 4, 5, 6 are mutually exclusive and exhaustive events.

Question. Four cards are drawn at random from pack of 52 playing cards. Find the probability of getting
(i) all face cards.
(ii) two red cards and two black cards.
(iii) one card from each suit.
Answer. Total number of possible outcomes = 52C4
(i) We know that, there are 12 face cards.
∴ Number of favourable outcomes = 12C4

Question. Six new employees, two of whom are married to each other, are to be assigned six desks that are lined up in a row. If the assignment of employees to desks is made randomly, what is the probability that the married couple will not have adjacent desks?
Answer. Let the couple occupied adjacent desks consider those two as 1.
There are (4 + 1) i.e. 5 persons to be assigned.
∴ Number of ways of assigning these five persons = 5! × 2!
Total number of ways of assigning 6 persons = 6!
∴ Probability that the couple has adjacent desk = 5! × 2! X 6!
Probability that the married couple will have non-adjacent desks = 1 − 1/3 = 2/3

Question. A bag contains 8 red and 5 white balls. Three balls are drawn at random. Find the probability that
(i) all the three balls are white.
(ii) all the three balls are red.
(iii) one ball is red and two balls are white.

Answer. Number of red balls = 8
and number of white balls = 5

Question. A bag contains 9 red, 7 white and 4 black balls. A ball is drawn at random. Find the probability that the ball drawn is
(i) red
(ii) white
(iii) not black
Answer. Total number of balls = 9 red + 7 white + 4 black = 20
⇒ Total number of possible outcomes, n(S) = 20
(i) Let E1 be the event of getting a red ball.
Then, n(E1) = 9

Question. Two dice are thrown together. What is the probability that sum of the numbers on the two faces is neither divisible by 3 nor by 4?
Answer. Let S be the sample space. Then, n(S) = 36.
Also, let A be the event that sum of numbers of two faces is divisible by 3 and B be the event that sum of numbers is divisible by 4 i.e. A be the event of getting the sum 3, 6, 9, 12 and B be the event of getting the sum 4, 8, 12.
Then,
A = {(1,2),(2,1),(2,4),(4,2),(3,3),(1,5), (5,1),(4,5),(5,4),
B = {( , ),( , ),( , ),( 2 2 1 3 3 1 2,6),(6,2), (5,3),(3,5),(4,4),(6,6)}

Question. The probability that a student will pass the final examination in both English and Hindi is 0.5 and the probability of passing neither is 0.1. If the probability of passing the English examination is 0.75, then what is the probability of passing the Hindi examination?
Answer. Let events H and E denote the students that passing Hindi and English examination, respectively.
Given, probability of passing both examinations
= P(H ∩ E) = 0. 5
and probability of passing neither examinations
= P(H’ ∩ E¢ ) = P(H U E)¢ = 0.1
Then, P(H U E) = 1 −P(H U E)’ = 1 − 0.1 = 0.9
Also, it is given, that probability of passing English examination,
P(E) = 0.75
Now, consider P(H U E) = P(H) + P(E) −P(H ∩ E)
⇒ 0.9 = P(H) + 0.75 − 0.5
⇒ 0.9 = P(H) + 0.25
⇒ P(H) = 0.9 − 0.25 = 0.65
Hence, probability of passing Hindi examination is 0.65.

Question. P and Q are two candidates seeking admission in NEET. The probability that P is selected is 0.5 and the probability that both P and Q are selected is at most 0.3. Then, find the probability of Q being selected.
Answer. Let A1 and A2 be two events defined as follows
A1 P = is selected, A2 Q = is selected
We have, P(A1) = 0.5 and P(A1 ∩ A2) ≤ 0.3
Now, P(A1 U A2) ≤ 1
⇒ P(A1) + P(A2) −P(A1 ∩ A2) ≤ 1
⇒ P(A2) ≤ 0.5 + P(A1 ∩ A2)
⇒ P(A2) ≤ 0.8

Question. A coin is tossed and a die is thrown. Find the probability that the outcome will be a head or a number greater than 4.
Answer. When a coin and a die are thrown, then sample space is
S = {H1,H2,H3,H4,H5,H6,T1, T2, T3, T4, T5, T6}
∴ n (S) = 12
Let A : Event of getting a head.
Þ A = {H1,H2,H3,H4,H5,H6}
∴ n (A) = 6

Question. An integer is chosen at random from first two hundred natural numbers. What is the probability that the integer chosen is divisible by 6 or 8?
Answer. Let S be the sample space associated with the given random experiment.
Then, n(S)=200
Now, let A be the event that the chosen number is divisible by
6 and B be the event that the chosen number is divisible by 8.
Then, A = {6,12,18,…,198}
and B = {8,16,24,…,192}
⇒ n(A) = 33 and n(B)=24
[to find n(A) and n(B), divide 198 by 6 and 192 by 8]
Also, A ∩ B = {24,48,72,…,192}
[∵ A ∩ B denotes the event that the chosen number is divisible by LCM (6,8)]

Question. A basket contains 20 apples and 10 oranges out of which 5 apples and 3 oranges are defective. If a person takes out 2 at random, then what is the probability that either both are apples or both are good?
Answer. Let S be the sample space associated with the given experiment.
Then, n(S) = 30C2.
[total number of fruits = 20 +10 = 30]
Now, let A be the event that both are apples and B be the event that both are good fruits. A ∩ B denotes the event that both are good apples.
Then, n(A) = 20C2 and n(B) = 22C2
[ number of good fruits =15 + 7 =22]
n (A ∩ B) =15C2
[ number of good apples =15]

Question. In an interview for a job in call center, 5 boys and 3 girls appeared. If 4 persons are to be selected at random from this group, then find the probability that 3 boys and 1 girl or 1 boy and 3 girls are selected.
Answer. Given, number of boys = 5 and number of girls = 3
Let A : Selection of 3 boys and 1 girl
B : Selection of 1 boy and 3 girls

Question. Four cards are drawn at a time from a pack of 52 playing cards. Find the probability of getting all the four cards of the same unit.
Answer. Let S be the sample space associated with the given random
experiment. Then, n (S) = 52C4.
Now, let A be the event of getting all spade cards,
B be the event getting all club cards,
C be the event getting all heart cards and
D be the event of getting all diamond cards.
Then, n(A) = 13C4, n(B) = 13C4, n(C) = 13C4, and n(D) = 13C4,

Question. Find the probability that, when a hand of 7 cards is drawn from a well-shuffled deck of 52 cards, it contains
(i) all kings. (ii) 3 kings. (iii) at least 3 kings.
Answer. Let S be the sample space associated with the given random experiment. Then, n(S) = 52C7
(i) Let A be the event that it contains all kings.
Then, n(A) = 4C4 × 48C3

Question. The accompanying Venn diagram shows three events, A, B and C and also the probabilities of the various intersections [for instance, P(A ∩ B) = 0.6].

Determine
(i) P (A) (ii) P (B ∩ C̅)
(iii) P (A U B) (iv) P (A ∩ B̅)
(v) P (B ∩ C)
(vi) Probability of exactly one of the three occurs.

(i) P(A) = 0.16 + 0.06 = 0.22
(ii) P(B ∩ C̅) = P(B) − P(B ∩ C)
= 0.06 + 0.20 + 0.10 − 0.10
= 0.06 + 0.20 = 0.26
(iii) P(A È B) = P(A) + P(B) − P(A ∩ B)
= 0.22 + 0.06 + 0.20 + 0.10 − 0.06 = 0.52
(iv) P(A ∩ B̅) = P(A) − P(A ∩ B)
= 0.16 + 0.06 − 0.06 = 0.16
(v) P(B ∩ C) = 0.10
(vi) P(exactly one of the three occurs)
= 0.16 + 0.20 + 0.29 = 0.65

Question. Two dice are thrown. Find
(i) odds in favour of getting the sum 5.
(ii) the odd against getting the sum 6.
Answer. Let S be the sample space associated with the given random experiment. Then, n(S)=36.
(i) Let E be the event of getting the sum 5. Then,
E={(1,4),(4,1),(2,3),(3,2)}
⇒ n (E) = 4

Question. A bag contains 5 white and 7 black balls and a man draw 4 balls at random. What are the odds against these being all black?
Answer. Given, total number of white balls = 5
and total number of black balls = 7
∴ Total number of balls = 12
Let S be the sample space associated with the given experiment. Then, n(S) =12C4.
Now, let E be the event of getting black balls only.
Then, n(E) = 7C4

Question. Three numbers are chosen from 1 to 20. Find the probability that they are not consecutive is

Question. Let A, B and C be three events such that P(A) = 0.3,
P(B) = 0.4, P(C) = 0.8, P(A ∩ B) = 0.08,
P(B ∩ C) = 0.28 and P(A ∩ B ∩ C) = 0.09. If
P(A U B U C)³ 0.8, then show that P(A ∩ C) lies in the interval [0.23,0.43].
Answer. We have, P(A U B U C) ≥ 0.8
∴ 0.8 ≤ P(A U B U C) ≤ 1
[ probability of occurrence of an event
is always less than or equal to 1]
⇒ 0.8 ≤ P(A)+P(B)+P(C) − P(A ∩ B)−P(B ∩ C)
−P(A ∩ C)+P(A ∩ B ∩ C) ≤ 1
⇒ 0.8 ≤ 0.3 + 0.4 + 0.8 − 0.08 − 0.28
−P(A ∩ C)+ 0.09 ≤ 1
⇒ 0.8 ≤ 1.23 −P(A ∩ C) ≤ 1
⇒ 0.8−1.23 ≤ −P(A ∩ C) ≤ 1 −1.23
[subtracting 1.23 from each terms]
⇒ −0.43 ≤ −P(A ∩ C) ≤ −0.23
⇒ 0.23 ≤ P(A ∩ C) ≤0.43
[multiplying each term by (−1)]
Hence, P(A ∩ C) lies in the interval [0.23, 0.43].

Question. Let A, B and C be three events. If the probability of occuring exactly one event out of A and B is 1 − x, out of B and C is 1 − 2x, out of C and A is 1 − x and that of occuring three events simultaneously is x 2, then prove that the probability that at least one out of A, B and C will occur is greater than 1/2.
Answer. We have, P(A) + P(B) −2P(A ∩ B) = 1 − x    …(i)
P(B) + P(C) −2P(B ∩ C) = 1 −2x                …(ii)
P(C) + P(A) −2P(C ∩ A) = 1 − x                 …(iii)
and P(A ∩B ∩ C) = x2                                …(iv)
On adding Eqs. (i), (ii) and (iii), we get
P(A) + P(B) + P(C) −P(A ∩B) −P(B ∩ C)

Question. If A and B are mutually exclusive events, such that
P(A) = 0.35 and P(B) = 0.45, then find
(i) P(A’ )
(ii) P(B’ )
(iii) P(A U B)
(iv) P(A ∩ B)
(v) P(A ∩ B’ )
(vi) P(A’ ∩ B’ )
Answer. We have, A and B are mutually exclusive events.
Also, it is given that P(A) = 0.35 and P(B) = 0.45
(i) P(A’ ) = 1−P(A) = 1−0.35 = 0.65
(ii) P(B’ ) = 1−P(B) = 1−0.45 = 0.55
(iii) P(A U B) = P(A) + P(B) = 0.35 + 0.45 = 0.8
[∵ A and B are mutually exclusive]
(iv) P(A ∩ B) = 0
[ A and B are mutually exclusive, therefore A ∩ B =∮]
(v) P(A ∩ B’ ) = P(A) − P(A ∩ B)
= P(A) − 0 = P(A) = 0.35
(vi) P(A’ ∩ B’ ) = P[(A U B)’ ] = 1−P(A U B) = 1 − 0.8 = 0.2

Question. A team of medical students doing their internship have to assist during surgeries at a city hospital. The probabilities of surgeries rated as very complex, complex, routine, simple or very simple are respectively, 0.15, 0.20, 0.31, 0.26 and 0.08.
Find the probabilities that a particular surgery will be rated
(i) complex or very complex.
(ii) neither very complex nor very simple.
(iii) routine or complex.
(iv) routine or simple.
Answer. Let E1, E2, E3, E4 and E5 be the event that surgeries are rated as very complex, complex, routine, simple or very simple, respectively.
∴ P (E1) = 0.15, P (E2) = 0.20, P (E3) = 0. 31,
P (E4) = 0.26, P (E5) = 0. 08
(i) P (complex or very complex) = P (E2 or E1) = P(E1 or E2)
= P (E1 U E2) = P (E1) + P (E2) − P (E1 ∩ E2)
= 0.15 + 0.20 − 0 [P (E1 ∩ E2) = 0]
[because all events are independent]
= 0.35
(ii) P (neither very complex nor very simple),
P (E’1 ∩ E’5 ) = P (E’1 U E’5)
= 1 − P (E1 U E5)
= 1 − [P (E1) + P (E5)]
= 1 − (0.15 + 0.08)
= 1 − 0.23
= 0.77
(iii) P (routine or complex) = P (E3 U E2) = P (E3) + P (E2)
= 0. 31 + 0.20 = 0. 51
(iv) P (routine or simple) = P (E3 U E4) = P (E3) + P (E4)
= 0. 31 + 0. 26 = 0. 57

Question. A sample space consists of 9 elementary outcomes E1 , E2, …, E9 whose probabilities are
P(E1 ) = P(E2) = 0.08, P (E3) = P(E4) = P(E5) = 0.1,
P(E6) = P(E7) = 0.2 and P(E8) =P(E9) = 0.07.
Suppose A = {E1 ,E5 ,E8}, B = {E2 ,E5 ,E8 ,E9}
(i) Calculate P(A), P(B) and P(A ∩ B).
(ii) Using the addition law of probability, calculate P (A U B).
(iii) List the composition of the event A U B and calculate P (A U B) by adding the probabilities of the elementary outcomes.
(iv) Calculate P (B) from P (B), also calculate P (B) directly from the elementary outcomes of B.

Question. Three coins are tossed once. Find the probability of getting
(iv) exactly two tails.
(v) no tail.
Answer. In random experiment of tossing three coins, the sample space is
S = {HHH,HHT,HTH, THH,HTT, THT, TTH, TTT}
⇒ n(S) = 8
(i) Let E1 be the event of getting 3 heads. Then, outcomes favourable to E1 is {HHH}.
Thus, n(E1) = 1

Question. On vacations, Veena visits four cities (A, B, C and D) in a random order. What is the probability that she visits
(i) A before B?
(ii) A before B and B before C?
(iii) A first and B last?
(iv) A either first or second?
(v) A just before B?
Answer. Number of ways, in which Veena can visit four cities A, B, C and D, is 4! i.e. 24.
∴ n(S) = 24
Clearly, sample space for this experiment is

Question. A bag contains 8 red, 3 white and 9 blackballs. If three balls are drawn at random, determine the probability that
(i) all the three balls are of blue colour.
(ii) all the balls are of different colours.
(iii) one is red and two are white.
Answer. Total number of balls = 8 + 3 + 9 = 20
Number of ways of selecting 3 balls out of 20 balls = 220C3
⇒ Total number of possible outcomes = 20C3
(i) Let E1 be the event of getting all balls of blue colour.
Then, n(E1) = 0
Hence, required probability

Question. If an integer from 1 through 1000 is chosen at random, then find the probability that the integer is a multiple of 2 or a multiple of 9.
Answer. Multiple of 2 from 1 to 1000 are 2, 4, 6, 8, …, 1000
Let n be the number of terms of above series.
∴ n th term = 1000
⇒ 2 + (n − 1) 2 = 1000
⇒ 2 + 2n − 2 = 1000
⇒ 2n = 1000
∴ n = 500
Since, the number of multiple of 2 are 500.
So, the multiple of 9 are 9, 18, 27, …, 999
Let m be the number of term in above series.
∴ m th term = 999
⇒ 9 + (m − 1) 9 = 999
⇒ 9 + 9 m − 9 = 999
⇒ 9 m = 999
∴ m = 111
Since, the number of multiple of 9 are 111. So, the multiple of 2 and 9 both are 18, 36, …, 990
Let p be the number of terms in above series.
∴ p th term = 990
⇒ 18 + (p − 1)18 = 990
⇒ 18 + 18p − 18 = 990
⇒ 18p = 990

Question. A die is thrown. Describe the following events
(i) A : a number less than 7.
(ii) B: a number greater than 7.
(iii) C: a multiple of 3.
(iv) D: a number less than 4.
(v) E : an even number greater than 4.
(vi) F : a number not less than 3.
Also, find A U B,A ∩ B,B ÈC,E ∩ F,D ∩ E,A −C, D −E,F’ and E ∩ F’.
Answer. When a die is thrown, then sample space
S = {1, 2, 3, 4, 5, 6}
(i) A : a number less than 7 = {1, 2, 3, 4, 5, 6}
(ii) B : a number greater than 7 = { } = f
(iii) C: a multiple of 3 = {3,6}
(iv) D: a number less than 4 = {1, 2, 3}
(v) E : an even number greater than 4 = {6}
(vi) F : a number not less than 3 = {3, 4, 5, 6}
Now, A U B = The elements which are in A or B or both
= {1, 2, 3, 4, 5, 6} U f
= {1, 2, 3, 4, 5, 6}
A ∩ B = The elements which are common in both
A and B = {1, 2, 3, 4, 5, 6} ∩ f = f
B U C = The elements which are in B or C or both
= { }U{3, 6} = {3, 6}
E ∩ F = The elements which are common in both
E and F = {6}∩{3, 4, 5, 6} = {6}
D ∩ E = The elements which are common in both
D and E = {1, 2, 3}∩{6} = f
A − C = The elements which are in A but not in C
= {1, 2, 3, 4, 5, 6} − {3, 6}
= {1, 2, 4, 5}
D − E = The elements which are inD but not in E
= {1, 2, 3} − {6}
= {1, 2, 3}
F’ = The elements which are not in F
= (S −F)
= {1, 2, 3, 4, 5, 6} − {3, 4, 5, 6} = {1, 2}
and E ∩F¢ = E ∩(S −F)
= E ∩ ({1, 2, 3, 4, 5, 6} − {3, 4, 5, 6})
= {6} ∩ {1, 2} = ∮

Question. Two dice are thrown. The events A, B and C are as follows.
A : getting an even number on the first die.
B : getting an odd number on the first die.
C : getting the sum of the numbers on the dice £ 5.
Decribe the events
(i) A’
(ii) not B
(iii) A or B
(iv) A and B
(v) A but not C
(vi) B or C

Question. A box contains 6 red, 4 white and 5 black balls. A person draws 4 balls from the box at random. Find the probability that among the balls drawn, there is atleast one ball of each colour.
Answer. Given, total number of red balls = 6
Total number of white balls = 4
and total number of black balls = 5
∴ Total number of balls = 15
Let S be the sample space associated with the random experiment. Then, n(S) = 15C4 = 1365
Here, atleast one ball of each colour can be drawn in one of the following ways
(i) 1 red, 1 white and 2 black balls
(ii) 2 red, 1 white and 1 black balls
(iii) 1 red, 2 white and 1 black balls
Now, let A be the event of getting 1 red, 1 white and 2 black
balls, B be the event of getting 2 red, 1 white and 1 black
balls and C be the event of getting 1 red, 2 white and 1 black balls.

Question. One of the four persons John, Rita, Aslam and Gurpreet will be promoted next month.
Consequently the sample space consists of four elementary outcomes S = {John promoted, Rita promoted, Aslam promoted, Gurpreet promoted}.
You are told that the chances of John’s promotion is same as that of Gurpreet. Rita’s chances of
promotion are twice as likely as Johns. Aslam’s chances are four times that of John.
(i) Determine
P (John promoted),
P (Rita promoted),
P (Aslam promoted),
P (Gurpreet promoted).
(ii) If A = {John promoted or Gurpreet promoted}, find P (A)
Answer. Let E1 = John promoted
E2 = Rita promoted
E3 = Aslam promoted
E4 = Gurpreet promoted
Given, sample space, S = {John promoted, Rita promoted,
Aslam promoted, Gurpreet promoted}
i.e. S = {E1,E2,E3,E4}
It is given that, chances of John’s promotion is same as that of Gurpreet.
P (E1) = P (E4)
Rita’s chances of promotion are twice as likely as John.
P(E2) = 2P(E1)
And Aslam’s chances of promotion are four times that of John.
P (E3) = 4P (E1)
Now, P (E1) + P (E2) + P (E3) + P (E4) = 1
⇒ P (E1) + 2P (E1) + 4P (E1) + P (E1) = 1
⇒  8P (E1) = 1

Question. Four candidates A, B, C and D have applied for the assignment to coach a school cricket team. If A is twice as likely to be selected as B and B and C are given about the same chance of being selected, while C is twice as likely to be selected as D, then what are the probabilities that
(i) C will be selected?
(ii) A will not be selected?
Answer. It is given that A is twice as likely to be selected as B.

Question. If the letters of the word ‘ASSASSINATION’ are arranged at random. Find the probability that
(i) four S’s come consecutively in the word.
(ii) two I’s and two N’s come together.
(iii) all A’s are not coming together.
(iv) no two A’s are coming together.
Answer. Total number of letters in the word ‘ASSASSINATION’ are 13.
Out of which 3A’s, 4S’s, 2 I’s, 2 N’s, 1 T and 1O.
(i) If four S’s come consecutively in the word, then we considers these 4 S’s as 1 group.
Now, the number of later’s is 10.

Question. A number is chosen at random from the numbers ranging from 1 to 50. What is the probability that the number chosen is a multiple of 2 or 3 or 10?
Answer. Let S be the sample space associated with the given random experiment. Then, n(S)=50.
Now, let A be the event that the chosen number is a multiple of 2, B be the event that the chosen number is a multiple of 3 and C be the event that the chosen number is a multiple of 10.
Then, A = {2,4,6,…,48, 50},
B = {3, 6, 9…, 48} and C = {10,20,…,50}
A ∩ B denotes the event that chosen number is multiple of 6= LCM(2,3)
B ∩ C denotes the event that chosen number is a multiple of 30 = LCM (3, 10)
A ∩ C denotes the event that chosen number is a multiple of 10 = LCM (2, 10) and A ∩ B ∩ C denotes the event that chosen number is a multiple of 30 = LCM (2, 3, 10)
∴ A ∩ B = {6,12,…,48},
B ∩ C = {30},A ∩ C = {10, 20…,50} and A ∩ B ∩ C = {30}
⇒ n(A) = 25, n(B) = 16, n(C) = 5, n(A ∩ B) = 8,
n(B ∩ C) = 1, n(A ∩ C) = 5 and n(A ∩ B ∩ C) = 1

Question. A card is drawn from a well-shuffled deck of 52 cards. Find
(i) the odds in favour of getting a face card.
(ii) the odds against getting a spade.
Answer. Let S be the sample space associated with the given random experiment. Then, n (S)=52.
(i) Let E be the event of getting of face card.
Then, n(E) = 12

Case Based Questions

Question. On her vacation, Sania visits four cities.
Delhi, Lucknow, Agra, Meerut in a random order.

On the basis of above information, answer the following questions.
(i) What is the probability that she visits Delhi before Lucknow?
(ii) What is the probability she visit Delhi before Lucknow and Lucknow before Agra?
(iii) What is the probability she visits Delhi either first or second?
Let the Sania visits four cities Delhi, Lucknow, Agra, Meerut are respectively D, L, A and M. Number of way’s in which Sania can visit four cities D, L, A and M is 4! i.e. 24.
∴ n(S) = 24
Clearly, sample space for this experiment is

iii) (c) Let E4 be the event that she visits Delhi (D) either first or second. Then,

Question. In a class of 60 students, 30 opted for NCC, 32 opted for NSS and 24 opted for both NCC and NSS. One of these students is selected at random.

Based on the above information, answer the following

(i) Find the probability that the student opted for NCC or NSS.
(ii) Find the probability that the student has opted for neither NCC nor NSS.
(iii) Find the probability that the student has opted NSS but not NCC.

Answer. Let A and B denote the events that, the selected students opted NCC and NSS, respectively.
Given, n(A) = 30, n(B) = 32
n(A ∩ B) = 24, n(S) = 60
[24 students opted for both NCC and NSS i.e. they are common in both]