NCERT Class 11 Mathematics Chapter 14 Probability

NCERT Class 11 Mathematics Chapter 14 Probability Solutions, NCERT Solutions For Class 11 Maths, NCERT Class 11 Mathematics Chapter 14 Probability Notes to each chapter is provided in the list so that you can easily browse throughout different chapter NCERT Class 11 Mathematics Chapter 14 Probability Notes and select needs one.

NCERT Class 11 Mathematics Chapter 14 Probability

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Also, you can read the CBSE book online in these sections Solutions by Expert Teachers as per (CBSE) Book guidelines. NCERT Class 11 Mathematics Chapter 14 Probability Textual Question Answer. These solutions are part of NCERT All Subject Solutions. Here we have given NCERT Class 11 Mathematics Chapter 14 Probability Solutions for All Subject, You can practice these here.

Probability

Chapter – 14

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?

Ans: When a die is rolled, the sample space is given by 

S = {1, 2, 3, 4, 5, 6} 

Accordingly, E = {4} and F = {2, 4, 6} 

It is observed that E ∩ F = {4} ≠ ⍉ 

Therefore, E and F are not mutually exclusive events.

2. A die is thrown. Describe the following events:

(i) A: a number less than 7.

Ans: A = {1, 2, 3, 4, 5, 6}.

(ii) B: a number greater than 7.

Ans: B = I|

(iii) C: a multiple of 3.

Ans: C = {3, 6}.

(iv) D: a number less than 4.

Ans: D = {1, 2, 3}.

(v) E: an even number greater than 4.

Ans: E = {6}.

(vi) F: a number not less than 3.

Ans: F = {3, 4, 5, 6}.

Also find A ∪ B, A ∩ B, B ∪ C E ∩ F D ∩ E A – C D – E, E ∩ F’ , F’

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?

Ans: When a pair of dice is rolled, the sample space will have 36 sample points

S = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)

(2, 1), (2, 2), .., (2, 6)

(3, 1), (3, 2)… (3,6)

(4, 1), (4, 2)… .. (4,6),

(5, 1), (5, 2). .. (5,6)

(6, 1), (6, 2)………..(6,6)}

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

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

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

Clearly, A ∩ B = ⍉

Hence, A and B are mutually exclusive events.

B ∩ C = ⍉

Hence, B and C are mutually exclusive events.

A ∩ C = {(4,5), (5, 4), (3, 6), (6, 3), (6, 6)} ≠ ⍉  

Hence, A and C are not mutually exclusive events.

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?

Ans: Event A and B; event A and C; event B and C; and event C and D are all mutually exclusive.

(ii) simple?

Ans: If an event has only one sample point of a sample space, it is called a simple event. Thus, A and C are simple events.

(iii) Compound?

Ans: If an event has more than one sample point of a sample space, it is called a compound event. Thus, B and D are compound events.

5. Three coins are tossed. Describe

(i) Two events which are mutually exclusive.

Ans: Two events that are mutually exclusive can be

A: getting no heads and B: getting no tails

This is because sets A = {TTT} and B = {HHH} are disjoint.

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

Ans: Three events that are mutually exclusive and exhaustive can be

A: getting no heads ⇒A = {TTT}

B: getting exactly one head ⇒B = {HTT, THT, TTH}

C: getting at least two heads ⇒ C = {HHH, HHT, HTH, THH}

This is because A ∩ B = B ∩ C = C ∩ A = ∅ and A ∪ B ∪ C = S

(iii) Two events, which are not mutually exclusive.

Ans: Two events that are not mutually exclusive can be

A: getting three heads ⇒A = {HHH}

B: getting at least 2 heads ⇒ B = {HHH, HHT, HTH, THH}

This is because A ∩ B= {HHH} ≠ ⍉

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

Ans: Two events which are mutually exclusive but not exhaustive can be

A: getting exactly one head ⇒A = {HTT, THT, TTH}

B: getting exactly one tail ⇒B = {HHT, HTH, THH}

This is because, A ∩ B = ⍉ but A ∪ B ≠ S

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

Ans: Three events that are mutually exclusive but not exhaustive can be

A: getting exactly three heads ⇒A = {HHH}

B: getting one head and two tails ⇒B = {HTT, THT, TTH}

C: getting one tail and two heads ⇒C = {HHT, HTH, THH}

This is because A ∩ B = B ∩ C = C ∩ A= ⍉ but A ∪ B ∪ C ≠ S.

6. Two dice are thrown. The events A, B and C are as follows: 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.

ribe the events

(i) A’.

Ans: 

(ii) not B.

Ans: 

(ii) A or B.

Ans:

(iv) A and B.

Ans: A and B = A ∩ B = ⍉.

(v) A but not C.

Ans: A but not C = A – C

(vi) B or C.

Ans: B or C = B ∪ C.

(vii) B and C.

Ans: B and C = B ∩ C = {1, 1), (1, 2), (1, 3), (1, 4), (3, 1), (3, 2)}.

(viii) A ∩ Β’∩ C’.

Ans: A ∩ Β’∩ C’ =

7. Refer to question 6 above, state true or false: (give reason for your answer)

(i) A and B are mutually exclusive.

Ans: A ∩ B = ⍉

Thus A and B are mutually exclusive events.

∴ True.

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

Ans: A ∩ B = ⍉ and A ∩ B = S

Thus A and B are mutually exclusive and exhaustive events.

∴ True.

(ii) A = B’.

Ans: 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)} 

= A

∴ True

(iv) A and C are mutually exclusive.

Ans: A ∩ c = {(2, 1), (2, 2), (2, 3), (4, 1)} = ⍉

Thus A and c are not mutually exclusive events ∴ False

(v) A and B’ are mutually exclusive.

Ans: A ∩ B’ = A ≠ ⍉

Thus A and B’ are not mutually exclusive events.

(vi) A’, B’, C are mutually exclusive and exhaustive.

Ans: Since A’ = BandB’ = A, A ∩ B = ⍉ 

B ∩ C = {(1, 1), (1, 2), (1, 3), 

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

≠ ⍉ 

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

= ⍉ 

Thus A’, B’ and C are not mutually exclusive. 

∴ False.

1. Which of the following can not be valid assignment of probabilities for outcomes of sample Space S = {ω₁,ω₂,ω₃,ω₄,ω₅,ω₆,ω₇}

Ans: 

(a)

Here, each of the numbers p(ωᵢ) is positive and less than 1.

Sum of probabilities

= p(ω₁)+p(ω₂)+p(ω₃)+p(ω₄)+p(ω₅)+p(ω₆)+p(ω₇) 

= 0.1 + 0.01 + 0.05 +0.03 + 0.01 + 0.2 + 0.6 

= 1

Thus, the assignment is valid.

(b) 

Here, each of the numbers p(ωᵢ) is positive and less than 1.

Sum of probabilities

= p(ω₁)+p(ω₂)+p(ω₃)+p(ω₄)+p(ω₅)+p(ω₆)+p(ω₇)

Thus, the assignment is valid.

(c)

Here, each of the numbers p(ωᵢ) is positive and less than 1.

Sum of probabilities

= p(ω₁)+p(ω₂)+p(ω₃)+p(ω₄)+p(ω₅)+p(ω₆)+p(ω₇) 

=0.1+0.2+0.3+0.4+0.5+0.6+0.7 

= 2.8 ≠ 1

Thus, the assignment is not valid.

(d) 

Here, p(ω₁) and p(ω₅) are negative.

Hence, the assignment is not valid.

(e) 

Here

Hence, the assignment is not valid.

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

Ans: When a coin is tossed twice, the sample space is given by

S = {HH, HT, TH, TT}

Let A be the event of the occurrence of at least one tail. 

Accordingly, A = {HT, TH, TT}

3. A die is thrown, find the probability of following events:

(i) A prime number will appear,

Ans: Let A be the event of the occurrence of a prime number. 

Accordingly, A = {2, 3, 5}

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

Ans: Let B be the event of the occurrence of a number greater than or equal to 3. Accordingly, B = {3, 4, 5, 6}

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

Ans: Let C be the event of the occurrence of a number less than or equal to one. Accordingly, C = {1}

(iv) A number more than 6 will appear,

Ans: Let D be the event of the occurrence of a number greater than 6. 

Accordingly, D = ⍉

(v) A number less than 6 will appear.

Ans: Let E be the event of the occurrence of a number less than 6.

Accordingly, E = {1, 2, 3, 4, 5}

4. A card is selected from a pack of 52 cards.

(a) How many points are there in the sample space? 

Ans: When a card is selected from a pack of 52 cards, the number of possible outcomes is 52 i.e., the sample space contains 52 elements.

Therefore, there are 52 points in the sample space.

(b) Calculate the probability that the card is an ace of spades.

Ans:  Let A be the event in which the card drawn is an ace of spades. 

Accordingly, n(A) = 1

(c) Calculate the probability that the card is 

(i) an ace.

Ans: Let E be the event in which the card drawn is an ace.

Since there are 4 aces in a pack of 52 cards, n(E) = 4

(ii) black card.

Ans: Let F be the event in which the card drawn is black.

Since there are 26 black cards in a pack of 52 cards, n(F) = 26

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.

Ans: Let A be the event in which the sum of numbers that turn up is 3. 

Accordingly, A = {(1, 2)}

(ii) 12.

Ans: Let B be the event in which the sum of numbers that turn up is 12.

Accordingly, B = {(6, 6)}

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?

Ans: There are four men and six women on the city council.

As one council member is to be selected for a committee at random, the sample space contains 10 (4+ 6) elements.

Let A be the event in which the selected council member is a woman.

Accordingly, n(A) = 6

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.

Ans: Here a coin is tossed four times. The sample space S is S = {HHHH, HHHT, HHTΗ, ΗΤΗΗ, HTTH, HTHT, HHTT, HTTT, THHH, THHT, THTH, TTHH, TTTH, TTHT, THTT, TTTT}

Amounts:

(i) for 4 heads 

= 1 +1 + 1 + 1

= Rs. 4 = winning Rs. 4

(ii) For 3 heads 1 tail 

1+1+1-1.50 

= Rs. 1.50 = winning Rs. 1.50

(iii) For 2 heads 2 tails 

= 1 + 1-1.50 – 1.50

= Rs. (-1) = losing Rs. 1 

(iv) For 1 head 3 tails 

= 1-1.50 -1.50 -1.50 

= Rs. (-3.50) = losing Rs. 3.50

(v) For 4 tails = -1.50 -1.50 -1.50 -1.50 

= losing Rs. 6 

Thus the sample space of amounts is 

= {4, 1.50, 1.50, 1.50, 1.50, -1, -1, -1, -1, -1, -1, -3.50, -3.50, -3.50,-6} 

Now P (winning Rs. 4) 1/16 

P (winning Rs. 1.50) = 4/16 = 1/4

P(Losing Rs. 1) = 6/16 = 3/8

P (Losing Rs. 3.50) = 4/16 = 1/4

P (Losing Rs. 6) = 1/16

8. Three coins are tossed once. Find the probability of getting

(i) 3 heads.

Ans: Let B be the event of the occurrence of 3 heads. Accordingly, B = {HHH}

∴ P(B) =

(ii) 2 heads.

Ans: Let C be the event of the occurrence of 2 heads. Accordingly, C = {HHT, HTH, THH}

∴ P(C) =

(iii) at least 2 heads.

Ans: Let D be the event of the occurrence of at least 2 heads. 

Accordingly, D= {HHH , HHT, HTH, THH}

∴ P(D) =

(iv) at most 2 heads.

Ans: Let E be the event of the occurrence of at most 2 heads.

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

∴ P(E) =

(v) no head.

Ans: Let F be the event of the occurrence of no head.

Accordingly, F = {TTT}

∴ P(F) = 

(vi) 3 tails.

Ans: Let G be the event of the occurrence of 3 tails.

Accordingly, G = {TTT}

∴ P(G) =

(vii) exactly two tails.

Ans: Let H be the event of the occurrence of exactly 2 tails.

Accordingly, H = {HTT, THT, TTH}

∴ P(H) =

(viii) no tail.

Ans: Let I be the event of the occurrence of no tail. 

Accordingly, I = {HHH}

∴ P(I) =

(ix) at most two tails.

Ans: Let J be the event of the occurrence of at most 2 tails. 

Accordingly, I= {HHH , HHT, HTH, THH, HTT, THT, TTH}

∴ P(J) = 

9. If is the probability of an event, what is the probability of the event ‘not A’.

Ans: It is given that P(A) = 

Accordingly, P(not A) = 1- P(A)

10. A letter is chosen at random from the word ‘ASSASSINATION’. Find the probability that letter is 

(i) a vowel.

Ans: There are 6 vowels in the given word

∴ Probability (vowel) = 6/13

(ii) a consonant.

Ans: There are 7 consonants in the given word.

∴ Probability (consonant) = 7/13

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.]

Ans: Total number of ways in which one can choose six different numbers from 1 to 20

Hence, there are 38760 combinations of 6 numbers.

Out of these combinations, one combination is already fixed by the lottery committee.

∴ Required probability of winning the prize in the game

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.

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

It is known that if E and F are two events such that E ⊂ F, then P(E) ≤ P(F). 

However, here, P(A ∩ B) > P(A). 

Hence, P(A) and P(B) are not consistently defined.

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

Ans: P(A) = 0.5, P(B) = 0.4, P(A ∪ B) = 0.8

It is known that if E and F are two events such that E ⊂ F, then P(E) ≤ P(F). 

Here, it is seen that P(A ∪ B) > P(A) and P(A ∪ B) > P(B). 

Hence, P(A) and P(B) are consistently defined.

13. Fill in the blanks in following table:

Ans: (i) Here,

We know that P(A∪B)=P(A)+P(B)-P(A∩B)

(ii) Here, P(A) = 0.35, P(A ∩ B) = 0.25, P(A ∪ B) = 0.6

We know that P(A ∪ B) = P(A) + P(B) – P(A ∩ B) 

∴ 0.6 = 0.35 + P(B) – 0.25 

⇒ P(B) = 0.6 – 0.35 + 0.25 

⇒ P(B) = 0.5

(iii) Here, P(A) = 0.5, P(B) = 0.35, P(A ∪ B) = 0.7

We know that P(A ∪ B) = P(A) + P(B) – P(A ∩ B) 

∴ 0.7 = 0.5 + 0.35 – P(A ∩ B) 

⇒ P(A ∩ B) = 0.5 + 0.35 – 0.7 

⇒ P(A ∩ B) = 0.15.

14. Given P(A) = 3/5 and P(B) = 1/5. Find P(A or B), if A and B are mutually exclusive events.

Ans: Here, P(A) = 3/5, P(B) = ⅕

For mutually exclusive events A and B, 

P(A or B) = P(A) + P(B)

∴ P(A or B)

15. If E and F are events such that P(E) = 1/4,P(F) = 1/2 and P(E and F) = 1/8, find

(i) P(E or F), 

Ans: We know that P(E or F) = P(E) + P(F) – P(E and F)

∴ P (E or F)

(ii) P(not E and not F). 

Ans: From (i) P(E or F) = P(E ∪ F) = 5/8

We have (E ∪ F)’ = ( E’ ∩ F’ ) [By De Morgan’s law] 

∴ P( E’ ∩ F’) = P(E ∪ F)’  

Now, P(E ∪ F)’ = 1 – P(E ∪ F) = 1 – 5/8 = 3/8 

∴ P(E’ ∩ F’)= 3/8 

Thus, P(not E and not F) = 3/8

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

Ans: It is given that P (not E or not F) = 0.25

i.e., P( E’ ∪ F’) = 0.25 

⇒ P( E ∩ F)’ = 0.25    [ E’ ∪ F’ =(E ∩ F)’ ]

Now, P( E ∩ F) = 1 – P(E ∩ F)’ 

⇒ P(E ∩ F)= 1 – 0.25 

⇒ P( E ∩ F)=0.75 ≠ 0 

⇒ E ∩ F ≠ ⍉ 

Thus, E and F are not mutually exclusive.

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),

Ans: P(notA) = 1 – P(A) = 1 – 0.42 = 0.58.

(ii) P(not B).

Ans: P(notB) = 1 – P(B) = 1 – 0.48 = 0.52.

(iii) P(A or B).

Ans: We know that P(A or B) = P(A) + P(B) – P(A and B)

∴ P(A or B) = 0.42 + 0.48 – 0.16 = 0.74.

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.

Ans: Let A be the event in which the selected student studies Mathematics and B be the event in which the selected student studies Biology.

Accordingly, P(A) = 40% =

P(B) = 30%

P(A and B) = 10%

We know that P(A or B) = P(A) + P(B) – P(A and B)

Thus, the probability that the selected student will be studying Mathematics or Biology is 0.6.

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 at least one of them is 0.95. What is the probability of passing both?

Ans: Let A be the event that the student passes the first examination and B be the event that the student passes the second examination. 

Then P(A) = 0.8 , P(B) = 0.7 

and P( a ∪ B) = 0.95

We know that 

P(A ∪ B) = P(A) + P(B) – P(A ∪ B) 

∴ 0.95 = 0.8 + 0.7 -P(A ∪ B) 

∴ 0.95 =1.5 – P(A ∩ B) 

∴ P(A ∩ B) = 1.5 – 0.95 = 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?

Ans: Let A and B be the events of passing English and Hindi examinations respectively. 

Accordingly, P(A and B) = 0.5, P(not A and not B) = 0.1, i.e., P( A’ ∩ B’)=0.1 

P(A) = 0.75 

Now, (A ∪ B)’ = ( A’ ∩ B’)  [De Morgan’s law]

∴ P(A ∪ B)’ = P( A’ ∩ B’) = 0.1 

P(A ∪ B)= 1- P(A ∪ B)’ = 1 – 0.1 = 0.9

We know that P(A or B) = P(A) + P(B) – P(A and B)

∴ 0.9 = 0.75 + P(B) – 0.5 

⇒ P(B) = 0.9 – 0.75 + 0.5 

⇒ P(B) = 0.65

Thus, 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.

Ans: We know that P(A or B) = P(A) + P(B) – P (A and B)

Thus, the probability that the selected student has opted for NCC or NSS is

(ii) The student has opted neither NCC nor NSS.

Ans: P(not A and not B)

= P(A’ and B’)

= P(A’ ∩ B’)

= P(A U B)’  [(A’ ∩ B’) = (A U B)’ (by De Morgan’s law)]

=1- P(A U B)

=1- P(A or B)

Thus, the probability that the selected students has neither opted for NCC nor NSS is

(iii) The student has opted NSS but not NCC.

Ans: The given information can be represented by a Venn diagram as

It is clear that

Number of students who have opted for NSS but not NCC

= n(B – A) = n(B) – n(A ∩ B) = 32 – 24 = 8

Thus, the probability that the selected student has opted for NSS but not for NCC =

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?

Ans: All the drawn marbles will be blue if we draw 5 marbles out of 20 blue marbles.

5 blue marbles can be drawn from 20 blue marbles in ²⁰C₅ ways.

∴ Probability that all marbles will be blue =

(ii) at least one will be green?

Ans: Number of ways in which the drawn marble is not green = ²⁰⁺¹⁰C₅ = ³⁰ C₅

∴ Probability that no marble is green =

∴ Probability that at least one marble is 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?

Ans: Number of ways of drawing 4 cards from 52 cards = ⁵² C₄

In a deck of 52 cards, there are 13 diamonds and 13 spades.

∴ Number of ways of drawing 3 diamonds and one spade = ¹³C₃ × ¹³C₁

Thus, the probability of obtaining 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).

Ans: Number faces with number ‘2’ = 3

(ii) P(1 or 3).

Ans: P(1 or 3) = P(not 2) = 1 – P(2)

(iii) P(not 3).

Ans: Number of faces with number ‘3′ = 1

Thus, P(not 3) = 1 – P(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.

Ans: If we buy one ticket, then

P (getting a prize) =

∴ P (not getting a prize) =

(b) two tickets.

Ans: If we buy two tickets, then 

Number of tickets not awarded = 10,000- 10 = 9990

P (not getting a prize) =

(c) 10 tickets?

Ans: If we buy 10 tickets, then

P (not getting a prize) =

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?

Ans: The two of us will enter the same section if both of us are among 40 students or among 60 students.

∴ Number of ways in which both of us enter the same section = ⁴⁰ C₂+ ⁶⁰C₂

∴ Probability that both of us enter the same section

(b) you both enter the different sections?

Ans: P(we enter different sections)

= 1 – P(we enter the same section)

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.

Ans: Let L₁, L₂, L₃ be three letters and E₁, E₂, and E₃ be their corresponding envelops respectively.

There are 6 ways of inserting 3 letters in 3 envelopes. These are as follows:

There are 4 ways in which at least one letter is inserted in a proper envelope.

Thus, the required probability 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 U B).

Ans: We know that P (A U B) = P (A) + P (B) – P (A ∩ B)

∴ P(A U B) = 0.54 + 0.69 – 0.35 = 0.88.

(ii) P( A’ ∩ B’).

Ans: A’ ∩ B’ = (AUB)’, [by De Morgan’s law ]

∴ P (A΄ ∩ B’) = P(A U B) = 1 – P (A U B) = 1 – 0.88 = 0.12

(iii) P(A ∩ B’ ).

Ans: P(A ∩ B)’ = P (A) – P (A ∩ B) 

= 0.54 – 0.35 

= 0.19

(iv) P(B ∩ A’ ).

Ans: We know that, P (B ∩ A’) = P (B) – P (A ∩ B)

∴ P (B ∩ A’) = 0.69 – 0.35 = 0.34

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:

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?

Ans: Let E be the event in which the spokesperson will be a male and F be the event in which the spokesperson will be over 35 years of age.

Since there is only one male who is over 35 years of age,

We know that P(E U F) = P(E) + P(F) -P(E ∩ F)

Thus, the probability that the spokesperson will either be a male or over 35 years of age is 4/5.

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?

Ans: When the digits are repeated

Since four-digit numbers greater than 5000 are formed, the leftmost digit is either 7 or 5.

The remaining 3 places can be filled by any of the digits 0, 1, 3, 5, or 7 as repetition of digits is allowed.

∴ Total number of 4-digit numbers greater than 5000 = 2 x 5 x 5 x 5 -1

= 250 – 1 = 249

[In this case, 5000 can not be counted; so 1 is subtracted]

A number is divisible by 5 if the digit at its units place is either 0 or 5.

∴ Total number of 4-digit numbers greater than 5000 that are divisible by 5 = 2 x 5 x 5 x 2 – 1 = 100 – 1 = 99

Thus, the probability of forming a number divisible by 5 when the digits are repeated is

(ii) the repetition of digits is not allowed?

Ans: When repetition of digits is not allowed

The thousands place can be filled with either of the two digits 5 or 7.

The remaining 3 places can be filled with any of the remaining 4 digits.

∴ Total number of 4-digit numbers greater than 5000 = 2 x 4 x 3 x 2

= 48

When the digit at the thousands place is 5, the units place can be filled only with 0 and the tens and hundreds places can be filled with any two of the remaining 3 digits.

∴ Here, number of 4-digit numbers starting with 5 and divisible by 5

= 3 x 2=6

When the digit at the thousands place is 7, the units place can be filled in two ways (0 or 5) and the tens and hundreds places can be filled with any two of the remaining 3 digits.

∴ Here, number of 4-digit numbers starting with 7 and divisible by 5 

= 1 x 2 x 3 x 2 = 12

∴ Total number of 4-digit numbers greater than 5000 that are divisible by 5 = 6 + 12 = 18

Thus, the probability of forming a number divisible by 5 when the repetition of digits is not allowed is 

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?

Ans: The number lock has 4 wheels, each labelled with ten digits i.e., from 0 to 9. 

Number of ways of selecting 4 different digits out of the 10 digits = ¹⁰C₄

There is only one number that can open the suitcase