14 . PROBABILITY

CBSE Class 11 Maths Chapter 14 Probability

Chapter 14. Probability

Class 11 Maths Chapter 14. Probability Exercise 14.1 Solutions :

1. A die is rolled. Let E be the event “die shows 4” and F be the event “die shows even number”. Are E and F mutually exclusive?

Solution : The sample space = {1 , 2 , 3 , 4 , 5 , 6}

Given , E : die shows 4 = {4}

F : die shows an even number = {2 , 4 , 6}

Hence, E and F events are not mutually exclusive.

2. 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 ∪ B , A ∩ B , B ∪ C , E ∩ F , D ∩ E , A – C , D – E , E ∩ F′ , F′

Solution : The sample space = { 1, 2 , 3 , 4 , 5 , 6}

(i) A: a number less than 7 .

A = {1, 2, 3, 4, 5, 6}

(ii) B: a number greater than 7 .

B =

(iii) C: a multiple of 3

C = {3, 6}

(iv) D: a number less than 4

D = {1, 2, 3}

(v) E: an even number greater than 4

E = {6}

(vi) F: a number not less than 3

F = {3, 4, 5, 6}

Now ,

E ∩ F′ = {6} ∩ {1 , 2 , 3 , 4 , 5 , 6 } – {3 , 4 , 5 , 6} = {6} ∩ {1,2} =

F′ = {1 , 2 , 3 , 4 , 5 , 6 } – {3 , 4 , 5 , 6} = {1 , 2}

3. An experiment involves rolling a pair of dice and recording the numbers that come up. Describe the following events:
A : the sum is greater than 8, B : 2 occurs on either die C : the sum is at least 7 and a multiple of 3. Which pairs of these events are mutually exclusive?

Solution : The sample space = {(1,1) , (1,2) , (1,3) , (1,4) , (1,5) , (1,6) , (2,1) , (2,2) , (2,3) , (2,4) , (2,5) , (2,6) , (3,1) , (3,2) , (3,3) , (3,4) , (3,5) , (3,6) , (4,1) , (4,2) , (4,3) , (4,4) , (4,5) , (4,6) , (5,1) , (5,2) , (5,3) , (5,4) , (5,5) , (5,6) , (6,1), (6,2) , (6,3) , (6,4) , (6,5) , (6,6)}

A : The sum is greater than 8.

A = {(3, 6), (4, 5), (4, 6), (5, 4), (5, 5), (5, 6), (6, 3), (6, 4), (6, 5), (6, 6)}

B : 2 occurs on either die.

B = {(1, 2), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 2), (4, 2), (5, 2), (6, 2)}

C : The sum is at least 7 and a multiple of 3.

C = {(1, 6), (2, 5), (3, 4), (3, 5), (3, 6), (4, 3), (4, 6), (5, 2), (5, 3), (5, 6), (6, 1), (6, 3), (6, 5), (6, 6)}

Therefore, A and B or B and C are mutually exclusive.

4. Three coins are tossed once. Let A denote the event ‘three heads show”, B denote the event “two heads and one tail show”, C denote the event” three tails show and D denote the event ‘a head shows on the first coin”. Which events are (i) mutually exclusive? (ii) simple? (iii) Compound?

Solution : The sample space = {HHH ,HHT ,HTT , HTH , THT ,THH , TTH ,TTT }

A: Three heads show .

A = {(HHH)}

B: Two heads and one tail show .

B = {(HHT) , (HTH) , (THH)}

C: Three tails show.

C = {(TTT)}

D: A head shows on the first coin.

D = {(HHH) , (HTT) , (HHT) , (THH)}

(i) Mutually exclusive events are events that cannot occur at the same time.

A and B , A and C , B and C , C and D , A , B and C are mutually exclusive .

(ii) Simple events are events that consist of only one outcome.

A is a simple event.

B is not a simple event because it consists of three outcomes.

C is a simple event.

D is not a simple event because it consists of three outcomes.

(iii) Compound events are events that consist of more than one outcome.

A is not a compound event.

B is a compound event.

C is not a compound event.

D is a compound event.

5. Three coins are tossed. Describe
(i) Two events which are mutually exclusive.
(ii) Three events which are mutually exclusive and exhaustive.
(iii) Two events, which are not mutually exclusive.
(iv) Two events which are mutually exclusive but not exhaustive.
(v) Three events which are mutually exclusive but not exhaustive.

Solution : The sample space = {HHH ,HHT ,HTT , HTH , THT ,THH , TTH ,TTT }

(i) Two events which are mutually exclusive:

A : Three heads show.

A = {HHH}

B: Three tails show .

B = {TTT}

A and B events are mutually exclusive because they cannot both occur in a single trial.

(ii) Three events which are mutually exclusive and exhaustive:

C: Exactly two heads and one tail show .

C = {(HHT), (HTH), (THH)}

A, B, and C events are mutually exclusive and exhaustive because one of them must occur in every trial.

(iii) Two events which are not mutually exclusive:

D: Two heads and one tail show .

(D = {(HHT), (HTH), (THH)}

E: A head shows on the first coin.

E = {(HTT), (HHT), (HTH)}

D and E events are not mutually exclusive because they can both occur if the outcome is (H, H, T).

(iv) Two events which are mutually exclusive but not exhaustive:

F: Three heads show .

F = {(H, H, H)}

G: Three tails show .

G = {(TTT)}

F and G events are mutually exclusive, but they are not exhaustive because they do not cover all possible outcomes.

(v) Three events which are mutually exclusive but not exhaustive:

H: Two heads and one tail show .

H = {(HHT), (HTH), (THH)}

I: Two tails and one head show .

I = {(TTH), (THT), (HTT)}

J: No heads show .

J = {(TTT)})

H, I, and J events are mutually exclusive but not exhaustive because they do not cover all possible outcomes.

6. 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.
Describe the events
(i) A′ (ii) not B (iii) A or B (iv) A and B (v) A but not C (vi) B or C (vii) B and C (viii) A ∩ B′ ∩ C′

A : Getting an even number on the first die.

A = {(2,1),(2,2),(2,3),(2,4),(2,5),(2,6),(4,1),(4,2),(4,3),(4,4),(4,5),(4,6),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)}

B : Getting an odd number on the first die.

B = {(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(3,1),(3,2),(3,3),(3,4),(3,5),(3,6),(5,1),(5,2),(5,3),(5,4),(5,5),(5,6)}

C : Getting the sum of the numbers on the dice ≤ 5.

C = {(1,1),(1,2),(1,3),(1,4),(2,1),(2,2),(2,3),(3,1),(3,2),(4,1)}

(i) A′: Not getting an even number on the first die.

A′= B = (1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(3,1),(3,2),(3,3),(3,4),(3,5),(3,6),(5,1),(5,2),(5,3),(5,4),(5,5),(5,6)}

(ii) B’ : Not getting an odd number on the first die.

B = {(2,1),(2,2),(2,3),(2,4),(2,5),(2,6),(4,1),(4,2),(4,3),(4,4),(4,5),(4,6),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)}

(iii) A∪B: Getting an even number on the first die or getting an odd number on the first die.

A∪B = {(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(2,1),(2,2),(2,3),(2,4),(2,5),(2,6),(3,1),(3,2),(3,3),(3,4),(3,5),(3,6),

(4,1),(4,2),(4,3),(4,4),(4,5),(4,6),(5,1),(5,2),(5,3),(5,4),(5,5),(5,6),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)}

(iv) A∩B : Getting an even number on the first die and getting an odd number on the first die.

A∩B = (empty set, because it is not possible to get both an even and an odd number on the first die simultaneously)

(v) A ∩ not C : Getting an even number on the first die but not getting the sum of the numbers on the dice ≤ 5.

A∩not C = A – C = {(2,4) , (2,5) , (2,6) , (4,2) , (4,4) , (4,5) , (4,6) ,(6,1) ,(6,2) ,(6,3) ,(6,4) , (6,5) , (6,6)}

(vi) B∪C: Getting an odd number on the first die or getting the sum of the numbers on the dice ≤ 5.

B∪C = {(1,1),(1,2),(1,3),(1,4), (1,5),(1,6), (2,1) ,(2,2),(2,3),(3,1),(3,2),(3 ,3) ,(3 , 4) ,(3 , 5) ,(3,6) ,(4,1), (5,1),(5,2) ,(5,3) ,(5,4),(5,5) ,(5,6)}

(vii) B∩C: Getting an odd number on the first die and getting the sum of the numbers on the dice ≤ 5.

B∩C = {(1,1),(1,2),(1,3),(1,4),(3,1),(3,2)}

(viii) A∩B′∩C′ : Getting an even number on the first die, not getting an odd number on the first die, and not getting the sum of the numbers on the dice ≤ 5.

A∩B′∩C′={(2,4) ,(2,5) ,(2,6) ,(4,2) ,(4,3) ,(4,4) ,(4,5) ,(4,6) ,(6,1) ,(6,2) , (6,3) ,(6,4) , (6,5) , (6,6)}

7. Refer to question 6 above, state true or false: (give reason for your answer)
(i) A and B are mutually exclusive
(ii) A and B are mutually exclusive and exhaustive
(iii) A = B′ (iv) A and C are mutually exclusive
(v) A and B′ are mutually exclusive.
(vi) A′, B′, C are mutually exclusive and exhaustive.

A: Getting an even number on the first die. A={(2,1),(2,2),(2,3),(2,4),(2,5),(2,6),(4,1),(4,2),(4,3),(4,4),(4,5),(4,6),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)}

B: Getting an odd number on the first die. B={(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(3,1),(3,2),(3,3),(3,4),(3,5),(3,6),(5,1),(5,2),(5,3),(5,4),(5,5),(5,6)}

C: Getting the sum of the numbers on the dice ≤ 5.

C={(1,1),(1,2),(1,3),(1,4),(2,1),(2,2),(2,3),(3,1),(3,2),(4,1)}

(i) A and B are mutually exclusive :

True. A and B events are mutually exclusive because an outcome cannot have both an even and an odd number on the first die simultaneously.

(ii) A and B are mutually exclusive and exhaustive :

True. A and B events are mutually exclusive, and together they cover all possible outcomes of getting either an even or an odd number on the first die.

(iii) A = B' :

True. The complement of B (not getting an odd number on the first die) is the event of getting an even number on the first die (A).

(iv) A and C are mutually exclusive :

False. A and C events are not mutually exclusive; they can both happen when the outcome is (2, 3) or (4, 1) .

(v) A and B' are mutually exclusive :

False . A and B' can both happen when the outcome is (2, 1) .

(vi) A', B', C are mutually exclusive and exhaustive :

False. A', B', and C events are not mutually exclusive because an outcome can belong to more than one of these events. Also, they are not exhaustive as they do not cover all possible outcomes.

Class 11 Maths Chapter 14. Probability Exercise 14.2 Solutions :

1. Which of the following can not be valid assignment of probabilities for outcomes of sample Space ?

Assignment
(a)	0.1	0.01	0.05	0.03	0.01	0.2	0.6
(b)
(c)	0.1	0.2	0.3	0.4	0.5	0.6	0.7
(d)	- 0.1	0.2	0.3	0.4	- 0.2	0.1	0.3
(e)

Solution : (a) Each of the number P( ) is positive and less than one.

Sum of probabilities = 0.10 + 0.01 + 0.05 + 0.03 + 0.01 + 0.2 + 0.6 = 1

Therefore, the assignment is valid .

(b) Each of the number P( ) is either 0 or 1.

Sum of the probabilities

Therefore, the assignment is valid .

Sum of probabilities = 0.1 + 0.2 + 0.3 + 0.4 + 0.5 + 0.6 + 0.7 = 2.8 > 1

Therefore, the assignment is not valid .

(d) Two of the probabilities P() and P() are negative .

Therefore, the assignment is not valid .

(e) Each of the number P( ) is either 0 or 1.

Sum of probabilities

Therefore, the assignment is not valid .

2. A coin is tossed twice, what is the probability that atleast one tail occurs?

Solution : The possible outcomes when tossing a coin twice are: HH, HT, TH, and TT.

let, E = {HT , TH , TT}

Here, n(S) = 4 and n(E) = 3

So, the probability of getting at least one tail in two coin tosses is 3/4 .

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

Solution : The sample space = {1 , 2 , 3 , 4 , 5 , 6}

i.e., total faces of die = 6

(i) A prime number will appear:

Total number of prime = 3 (2, 3, 5) .

P(a prime number)

(ii) A number greater than or equal to 3 will appear:

Numbers greater than or equal to 3 are 3, 4, 5, 6.

There are four such numbers.

P(A number greater than or equal to 3)

(iii) A number less than or equal to one will appear:

Numbers less than or equal to one is only 1.

P(A number less than or equal to one)

(iv) A number more than 6 will appear:

There is no number more than 6 on a standard six-sided die.

P(A number more than 6)

(v) A number less than 6 will appear:

Numbers less than 6 are 1, 2, 3, 4, 5. There are five such numbers.

P(A number less than 6)

4. A card is selected from a pack of 52 cards.
(a) How many points are there in the sample space?
(b) Calculate the probability that the card is an ace of spades.
(c) Calculate the probability that the card is (i) an ace (ii) black card.

Solution : (a) The sample space is the set of all possible outcomes when selecting a card from a deck of 52 cards.

(b) Number of favorable outcomes (ace of spades) = 1

Total number of possible outcomes: 52

P(getting the card is an ace of spades)

Number of favorable outcomes (aces): 4 (there are four aces in a deck)

P(getting an ace)

(ii) Total number of possible outcomes: 52

Number of favorable outcomes (black cards): 26 (there are two black suits, clubs, and spades, each with 13 cards)

P(getting a black card)

5. A fair coin with 1 marked on one face and 6 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

Solution : For the coin = {1, 6} (1 for the marked face, 6 for the other face)

For the die = {1, 2, 3, 4, 5, 6}

Total number of favorable outcome = 6 + 6 = 12

(i) Sum of 3 : Favorable outcomes = 1 {(1, 2)}

P(sum of 3)

(ii) Sum of 12: Favorable outcomes = 1 {(6, 6)}

P(Sum of 12)

6. There are four men and six women on the city council. If one council member is selected for a committee at random, how likely is it that it is a woman?

Solution : Number of women = 6

Total number of council members = 4 (men) + 6 (women) = 10

P(a woman)

Therefore, the probability that a randomly selected council member is a woman is .

7. A fair coin is tossed four times, and a person win Re 1 for each head and lose Rs 1.50 for each tail that turns up.
From the sample space calculate how many different amounts of money you can have after four tosses and the probability of having each of these amounts.

Solution : When a fair coin is tossed four times, there are possible outcomes in the sample space.

Each outcome can be represented as a sequence of heads (H) and tails (T), where each H represents a win of Re 1 and each T represents a loss of Rs 1.50.

Four heads (HHHH):

4 × Re1 = Re4 × Re1 = Re4

Three heads, one tail (HHHT, HHTH, HTHH, THHH):

3 × Re1 – 1 × Rs1.50 = Re1.50

Two heads, two tails (HHTT, HTHT, TTHH, THHT, THTH, HTTH):

2 × Re1 − 2 × Rs1.50 = − Re1

One head, three tails (THHT, THTH, HTTH, HTHT):

1 × Re1 – 3 × Rs1.50 = − Rs3

Four tails (TTTT):

4 × Rs1.50 = − Rs6

So, there are five different amounts of money that can be obtained after four tosses:

Re 4 , Re 1.50 , – Re 1 , – Rs 3 , – Rs 6 .

P( winning Re 4)

P(winning Re1.50)

P(losing Re1)

P(losing Rs3)

P(losing Rs 6)

8. Three coins are tossed once. Find the probability of getting
(i) 3 heads (ii) 2 heads (iii) atleast 2 heads (iv) atmost 2 heads (v) no head (vi) 3 tails (vii) exactly two tails (viii) no tail (ix) atmost two tails

Solution : The sample space = {HHH ,HHT ,HTT , HTH , THT ,THH , TTH ,TTT }

n(S) = 8

(i) Probability of getting 3 heads:

E = {HHH}

Here, n(E) = 1

P(3 heads)

(ii) Probability of getting 2 heads:

E = {HHT, HTH, THH }

Here, n(E) = 3

P(2 heads)

(iii) Probability of getting at least 2 heads:

E = {HHT, HTH, THH , HHH}

Here, n(E) = 4

P( atleast 2 heads)

(iv) Probability of getting at most 2 heads:

E = { HHT ,HTT , HTH , THT ,THH , TTH ,TTT }

Here, n(E) = 7

P( atmost 2 heads)

(v) Probability of getting no head :

E = {TTT}

Here, n(E) = 1

P( no heads)

(vi) Probability of getting 3 tails: E = {TTT}

Here, n(E) = 1

P(3 tails)

(vii) Probability of getting exactly two tails: E = {HTT, THT, TTH}

Here, n(E) = 3

P(2 tails)

(viii) Probability of getting no tail (all heads): E = {HHH}

Here, n(E) = 1

P(no tails)

(ix) Probability of getting at most two tails : E = {HHH , HHT ,HTT , HTH , THT ,THH , TTH}

Here, n(E) = 7

P(at most two tails)
9. If is the probability of an event, what is the probability of the event ‘not A’.

Solution : Given, E be any event .

We have,

10. A letter is chosen at random from the word ‘ASSASSINATION’. Find the probability that letter is (i) a vowel (ii) a consonant .

Solution : Total number of words = 13

(i) Vowels in the word are A, I, O (3 vowels).

The number of vowels = 6

P(a vowels)

(ii) Consonants in the word are S, S, S, S, N, T, T (7 consonants).

Total number of consonants = 7

P(a consonant)

11. 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.]

Solution : The total number of ways to choose 6 numbers out of 20

Since there is only one winning combination (the one fixed by the lottery committee), then

P( winning the prize)

12. Check whether the following probabilities P(A) and P(B) are consistently defined
(i) P(A) = 0.5 , P(B) = 0.7 , P(A ∩ B) = 0.6
(ii) P(A) = 0.5 , P(B) = 0.4 , P(A ∪ B) = 0.8

Solution : (i) P(A) = 0.5 , P(B) = 0.7 , P(A ∩ B) = 0.6

If P(A ∩ B) is less than or equal to P(A) and P(B) then it is consistent .

We have, P(A ∩ B) > P(A) or P(B) > P(A ∩ B)

So , P(A) and P(B) are not consistent .

(ii) P(A) = 0.5 , P(B) = 0.4 , P(A ∪ B) = 0.8

We have, P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

P(A ∩ B) = P(A) + P(B) – P(A ∪ B)

= 0.5 + 0.4 – 0.8 = 0.9 – 0.8 = 0.1 < P(A) and P(B) .

So , P(A) and P(B) are consistent .

13. Fill in the blanks in following table:
        P(A)     P(B)      P(A ∩ B)      P(A ∪ B)
(i)                                         …….
(ii) 0.35    …….         0.25              0.6
(iii) 0.5       0.35        ……..             0.7

Solution : (i) Here,

We have ,

(ii) Here, , ,

We have,

(iii) Here, , ,

We have,

14. Given and . Find P(A or B), if A and B are mutually exclusive events.

Solution : Since , A and B are mutually exclusive events.

We have ,

15. If E and F are events such that and , find (i) P(E or F), (ii) P(not E and not F).

Solution : Here,

We have,

(ii) We have,

16. Events E and F are such that P(not E or not F) = 0.25, State whether E and F are mutually exclusive.

Solution : Given , P(not E or not F) = 0.25

We have,

17. A and B are events such that P(A) = 0.42, P(B) = 0.48 and P(A and B) = 0.16. Determine (i) P(not A), (ii) P(not B) and (iii) P(A or B)

Solution : Here, and

(i) We have,

(ii) We have,

(iii) We have,

18. In Class XI of a school 40% of the students study Mathematics and 30% study Biology. 10% of the class study both Mathematics and Biology. If a student is selected at random from the class, find the probability that he will be studying Mathematics or Biology.

Solution : Let A be the event that a student is studying Mathematics, and B be the event that a student is studying Biology.

, and

We have,

Therefore, the probability that a randomly selected student from the class is studying Mathematics or Biology is 0.60 or 60%.

19. In an entrance test that is graded on the basis of two examinations, the probability of a randomly chosen student passing the first examination is 0.8 and the probability of passing the second examination is 0.7. The probability of passing atleast one of them is 0.95. What is the probability of passing both?

Solution : Let A be the event of passing the first examination, and B be the event of passing the second examination.

Here, P(A) = 0.8 , P(B) = 0.7 and P(A∪B) = 0.95

We have, P(A∪B) = P(A)+P(B) – P(A∩B)

P(A∩B) = P(A) +P(B) – P(A∪B)

= 0.8 + 0.7 – 0.95

= 1.5 – 0.95 = 0.55

Therefore, the probability of passing both examinations is 0.55 .

20. 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 the probability of passing the Hindi examination?

Solution: Let E be the event of passing the English examination and H be the event of passing the Hindi examination.

Here,

We have,

Therefore, the probability of passing the Hindi examination is 0.65 .

21. In a class of 60 students, 30 opted for NCC, 32 opted for NSS and 24 opted for both NCC and NSS. If one of these students is selected at random, find the probability that
(i) The student opted for NCC or NSS.
(ii) The student has opted neither NCC nor NSS.
(iii) The student has opted NSS but not NCC.

Solution: Let A be the event that a student opted for NCC and B be the event that a student opted for NSS.

Number of students who opted for NCC, n(A)=30

Number of students who opted for NSS, n(B)=32

Number of students who opted for both NCC and NSS, n(A∩B)=24

Total number of students = 60

and

(i) We have,

(ii) The probability that the student has opted for neither NCC nor NSS :

(iii) The probability that the student has opted for NSS but not NCC :

Class 11 Maths Chapter 14. Probability Miscellaneous Exercise Solutions :

1. 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) atleast one will be green?

Solution: Total number of marbles = 10 +20 +30 = 60

The total number of ways to draw 5 blue marbles out of 60

(i) Probability that all marbles drawn are blue:

The number of ways to draw 5 blue marbles out of 20

P(all marbles are blue)

(ii) Probability that at least one marble drawn is green:

P(At least one green)
2. 4 cards are drawn from a well – shuffled deck of 52 cards. What is the probability of obtaining 3 diamonds and one spade?

Solution : Total number of cards in deck = 52 .

The total number of ways to draw 4 cards out of 52

The number of ways to draw 3 diamonds out of 13

The number of ways to draw 1 spade out of the 13 spades

P(getting 3 diamonds and one spade)

3. A die has two faces each with number ‘1’, three faces each with number ‘2’ and one face with number ‘3’. If die is rolled once, determine : (i) P(2) (ii) P(1 or 3) (iii) P(not 3)

Solution : The total number of faces on the die = 2 (with '1') + 3 (with '2') + 1 (with '3') = 6 faces.

(i) Since, three faces with the number '2'

P(getting 2)

(ii) There are two faces with '1' and one face with '3'.

The total number of favorable outcomes is 2 + 1 = 3.

P(getting 1 or 3)

(iii) There are two faces with '1' and three faces with '2'.

The total number of faces that are not '3' is 2 + 3 = 5

P(not 3)

4. In a certain lottery 10,000 tickets are sold and ten equal prizes are awarded. What is the probability of not getting a prize if you buy (a) one ticket (b) two tickets (c) 10 tickets.

Solution : The total number of tickets is 10,000.

(a) The number of tickets without a prize is 10000 – 10 = 9990

P(not getting a prize with one ticket)

(b) P(not getting a prize with two tickets)

(c)

P(not getting a prize with 10 tickets)

5. Out of 100 students, two sections of 40 and 60 are formed. If you and your friend are among the 100 students, what is the probability that
(a) you both enter the same section?
(b) you both enter the different sections?

Solution : Let ,the two sections are A and B.

Total number of students in two sections = 100 .

(a) Probability of both in Section A:

For you, the probability

For your friend, the probability (as there are now 99 students left in the pool after you've been placed).

So, the combined probability

Probability of both in Section B :

For you, the probability

For your friend, the probability

So, the combined probability

P(you both enter the same section)

(b) Probability of you in Section A and your friend in Section B:

For you, the probability

For your friend, the probability (as there are 60 students left in Section B after you've been placed).

So, the combined probability

Probability of you in Section B and your friend in Section A:

For you, the probability

For your friend, the probability

So, the combined probability

P(both in different sections)

6. Three letters are dictated to three persons and an envelope is addressed to each of them, the letters are inserted into the envelopes at random so that each envelope contains exactly one letter. Find the probability that at least one letter is in its proper envelope.

Solution : The total number of ways to arrange three letters in three envelopes

There are 2 ways in which none of the letters is in its proper envelope.

Therefore, the probability of having no letters in their proper envelopes

The probability of having at least one letter in its proper envelope

So, the probability that at least one letter is in its proper envelope is .

7. A and B are two events such that P(A) = 0.54, P(B) = 0.69 and P(A ∩ B) = 0.35. Find (i) P(A ∪ B) (ii) P(A' ∩ B.) (iii) P(A ∩ B') (iv) P(B ∩ A')

Solution : Here, P(A) = 0.54, P(B) = 0.69 and P(A ∩ B) = 0.35

(i) We have,

(ii) We have,

(iii) We have,

(iv) We have,

8. From the employees of a company, 5 persons are selected to represent them in the managing committee of the company. Particulars of five persons are as follows:
S. No.       Name          Sex         Age in years
1.            Harish            M                30
2.            Rohan            M                33
3.           Sheetal         F                 46
4.             Alis               F                28
5.            Salim             M                41
A person is selected at random from this group to act as a spokesperson. What is the probability that the spokesperson will be either male or over 35 years?

Solution : Total number of individuals = 5

Let, the number of individuals who are male (A) = 3 (Harish, Rohan, Salim)

The number of individuals who are over 35 years old (B) = 2 (Rohan, Salim)

and the number of individuals who are both male and over 35 years old (A∩B) = 1 ( Salim)

So, and

We have,

So, the probability that the spokesperson will be either male or over 35 years old is .

9. If 4-digit numbers greater than 5,000 are randomly formed from the digits 0, 1, 3, 5, and 7, what is the probability of forming a number divisible by 5 when, (i) the digits are repeated? (ii) the repetition of digits is not allowed?

Solution : (i) When the digits are repeated:

A 4-digit number greater than 5,000 can be formed in the following way:

The first digit can be any of the digits 5, 7 (2 choices).

The remaining three digits can be any of the digits 0, 1, 3, 5, 7 (5 choices each).

Using the fundamental principle of counting, we have

the total number of ways to form a 4-digit number greater than 5,000 with repeated digits = 5 × 5 × 5 + 5 × 5 × 5 = 125 +125 = 250

For a number to be divisible by 5, the last digit must be either 0 or 5.

So, the number of ways to form a 4-digit number greater than 5,000, with repeated digits, and divisible by 5 = 5 × 5 + 5 × 5 + 5 × 5 + 5 × 5 = 100 .

P(the digits are repeated)

(ii) When the repetition of digits is not allowed:

The first digit can be either 5 or 7 (2 choices).

The second digit can be any of the remaining digits (4 choices).

The third digit can be any of the remaining digits (3 choices).

The fourth digit can be any of the remaining digits (2 choices).

Using the fundamental principle of counting, we have

The total number of ways to form a 4-digit number greater than 5,000 without repeated digits = 4 × 3 × 2 + 4 × 3 × 2 = 48 .

For a number to be divisible by 5, the last digit must be either 0 or 5.

So, the number of ways to form a 4-digit number greater than 5,000, without repeated digits, and divisible by 5 = 3 × 2 + 3 × 2 + 3 × 2 = 18 .

P(the repetition of digits is not allowed)

10. The number lock of a suitcase has 4 wheels, each labelled with ten digits i.e., from 0 to 9. The lock opens with a sequence of four digits with no repeats. What is the probability of a person getting the right sequence to open the suitcase?

Solution : Since the digits cannot repeat .

So, the number of ways to choose the first digit is 10.

The number of ways to choose the second digit is 9 (since it cannot be the same as the first digit).

The number of ways to choose the third digit is 8.

The number of ways to choose the fourth digit is 7.

Using the fundamental principle of counting , we have

The total number of ways to choose a sequence of four digits = 10 × 9 × 8 × 7 = 5040

The number of possible outcome = 1

P(right sequence)