CBSE Class 11 Mathematics Probability Worksheet

Question. Out of 11 consecutive natural numbers if three numbers are selected at random (without repetition), then the probability that they are in A.P. with positive common difference, is:
(a) 15/101
(b) 5/101
(c) 5/33
(d) 10/99
Answer : C

Question. If six students, including two particular students A and B, stand in a row, then the probability that A and B are separated with one student in between them is
(a) 8/15
(b) 4/15
(c) 2/15
(d) 1/15
Answer : B

Question. Two different families A and B are blessed with equal number of children. There are 3 tickets to be distributed amongst the children of these families so that no child gets more than one ticket. If the probability that all the tickets go to the children of the family B is 1/12, then the number of children in each family is?
(a) 4
(b) 6
(c) 3
(d) 5
Answer : D

Question. If 10 different balls are to be placed in 4 distinct boxes at random, then the probability that two of these boxes contain exactly 2 and 3 balls is :
(a) 965/2¹¹
(b) 965/2¹⁰
(c) 945/2¹⁰
(d) 945/2¹¹
Answer : C

Question. A set S contains 7 elements. A non-empty subset A of S and an element x of S are chosen at random. Then the probability that xÎA is:
(a) 1/2
(b) 64/127
(c) 63/128
(d) 31/128
Answer : B

Question. If three of the six vertices of a regular hexa on are chosen at random, then the probability that the triangle formed with these chosen vertices is equilateral is :
(a) 1/10
(b) 1/5
(c) 3/10
(d) 3/20
Answer : A

Question. If A and B are two events such that P(A∪B) = P(A∩B) , then the incorrect statement amongst the following statements is:
(a) A and B are equally likely
(b) P(A∩B') = 0
(c) P(A'∩B) = 0
(d) P(A) + P(B) = 1
Answer : D

Question. Five horses are in a race. Mr. A selects two of the horses at random and bets on them. The probability that Mr. A selected the winning horse is
(a) 2/5
(b) 4/5
(c) 3/5
(d) 1/5
Answer : A

Question. A number x is chosen at random from the set {1, 2, 3, 4, ...., 100}. Define the event: A = the chosen number x satisfies (x - 10) (x - 50) / (x - 30) ≥ 0
Then P (A) is:
(a) 0.71
(b) 0.70
(c) 0.51
(d) 0.20
Answer : A

Question. There are two balls in an urn. Each ball can be either white or black. If a white ball is put into the urn and there after a ball is drawn at random from the urn, then the probability that it is white is
(a) 1/4
(b) 2/3
(c) 1/5
(d) 1/3
Answer : B

Question. A box 'A' contanis 2 white, 3 red and 2 black balls. Another box 'B' contains 4 white, 2 red and 3 black balls. If two balls are drawn at random, without replacement, from a randomly selected box and one ball turns out to be white while the other ball turns out to be red, then the probability that both balls are drawn from box 'B' is
(a) 7/16
(b) 9/32
(c) 7/8
(d) 9/16
Answer : A

Question. Let A and B be two events such that the probability that exactly one of them occurs is 2/5 and the probability that A or B occurs is 1/2, then the probability of both of them occur together is:
(a) 0.02
(b) 0.20
(c) 0.01
(d) 0.10
Answer : D

Question. The probabilities of three events A, B and C are given by P(A) = 0.6, P(B) = 0.4 and P(C) = 0.5. If P (A∪B) = 0.8, P (A∩C) = 0.3, P (A∩B∩C) = 0.2, P (B∩C) = b and P (A∪B∪C) = a, where 0.85 ≤ a ≤ 0.95 , then b lies in the interval:
(a) [0.35, 0.36]
(b) [0.25, 0.35]
(c) [0.20, 0.25]
(d) [0.36, 0.40]
Answer : B

Question. Let X and Y are two events such that P(X ∪ Y ) = P( X ∩ Y ).
Statement 1: P( X ∩ Y ') = P( X '∩ Y ) = 0
Statement 2: P( X) + P(Y) = 2P( X ∩ Y)
(a) Statement 1 is false, Statement 2 is true.
(b) Statement 1 is true, Statement 2 is true, Statement 2 is not a correct explanation of Statement 1.
(c) Statement 1 is true, Statement 2 is false.
(d) Statement 1 is true, Statement 2 is true; Statement 2 is a correct explanation of Statement 1.
Answer : B

Question. In a class of 60 students, 40 opted for NCC, 30 opted for NSS and 20 opted for both NCC and NSS. If one of these students is selected at random, then the probability that the student selected has opted neither for NCC nor for NSS is :
(a) 1/6
(b) 1/3
(c) 2/3
(d) 5/6
Answer : A

Question. Let S = {1, 2, ....., 20}. A subset B of S is said to be “nice”, if the sum of the elements of B is 203. Than the probability that a randomly chosen subset of S is “nice” is :
(a) 7/2²⁰
(b) 5/2²⁰
(c) 4/2²⁰
(d) 6/2²⁰
Answer : B

Question. If the lengths of the sides of a triangle are decided by the three throws of a single fair die, then the probability that the triangle is of maximum area given that it is an isosceles triangle, is :
(a) 1/21
(b) 1/27
(c) 1/15
(d) 1/26
Answer : B

Question. For three events A, B and C, P(Exactly one of A or B occurs) = P(Exactly one of B or C occurs) = P(Exactly one of C or A occurs) = 1/4 and P(All the three events occur simultaneously) = 1/16.
Then the probability that at least one of the events occurs, is :
(a) 3/16
(b) 7/32
(c) 7/16
(d) 764
Answer : C

Question. Four numbers are chosen at random (without replacement) from the set {1, 2, 3, ...20}.
Statement -1: The probability that the chosen numbers when arranged in some order will form an AP is 1/85
Statement -2 : If the four chosen numbers form an AP, then the set of all possible values of common difference is (±1,±2,±3,±4,±5) .
(a) Statement -1 is true, Statement -2 is true; Statement -2 is not a correct explanation for Statement -1
(b) Statement -1 is true, Statment -2 is false
(c) Statement -1 is false, Statment -2 is true.
(d) Statement -1 is true, Statement -2 is true ; Statement - 2 is a correct explanation for Statement -1.
Answer : B

Question. An urn contains nine balls of which three are red, four are blue and two are green. Three balls are drawn at random without replacement from the urn. The probability that the three balls have different colours is
(a) 2/7
(b) 1/21
(c) 2/23
(d) 1/3
Answer : A

Question. From a group of 10 men and 5 women, four member committees are to be formed each of which must contain at least one woman. Then the probability for these committees to have more women than men, is :
(a) 21/220
(b) 3.11
(c) 1.11
(d) 2.23
Answer : C

Question. If 12 identical balls are to be placed in 3 identical boxes, then the probability that one of the boxes contains exactly 3 balls is :
(a) 220(1/3)¹²
(b) 22(1/3)¹¹
(c) 55/3(2/3)¹¹
(d) 55(2/3)¹⁰
Answer : C

Question. A die is thrown. Let A be the event that the number obtained is greater than 3. Let B be the event that the number obtained is less than 5. Then P(A∪B) is
(a) 3/5
(b) 0
(c) 1
(d) 2/5
Answer : C

Question. A and B are events such that P(A ∪ B)=3/4, P(A ∩ B)=1/4, P(A) = 2/3 then P ( A ∩ B) is 
(a) 5/12
(b) 3/8
(c) 5/8
(d) 1/4
Answer : A

Question 1. Find the probability that a leap year, selected at random, will contain 53 Sundays.
Answer: A leap year has 366 days.
Dividing 366 by 7 gives 52 weeks and 2 remaining days.
These 2 remaining consecutive days can be: (Sunday, Monday), (Monday, Tuesday), (Tuesday, Wednesday), (Wednesday, Thursday), (Thursday, Friday), (Friday, Saturday), or (Saturday, Sunday).
Total possible outcomes for the remaining 2 days = 7.
Favorable outcomes for getting 53 Sundays are (Sunday, Monday) and (Saturday, Sunday), which is 2 outcomes.
Probability = \(\frac{2}{7}\)
In simple words: A leap year has 2 extra days, and out of 7 possible pairs of consecutive days, exactly 2 pairs contain a Sunday.

Exam Tip: For a non-leap year with 365 days, the probability of 53 Sundays is \(\frac{1}{7}\) because there is only 1 extra day.

Question 2. A box contains 10 bulbs of which 3 are defective. If a random sample of 5 bulbs is drawn, find the probability that the sample contains (i) exactly one defective bulb (ii) exactly 2 defective bulbs (iii) no defective bulbs.
Answer: Total bulbs = 10, Defective bulbs = 3, Non-defective bulbs = 7.
Total ways to draw 5 bulbs = \(\binom{10}{5} = \frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1} = 252\).
(i) Probability of exactly 1 defective bulb:
Ways to choose 1 defective from 3 and 4 non-defective from 7 = \(\binom{3}{1} \times \binom{7}{4} = 3 \times 35 = 105\).
Probability = \(\frac{105}{252} = \frac{5}{12}\)
(ii) Probability of exactly 2 defective bulbs:
Ways to choose 2 defective from 3 and 3 non-defective from 7 = \(\binom{3}{2} \times \binom{7}{3} = 3 \times 35 = 105\).
Probability = \(\frac{105}{252} = \frac{5}{12}\)
(iii) Probability of no defective bulbs:
Ways to choose 0 defective from 3 and 5 non-defective from 7 = \(\binom{3}{0} \times \binom{7}{5} = 1 \times 21 = 21\).
Probability = \(\frac{21}{252} = \frac{1}{12}\)
In simple words: Use combinations to find the number of ways to pick the required number of defective and non-defective bulbs, then divide by the total possible ways to draw 5 bulbs.

Exam Tip: Always verify that the sum of probabilities for all mutually exclusive outcomes equals 1 (\(\frac{5}{12} + \frac{5}{12} + \frac{2}{12} = \frac{12}{12} = 1\)).

Question 3. If A and B are two mutually exclusive events and \(P(A) = \frac{1}{4}, P(B) = \frac{2}{5}, P(A \cup B) = \frac{1}{2}\) then Find (i) \(P(A \cap B)\) (ii) \(P(A \cap B')\) (iii) \(P(B - A)\)
Answer:
(i) Since A and B are mutually exclusive events, \(P(A \cap B) = 0\).
(ii) Using set theory, \(P(A \cap B') = P(A) - P(A \cap B) = \frac{1}{4} - 0 = \frac{1}{4}\).
(iii) \(P(B - A) = P(B) - P(A \cap B) = \frac{2}{5} - 0 = \frac{2}{5}\).
In simple words: Mutually exclusive events cannot happen at the same time, meaning their intersection probability is zero.

Exam Tip: Remember the fundamental identity \(P(B - A) = P(B) - P(A \cap B)\) for any two events A and B.

Question 4. In single throw of three dice, find the probability of getting a total of 17 or 18.
Answer: Total number of outcomes when three dice are thrown = \(6^3 = 216\).
Let E be the event of getting a total of 17 or 18.
Favorable outcomes for a total of 17: (5, 6, 6), (6, 5, 6), (6, 6, 5) -> 3 outcomes.
Favorable outcomes for a total of 18: (6, 6, 6) -> 1 outcome.
Total favorable outcomes = \(3 + 1 = 4\).
Probability = \(\frac{4}{216} = \frac{1}{54}\)
In simple words: List out all combinations of numbers on three dice that sum up to 17 or 18 and divide by 216.

Exam Tip: For large sample spaces with dice, systematic listing starting from the highest numbers prevents missing any outcomes.

Question 5. The letter of word "SOCIETY" are placed at random in a row. What is probability that 3 vowels came together?
Answer: The word SOCIETY has 7 letters: S, O, C, I, E, T, Y.
Vowels in SOCIETY are O, I, E (3 vowels, all distinct).
Consonants are S, C, T, Y (4 consonants, all distinct).
Total number of letters = 7.
Total arrangements of 7 letters = \(7! = 5040\).
Treat the 3 vowels (O, I, E) as a single block. Along with the 4 consonants, we have 5 units to arrange.
Number of ways to arrange these 5 units = \(5! = 120\).
The 3 vowels within their block can be arranged among themselves in \(3! = 6\) ways.
Total favorable arrangements = \(5! \times 3! = 120 \times 6 = 720\).
Probability = \(\frac{720}{5040} = \frac{1}{7}\)
In simple words: Group the vowels together as one single item, find the arrangements, and then multiply by the internal arrangements of the vowels.

Exam Tip: Whenever items must come together, treat them as a single super-item using the block method.

Question 6. A card is drawn at random from well-shuffled deck of 52 cards. Find probability that it is neither ace, nor a king?
Answer: Total number of cards = 52.
Number of aces in a deck = 4.
Number of kings in a deck = 4.
Since aces and kings are mutually exclusive cards, total number of face/target cards to exclude = \(4 + 4 = 8\).
Number of cards that are neither ace nor king = \(52 - 8 = 44\).
Probability = \(\frac{44}{52} = \frac{11}{13}\)
In simple words: Subtract the 4 aces and 4 kings from the total 52 cards to get the remaining favorable cards.

Question 7. In a lottery of 50 tickets numbered from 1 to 50, 2 tickets are drawn simultaneously. Find the probability that :-
(i) both the tickets have prime numbers on it
(ii) None of the tickets drawn has a prime number on it.
Answer: Total tickets = 50.
Total ways to draw 2 tickets = \(\binom{50}{2} = \frac{50 \times 49}{2} = 1225\).
Prime numbers between 1 and 50 are: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47 (Total 15 prime numbers).
Non-prime numbers = \(50 - 15 = 35\).
(i) Probability that both tickets have prime numbers:
Favorable ways = \(\binom{15}{2} = \frac{15 \times 14}{2} = 105\).
Probability = \(\frac{105}{1225} = \frac{21}{245}\)
(ii) Probability that none of the tickets has a prime number:
Favorable ways = \(\binom{35}{2} = \frac{35 \times 34}{2} = 595\).
Probability = \(\frac{595}{1225} = \frac{17}{35}\)
In simple words: Count how many prime and non-prime numbers exist from 1 to 50, then use combinations to compute the required probabilities.

Exam Tip: Be careful to note that 1 is neither prime nor composite when counting primes up to 50.

Question 8. Out of the students attending a lecture, 50% could not see what was written on the board and 40% could not hear what the lecturer was saying. Most unfortunate 30% fell into both of these categories. What is the probability that a student picked at random was able to hear and see satisfactorily.
Answer: Let S be the event that a student cannot see the board, so \(P(S) = 0.50\).
Let H be the event that a student cannot hear the lecturer, so \(P(H) = 0.40\).
Given that 30% fell into both categories, \(P(S \cap H) = 0.30\).
Probability that a student cannot see or cannot hear (at least one problem) = \(P(S \cup H) = P(S) + P(H) - P(S \cap H)\)
\(P(S \cup H) = 0.50 + 0.40 - 0.30 = 0.60\).
The probability that a student was able to hear and see satisfactorily is the complement of having at least one problem:
\(P(\text{Satisfactory}) = 1 - P(S \cup H) = 1 - 0.60 = 0.40 = \frac{2}{5}\).
In simple words: Find the percentage of students having at least one difficulty and subtract it from 100% to find those with no difficulties.

Exam Tip: Use De Morgan's laws and set union formulas carefully when dealing with 'neither-nor' or 'both' percentages.

Question 9. If A, B and C are mutually exclusive and exhaustive events and it is known that \(P(A \cup B) = 0.63\) Calculate \(P(C)\).
Answer: Since A, B, and C are mutually exclusive and exhaustive events, their union covers the entire sample space:
\(P(A \cup B \cup C) = 1\) and \(P(A \cup B) + P(C) = 1\).
Given \(P(A \cup B) = 0.63\):
\(P(C) = 1 - P(A \cup B) = 1 - 0.63 = 0.37\).
In simple words: Exhaustive events sum up to 1, so subtract the probability of A union B from 1 to find the probability of C.

Exam Tip: Exhaustive means their combined union is the entire sample space S, so \(P(S) = 1\).

Question 10. The probability that a student will pass final examination in both Hindi & Eng is 0.5 & probability of passing neither is 0.1. If probability of passing Eng examination is 0.75, what is probability of passing Hindi examination. (Ans. 0.65)
Answer: Let H be passing Hindi and E be passing English.
Given: \(P(H \cap E) = 0.5\), \(P(\text{neither}) = P(H' \cap E') = 0.1\), and \(P(E) = 0.75\).
Using De Morgan's law, probability of passing at least one exam is:
\(P(H \cup E) = 1 - P(H' \cap E') = 1 - 0.1 = 0.9\).
Using the addition theorem of probability:
\(P(H \cup E) = P(H) + P(E) - P(H \cap E)\)
\(0.9 = P(H) + 0.75 - 0.5\)
\(0.9 = P(H) + 0.25\)
\(P(H) = 0.9 - 0.25 = 0.65\).
In simple words: Find the union probability from the 'neither' probability, then use the standard probability addition formula to solve for P(H).

Exam Tip: The relation \(P(A' \cap B') = 1 - P(A \cup B)\) is extremely useful for solving problems involving 'neither' conditions.

Question 11. A box contains 9 red, 7 white and 4 black balls. If two balls drawn at random, find the probability that :-
(i) both balls are red
(ii) 1 ball is white
(iii) balls are of same colour
(iv) 1 is white and other red.
Answer: Total red = 9, white = 7, black = 4. Total balls = \(9 + 7 + 4 = 20\).
Total ways to draw 2 balls = \(\binom{20}{2} = \frac{20 \times 19}{2} = 190\).
(i) Both balls are red:
Favorable ways = \(\binom{9}{2} = \frac{9 \times 8}{2} = 36\).
Probability = \(\frac{36}{190} = \frac{18}{95}\)
(ii) 1 ball is white (and 1 non-white):
Favorable ways = \(\binom{7}{1} \times \binom{13}{1} = 7 \times 13 = 91\).
Probability = \(\frac{91}{190}\)
(iii) Balls are of same colour (both red, both white, or both black):
Favorable ways = \(\binom{9}{2} + \binom{7}{2} + \binom{4}{2} = 36 + 21 + 6 = 63\).
Probability = \(\frac{63}{190}\)
(iv) 1 is white and other red:
Favorable ways = \(\binom{7}{1} \times \binom{9}{1} = 7 \times 9 = 63\).
Probability = \(\frac{63}{190}\)
In simple words: Break down each requirement into combinations of specific colored balls and divide by the total number of ways to draw 2 balls from 20.

Exam Tip: For 'same color', compute combinations for each color separately and add them up.

Question 12. One number is chosen from numbers 1 to 200. What is the probability that it is divisible by 4 or 6.
Answer: Total numbers = 200.
Let A be the set of numbers divisible by 4, and B be the set of numbers divisible by 6.
Number of multiples of 4 up to 200: \(n(A) = \lfloor \frac{200}{4} \rfloor = 50\).
Number of multiples of 6 up to 200: \(n(B) = \lfloor \frac{200}{6} \rfloor = 33\).
Numbers divisible by both 4 and 6 are multiples of their LCM(4, 6) = 12.
Number of multiples of 12 up to 200: \(n(A \cap B) = \lfloor \frac{200}{12} \rfloor = 16\).
Using the inclusion-exclusion principle:
\(n(A \cup B) = n(A) + n(B) - n(A \cap B) = 50 + 33 - 16 = 67\).
Probability = \(\frac{67}{200}\)
In simple words: Count multiples of 4 and 6, then subtract the multiples of their LCM (12) to avoid double counting.

Exam Tip: Always use the Least Common Multiple (LCM) to find the intersection of divisibility conditions.

Question 13. The probability that a person will get an electric contract is \(\frac{2}{5}\) and the probability that he will not get plumbing contract is \(\frac{4}{7}\). If the probability of getting atleast one contract is \(\frac{2}{3}\), what is the probability that he will get both?
Answer: Let E be getting the electric contract and P be getting the plumbing contract.
Given: \(P(E) = \frac{2}{5}\).
Given probability of not getting plumbing contract is \(\frac{4}{7}\), so probability of getting plumbing contract is \(P(P) = 1 - \frac{4}{7} = \frac{3}{7}\).
Probability of getting at least one contract is \(P(E \cup P) = \frac{2}{3}\).
We need to find the probability of getting both: \(P(E \cap P)\).
Using the addition theorem:
\(P(E \cup P) = P(E) + P(P) - P(E \cap P)\)
\(\frac{2}{3} = \frac{2}{5} + \frac{3}{7} - P(E \cap P)\)
Find common denominator for fractions on the right side (LCM = 105):
\(\frac{2}{5} = \frac{42}{105}\), \(\frac{3}{7} = \frac{45}{105}\), \(\frac{2}{3} = \frac{70}{105}\).
\(\frac{70}{105} = \frac{42}{105} + \frac{45}{105} - P(E \cap P)\)
\(\frac{70}{105} = \frac{87}{105} - P(E \cap P)\)
\(P(E \cap P) = \frac{87}{105} - \frac{70}{105} = \frac{17}{105}\).
In simple words: Convert the given 'not getting' probability into a 'getting' probability, then apply the standard union formula to solve for the intersection.

Exam Tip: Pay close attention when problems state 'probability of not getting'—always convert to complementary success probabilities first.

Question 14(i). A and B are 2 events such that \(P(A) = 0.42, P(B) = 0.48\) and \(P(A \text{ and } B) = 0.16\). Determine (a) \(P(\text{not } A)\) (b) \(P(A \text{ or } B)\)
Answer:
(a) \(P(\text{not } A) = 1 - P(A) = 1 - 0.42 = 0.58\).
(b) \(P(A \text{ or } B) = P(A \cup B) = P(A) + P(B) - P(A \cap B) = 0.42 + 0.48 - 0.16 = 0.74\).
In simple words: Complementary probability gives 'not A', and the addition formula gives 'A or B'.

Exam Tip: 'A or B' translates to union (\(\cup\)) and 'A and B' translates to intersection (\(\cap\)).

Question 14(ii). E and F are two events such that \(P(E) = 0.4, P(F) = 0.5, P(E \cup F) = 0.6\), find \(P(E \cap F)\).
Answer: Using the addition formula:
\(P(E \cup F) = P(E) + P(F) - P(E \cap F)\)
\(0.6 = 0.4 + 0.5 - P(E \cap F)\)
\(0.6 = 0.9 - P(E \cap F)\)
\(P(E \cap F) = 0.9 - 0.6 = 0.3\).
In simple words: Rearrange the standard probability addition formula to solve for the intersection term.

Exam Tip: Always substitute known values into the equation before isolating the unknown variable.

Question 14(iii). \(P(E) = 0.60, P(E \text{ or } F) = 0.85, P(E \text{ and } F) = 0.42\) Find \(P(F)\).
Answer: Using the addition formula:
\(P(E \cup F) = P(E) + P(F) - P(E \cap F)\)
\(0.85 = 0.60 + P(F) - 0.42\)
\(0.85 = 0.18 + P(F)\)
\(P(F) = 0.85 - 0.18 = 0.67\).
In simple words: Substitute the given union, individual, and intersection probabilities to find P(F).

Exam Tip: Double-check decimal arithmetic when subtracting to avoid careless calculation errors.

Question 15. In a single throw of two dice, find the probability that neither a doublet not a total of 9 will appear.
Answer: Total outcomes when two dice are thrown = \(6^2 = 36\).
Let D be the event of getting a doublet: {(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)} -> 6 outcomes.
Let S be the event of getting a total of 9: {(3,6), (4,5), (5,4), (6,3)} -> 4 outcomes.
Note that there is no common outcome between doublets and a sum of 9 (none of the doublets sum to 9).
Thus, the number of favorable outcomes for either a doublet or a sum of 9 is \(6 + 4 = 10\).
Probability of getting a doublet or a sum of 9 = \(\frac{10}{36} = \frac{5}{18}\).
Therefore, the probability that neither a doublet nor a sum of 9 will appear is the complement:
\(1 - \frac{5}{18} = \frac{13}{18}\).
In simple words: Find the probability of getting either a doublet or a total of 9, then subtract it from 1 to find the probability of neither.

Exam Tip: Using the complement rule is much faster than manually counting all 26 remaining outcomes that are neither doublets nor sum to 9.

Q.1 A coin is tossed and a die is thrown. Find the probability that the outcome will be a head and a number greater than 4.

Q.2 In a class of 60 students, 32 like Maths, 30 like Biology and 24 like both Maths and Biology. If one of these students is selected at random, find the probability that the selected student.
(a) likes Maths or Biology
(b) likes neither Maths nor Biology
(c) likes Maths but not Biology.

Q.3 A fair coin with 1 marked on one face and 4 on the other and a fair die are both tossed, write the sample of the experiment.

Q.4 Give an example of a sure event and an impossible event.

Q.5 A box contains 10 red marbles, 20 blue marbles and 30 green marbles, 5 marbles are drawn from the box, what is the probability that (i) all will be blue? (ii) at least one will be green?

Q.6 A die is thrown, find the probability of the following events :
(i) A prime number will appear.
(ii) A number less than 6 will appear.
(iii) A number greater than or equal to 3 will appear.

Q.7 Three coins are tossed. Describe :
(i) two events which are mutually exclusive.
(ii) three events which are mutually exclusive and exhaustive.

Q.8 If 2/11 is the probability of an event, what is the probability of the event ‘not A’.

Q.9 Four cards are drawn at random from a pack of 52 playing cards. Find the probability of getting:
(a) all the four cards of the same suit.
(b) two red cards and two black cards. (
c) all cards of the same color. (d) one card from each suit.

Q.10 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, what is probability of passing the Hindi examination

Q.11 Two dice are thrown and the sum of the numbers which come up on the dice is noted. Let us consider the following events associated with this experiment
A : ‘the sum is even’
B : ‘the sum is multiple of 3’
C : ‘the sum is less than 4’
D : ‘the sum is greater than 11’
Which pairs of these events are mutually exclusive?

Q.12 One card is drawn from a well shuffled deck of 52 cards. If each outcome is equally likely, calculate the probability that the card will be (i) a diamond (ii) not an ace.

Q.13 Three dice are thrown simultaneously. Find the probability that:
(a) all of them show the same face.
(b) all show different faces. (
c) two of them show the same face.

Q.14 A bag contains 5 white and 3 black balls. Four balls are successively drawn out without replacement. What is the probability that they are alternatively of different colours?

Q.15 Two students Anil and Ashima appeared in an examination. The probability that Anil will qualify the examination is 0.05 and that Ashima will qualify the examination is 0.10. The probability that both will qualify the examination is 0.02. Find the probability that (i) both Anil and Ashima will not qualify the examination (ii) at least one of them will not qualify the examination, and (iii) only one of them will qualify the examination.

Q.16 Tickets are numbered from 1 to 25. They are well shuffled and a ticket drawn at random . What is the probability that the drawn ticket has a prime number?

Q.17 The probability that a person visiting a doctor will have his blood test done is 0.75 and the probabilitythat he will be admitted is 0.30. The probability that he will have his blood test done or be admitted is 0.45. Find the probability that a person visiting the doctor will have his blood test done and be admitted? 

Q.18 Find the probability that in a random arrangement of the word ‘society’ all the three vowels come together. 

Q.19 In a lottery, a person choses six different natural numbers at random from 1 to 20, and if these six numbers match with the six numbers already fixed by the lottery committee, he wins the prize. What is the probability of winning the prize in the game? [Hint : Order of the numbers is not important]

Q.20 Find the probability that a leap year selected at random will contain 53 Mondays.