## NCERT Solutions Class 11 |

Assignment | ω_{1} | ω_{2} | ω_{3} | ω_{4} | ω_{5} | ω_{6} | ω_{7} | |
---|---|---|---|---|---|---|---|---|

(a) | 0.1 | 0.01 | 0.05 | 0.03 | 0.01 | 0.2 | 0.6 | |

(b) | 1/7 | 1/7 | 1/7 | 1/7 | 1/7 | 1/7 | 1/7 | |

(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) | 1/14 | 2/14 | 3/14 | 4/14 | 5/14 | 6/14 | 15/14 |

**SOLUTION**

**(a)** Each of the outcomes have a positive probability which is less than 1.

Sum of all probabilities = 0.1 + 0.01 + 0.05 + 0.03 + 0.01 + 0.2 + 0.6 = 1

This is a valid assignment of probabilities.

**(b)** Each of the outcomes have a positive probability which is less than 1.

Sum of all probabilities = 1/7 + 1/7 + 1/7 + 1/7 + 1/7 + 1/7 + 1/7 = 7/7 = 1

This is a valid assignment of probabilities.

**(c)** Each of the outcomes have a positive probability which is less than 1.

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

This assignment of probabilities is not valid.

**(d)** Every outcome does not have a positive probability which is less than 1.

This assignment of probabilities is not valid.

**(e)** Each of the outcomes have a positive probability which is less than 1.

Sum of all probabilities = 1/14 + 2/14 + 3/14 + 4/14 + 5/14 + 6/14 + 15/14 = 36/14 ≠ 1

This assignment of probabilities is not valid.

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

**SOLUTION**

Each toss of the coin can show Heads or Tails. So,

Sample space when a coin is tossed twice = S = {HH, HT, TH, TT} ⇒ 4 outcomes

P (At least one tail occurs) = Favorable Outcomes/Total Number of outcomes

= 3/4 = 0.75

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

**A prime number will appear,****A number greater than or equal to 3 will appear,****A number less than or equal to one will appear,****A number more than 6 will appear,****A number less than 6 will appear**

**SOLUTION**

Sample space when a die is thrown = S = {1, 2, 3, 4, 5, 6} ⇒ 6 outcomes

**(i)** Sample space of Prime numbers in S = {2, 3, 5} ⇒ 3 outcomes

P (A prime number will appear) = 3/6 = 0.5

**(ii)** Sample space of Numbers greater than or equal to 3 in S = {3, 4, 5, 6} ⇒ 4 outcomes

P (A number greater than or equal to 3 will appear) = 4/6 = 0.66

**(iii)** Sample space of Numbers less than or equal to one in S = {1} ⇒ 1 outcome

P (A number less than or equal to one will appear) = 1/6 = 0.16

**(iv)** Sample space of Numbers greater than 6 in S = φ ⇒ 0 outcomes

P (A number more than 6 will appear) = 0/6 = 0

**(v)** Sample space of Numbers smaller than 6 in S = {1, 2, 3, 4, 5, 6} ⇒ 6 outcomes

P (A number less than 6 will appear) = 6/6 = 1

**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 deck consists of 52 cards. So, number of points in the sample space = 52

**(b)** Number of Ace of Spades in the deck = 1

P (getting an ace of spades) = 1/52

**(c) (i)** Number of Ace cards in the deck = 4

P (getting an Ace) = 4/52 = 1/13

**(ii)** Number of black cards in the deck = 26

P (getting a black card) = 26/52 = 1/2

**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**

Sample space when the coin is tossed = {1, 6}

Sample space when the die is thrown = {1, 2, 3, 4, 5, 6}

Sample space of throwing the coin and the dice = 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)}

Total possible outcomes = 12

**(i)** Outcomes where the sum of numbers that show up is 3 = {(1, 2)}

Favorable outcomes = 1

P (sum of the numbers is 3) = 1/12

**(ii)** Outcomes where the sum of numbers that show up is 12 = {(6, 6)}

Favorable outcomes = 1

P (sum of the numbers is 12) = 1/12

**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**

Total number of possible outcomes = Number of council members = 4 + 6 = 10

Number of women in the council = 6

P (a woman is selected) = 6/10 = 0.6

**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.**

**SOLUTION**

Sample space when 4 coins are tossed = S

= {

HHHH, HHHT, HHTH, HTHH,

THHH, HHTT, HTHT, THHT,

HTTH, THTH, TTHH, TTTH,

TTHT, THTT, HTTT, TTTT

}

CASE I: For 4 heads

Amount of money = Rs 1 + 1 + 1 + 1 = Rs 4

So, Rs 4 is won.

CASE II: For 3 heads and 1 tails

Amount of money = Rs 1 + 1 + 1 - 1.50 = Rs 1.50

So, Rs 1.50 is won.

CASE III: For 2 heads and 2 tails

Amount of money = Rs 1 + 1 - 1.50 - 1.50 = - Rs 1

So, Rs 1 is lost.

CASE IV: For 1 heads and 3 tails

Amount of money = Rs 1 - 1.50 - 1.50 - 1.50 = - Rs 3.50

So, Rs 3.50 is lost.

CASE V: For 4 tails

Amount of money = Rs -1.50 - 1.50 - 1.50 - 1.50 = - Rs 6

So, Rs 6 is lost.

Sample space for the amounts = {4, 1.50, 1.50, 1.50, 1.50, -1, -1, -1, -1, -1, -1, -3.50, -3.50, -3.50, -3.50, -6} ⇒ 16 outcomes

P (Rs 4 is won) = 1/16

P (Rs 1.50 is won) = 4/16

P (Rs 1 is lost) = 6/16

P (Rs 3.50 is lost) = 4/16

P (Rs 6 is lost) = 1/16

**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**

Sample space when 3 coins are tossed = S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}

Total possible outcomes = 8

**(i)** P (getting 3 heads) = 1/8

**(ii)** P (getting 2 heads) = 3/8

**(iii)** P (getting at least 2 heads) = 4/8

**(iv)** P (getting at most 2 heads) = 7/8

**(v)** P (getting no heads) = P (getting 3 tails) = 1/8

**(vi)** P (getting 3 tails) = 1/8

**(vii)** P (getting exactly two tails) = 3/8

**(viii)** P (getting no tails) = P (getting 3 heads) = 1/8

**(ix)** P (at most two tails) = 7/8

**9. If 2/11 is the probability of an event, what is the probability of the event 'not A'.**

**SOLUTION**

We know that

P (A) + P (not A) = 1

So,

2/11 + P (not A) = 1

P (not A) = 1 - 2/11

P (not A) = (11 - 2)/11

P (not A) = 9/11

**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 letters in 'ASSASSINATION' = 13

**(i)** Number of vowels in 'ASSASSINATION' = 6

P (chosen letter is a vowel) = 6/13

**(ii)** Number of consonants in 'ASSASSINATION' = 7

P (chosen letter is a consonant) = 7/13

**11. In a lottery, a person chooses 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**

Total numbers in the draw = 20

Number of ways to select 6 numbers from the given 20 = ^{20}C_{6}

Only 1 of ^{20}C_{6} ways of selecting the six numbers would result in a win. Therefore,

P (winning the prize) = 1/^{20}C_{6} = 6! 14!/20!

= (6 × 5 × 4 × 3 × 2) 14!/(20 × 19 × 18 × 17 × 16 × 15 × 14!)

= 1/38760

**12. Check whether the following probabilities P(A) and P(B) are consistently defined**

**P(A) = 0.5, P(B) = 0.7, P(A ∩ B) = 0.6****P(A) = 0.5, P(B) = 0.4, P(A ∪ B) = 0.8**

**SOLUTION**

**(i)** P (A ∩ B) = 0.6 which is greater than P (A) 0.5. This is not possible.

Therefore, the given probabilities are not defined consistently.

**(ii)** We know that

P (A ∪ B) = P (A) + P (B) - P (A ∩ B)

0.8 = 0.5 + 0.4 - P (A ∩ B)

P (A ∩ B) = 0.1 which is less than P (A) and P (B).

Therefore, the given probabilities are defined consistently.

**13. Fill in the blanks in following table:**

P (A) | P (B) | P (A ∪ B) | P (A ∩ B) | |
---|---|---|---|---|

(i) | 1/3 | 1/5 | 1/15 | ... |

(ii) | 0.35 | ... | 0.25 | 0.6 |

(iii) | 0.5 | 0.35 | ... | 0.7 |

**SOLUTION**

**(i)** P (A) = 1/3, P (B) = 1/5 and P (A ∩ B) = 1/15

We know that

P (A ∪ B) = P (A) + P (B) - P (A ∩ B)

P (A ∪ B) = 1/3 + 1/5 - 1/15

P (A ∪ B) = (5 + 3)/15 - 1/15

P (A ∪ B) = (8 - 1)/15

P (A ∪ B) = 7/15

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

We know that

P (A ∪ B) = P (A) + P (B) - P (A ∩ B)

P (B) = P (A ∪ B) - P (A) + P (A ∩ B)

P (B) = 0.25 - 0.35 + 0.6

P (B) = 0.5

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

We know that

P (A ∪ B) = P (A) + P (B) - P (A ∩ B)

P (A ∩ B) = P (A) + P (B) - P (A ∪ B)

P (A ∩ B) = 0.5 + 0.35 - 0.7

P (A ∩ B) = 0.15

Hence, the completed table will be:

P (A) | P (B) | P (A ∪ B) | P (A ∩ B) | |
---|---|---|---|---|

(i) | 1/3 | 1/5 | 1/15 | 7/15 |

(ii) | 0.35 | 0.5 | 0.25 | 0.6 |

(iii) | 0.5 | 0.35 | 0.15 | 0.7 |

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

**SOLUTION**

P (A) = 3/5 and P (B) = 1/5

Since, A and B are mutually exclusive. Therefore, P (A ∩ B) = 0

We know that

P (A ∪ B) = P (A) + P (B) - P (A ∩ B)

P (A or B) = 3/5 + 1/5 - 0

P (A or B) = 4/5

**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), (ii) P(not E and not F).**

**SOLUTION**

P(E) = 1/4, P(F) = 1/2 and P(E ∩ F) = 1/8

**(i)** We know that

P (A ∪ B) = P (A) + P (B) - P (A ∩ B)

So,

P (E or F) = P (E) + P (E) - P (E and F)

P (E or F) = 1/4 + 1/2 - 1/8

P (E or F) = (2 + 4 - 1)/8

P (E or F) = 5/8

We know that

P (A ∪ B) = P (A) + P (B) - P (A ∩ B)

**(ii)** P (not E and not F) = P (E' ∩ F') = P (E ∪ F)' = 1 - P (E ∪ F)

= 1 - 5/8

= (8 - 5)/8

= 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.**

**SOLUTION**

P (not E or not F) = 0.25

P (E' ∪ F') = 0.25

P (E ∩ F)' = 0.25

1 - P (E ∩ F) = 0.25

P (E ∩ F) = 1 - 0.25

P (E ∩ F) = 0.75 ≠ 0

Therefore, E and F are not mutually exlcusive.

**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**

P (A) = 0.42, P (B) = 0.48 and P (A ∩ B) = 0.16

**(i)** P (not A) = P (A') = 1 - P (A)

P (A') = 1 - 0.42

P (A') = 0.58

**(ii)** P (not B) = P (B') = 1 - P (B)

P (B') = 1 - 0.48

P (B') = 0.52

**(iii)** We know that

P (A ∪ B) = P (A) + P (B) - P (A ∩ B)

P (A or B) = 0.42 + 0.48 - 0.16

P (A or B) = 0.84

**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 the event that a student is studying Mathematics be denoted by 'M' and the event that a student is studying Biology be denoted by 'B'.

P (Student is studying Mathematics) = P (M) = 40/100 = 0.4

P (Student is studying Biology) = P (B) = 30/100 = 0.3

P (Student is studying both Mathematics and Biology) = P (M ∩ B) = 10/100 = 0.1

P (a student is studying Mathematics or Biology) = P (M ∪ B)

We know that

P (M ∪ B) = P (M) + P (B) - P (M ∩ B)

P (M ∪ B) = 0.4 + 0.3 - 0.1

P (M ∪ B) = 0.6

Hence, the probability that he 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 atleast one of them is 0.95. What is the probability of passing both?**

**SOLUTION**

Let the event that a chosen student passes the first examination be denoted by 'A' and the event that a chosen student passes the second examination be denoted by 'B'.

P (Student passes the first examination) = P (A) = 0.8

P (Student passes the second examination) = P (B) = 0.7

P (Student passes at least one of the two examinations) = P (A ∪ B) = 0.95

P (Student passes both the examinations) = P (A ∩ B)

We know that

P (A ∪ B) = P (A) + P (B) - P (A ∩ B)

P (A ∩ B) = P (A) + P (B) - P (A ∪ B)

P (A ∩ B) = 0.8 + 0.7 - 0.95

P (A ∩ B) = 0.55

Hence, 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 the event that a student will pass the final examination in English be denoted by 'E' and the event that a student will pass the final examination is Hindi be denoted by 'H'.

P (Student passes the English examination) = P (E) = 0.75

P (Student passes both English and Hindi examinations) = P (E ∩ H) = 0.5

P (Student passes neither examination) = P (E' ∩ H') = P (E ∪ H)' = 0.1

P (E ∪ H)' = 0.1

1 - P (E ∪ H) = 0.1

P (E ∪ H) = 0.9

We know that

P (E ∪ H) = P (E) + P (H) - P (E ∩ H)

P (H) = P (E ∪ H) - P (E) + P (E ∩ H)

P (H) = 0.9 - 0.75 + 0.5

P (H) = 0.65

Hence, 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**

**The student opted for NCC or NSS.****The student has opted neither NCC nor NSS.****The student has opted NSS but not NCC.**

**SOLUTION**

Let the event that a student opted for NCC be denoted by 'A' and the event that a student opted for NSS be denoted by 'B'.

P (Student opted for NCC) = P (A) = 30/60 = 1/2

P (Student opted for NSS) = P (B) = 32/60 = 8/15

P (Student opted for both NCC and NSS) = P (A ∩ B) = 24/60 = 2/5

We know that

P (A ∪ B) = P (A) + P (B) - P (A ∩ B)

**(i)** P (Student has opted for NCC or NSS) = P (A ∪ B)

P (A ∪ B) = 1/2 + 8/15 - 2/5

P (A ∪ B) = (15 + 16 - 12)/30

P (A ∪ B) = 19/30

**(ii)** P (Student has opted for neither NCC nor NSS) = P (A' ∩ B')

= P (A ∪ B)' = 1 - P (A ∪ B)

= 1 - 19/30

= 11/30

**(iii)** Number of students that opted for NSS but not NCC = n (B) - n (B ∩ A)

= 32 - 24 = 8

P (Student has opted NSS but not NCC) = 8/60 = 2/15

**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**

Number of red marbles in the box = 10

Number of blue marbles in the box = 20

Number of green marbles in the box = 30

Total number of marbles in the box = 10 + 20 + 30 = 60

Number of ways of drawing 5 marbles from the box = ^{60}C_{5}

**(i)** All the drawn marbles will be blue if the 5 marbles are drawn from the 20 blue marbles.

Number of ways of drawing 5 marbles from 20 blue marbles = ^{20}C_{5}

P (All marbles will be blue) = ^{20}C_{5}/^{60}C_{5}

**(ii)** A marble will not be green if it is drawn from (10 + 20) red and blue marbles.

Number of ways of drawing 5 marbles from the 30 non-green marbles = ^{30}C_{5}

P (No marbles are green) = ^{30}C_{5}/^{60}C_{5}

P (At least one marble is green) = 1 - P (No marbles are green)

= 1 - ^{30}C_{5}/^{60}C_{5}

**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**

Number of cards in a deck = 52

Number of cards to be drawn = 4

So, total number of ways to select 4 cards from the deck = ^{52}C_{4}

There are 13 diamond cards in a deck of 52 cards. In order to draw a diamond, the card needs to be drawn from these 13 cards.

Number of ways to draw 3 diamonds from 13 = ^{13}C_{3}

There are 13 spade cards in a deck of 52 cards. In order to draw a spade, the card needs to be drawn from these 13 cards.

Number of ways to draw 1 spade from 13 = ^{13}C_{1}

Therefore,

Number of ways of drawing 3 diamonds and 1 spade card from a deck = ^{13}C_{3} × ^{13}C_{1}

P (Drawing 3 diamonds and 1 spade) = (^{13}C_{3} × ^{13}C_{1})/^{52}C_{4}

**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**

Number of faces on the die with number '1' = 2

Number of faces on the die with number '2' = 3

Number of faces on the die with number '3' = 1

Total number of faces on a die = 6

**(i)** P (2) = 3/6 = 1/2

**(ii)** P (1 or 3) = P (1) + P (3) = 2/6 + 1/6 = 3/6 = 1/2

Alternatively,

P (1 or 3) = P (not 2) = 1 - P (2) = 1 - 1/2 = 1/2

**(iii)** P (not 3) = 1 - P (3)

= 1 - 1/6 = 5/6

**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**

Number of lottery tickets sold = 10,000

Number of prizes awarded = 10

**(a)** P (getting a prize with one ticket) = 10/10000

= 1/1000

P (not getting a prize with one ticket) = 1 - P (getting a prize with one ticket)

= 1 - 1/1000 = 999/1000

**(b)** Number of tickets that don't get awarded the prize = 10000 - 10 = 9990

In order for us to not get the prize, our tickets have to be from the 9990 not awarded tickets.

Number of ways to select 2 tickets from 9990 = ^{9990}C_{2}

Total number of ways to select 10 winning tickets from 10000 = ^{10000}C_{10}

P (not getting a prize with two tickets) = ^{9990}C_{2}/^{10000}C_{10}

**(c)** In order for us to not get the prize, our tickets have to be from the 9990 not awarded tickets.

Number of ways to select 10 tickets from 9990 = ^{9990}C_{10}

Total number of ways to select 10 winning tickets from 10000 = ^{10000}C_{10}

P (not getting a prize with ten tickets) = ^{9990}C_{10}/^{10000}C_{10}

**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**

Number of students = 100

Me and my friend are part of these 100 students.

Number of ways to select us 2 students from 100 = ^{100}C_{2}

**(a)** In order for both of us to be selected in the same section, we both need to be selected from the 40 students or 60 students.

Number of ways to select 2 students from 40 = ^{40}C_{2}

Number of ways to select 2 students from 60 = ^{60}C_{2}

P (we enter the same section) = (^{40}C_{2} + ^{60}C_{2})/^{100}C_{2}

= (40!/2!38! + 60!/2!58!)/(100!/1!99!)

= 5100/9900 = 17/33

**(b)** P (we enter different sections) = 1 - P (we enter same section)

= 1 - 17/33

= 16/33

**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**

Let the letters be L_{1}, L_{2}, and L_{3} and their respective envelopes be E_{1}, E_{2}, and E_{3}.

Then the Sample space will be S = {

L_{1}E_{1}, L_{2}E_{3}, L_{3}E_{2},

L_{2}E_{2}, L_{1}E_{3}, L_{3}E_{1},

L_{3}E_{3}, L_{1}E_{2}, L_{2}E_{1},

L_{1}E_{1}, L_{2}E_{2}, L_{3}E_{3},

L_{1}E_{2}, L_{2}E_{3}, L_{3}E_{1},

L_{1}E_{3}, L_{2}E_{1}, L_{3}E_{2},

} ⇒ 6 ways

Out of the 6 possible ways, 4 have at least one letter inserted in the proper envelope.

P (At least one letter is inserted in the proper envelope) = 4/6

**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**

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

**(i)** We know that

P (A ∪ B) = P (A) + P (B) - P (A ∩ B)

P (A ∪ B) = 0.54 + 0.69 - 0.35

P (A ∪ B) = 0.88

**(ii)** P (A' ∩ B') = P (A ∪ B)' = 1 - P (A ∪ B)

= 1 - 0.88

= 0.12

**(iii)** P (A ∩ B') = P (A) - P (A ∩ B)

= 0.54 - 0.35

= 0.19

**(iv)** P (B ∩ A') = P (B) - P (A ∩ B)

= 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:**

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**

Number of people = 5

Let the event that a spokesperson will be male be denoted by 'A' and the event that spokesperson will be of age over 35 years be denoted by 'B'.

P (A) = 3/5

P (B) = 2/5

P (A ∩ B) = 1/5

We know that

P (A ∪ B) = P (A) + P (B) - P (A ∩ B)

P (A ∪ B) = 3/5 + 2/5 - 1/5

P (A ∪ B) = 4/5

Hence, probability that spokesperson will 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? (ii) the repetition of digits is not allowed?**

**SOLUTION**

**(i)** The numbers need to be greater than 5000. So, only 7 and 5 are possible at the thousand's place digit.

Number of ways to form a number greater than 5000 with given digits = 2 × 5 × 5 × 5 - 1

= 249

In order for the number to be divisible by 5, the unit's place digit has to be 0 or 5.

Number of ways to form a number greater than 5000, divisible by 5 = 2 × 5 × 5 × 2 - 1

= 99

P (Number formed is divisible by 5) = 99/249

**(ii)** The thousand's place can be filled by 5 or 7.

The 3 remaining places can be filled by the remaining digits.

Number of ways to form 4-digit number greater than 5000 = 2 × 4 × 3 × 2 = 48

Number of ways to form a number greater than 5000, divisible by 5 that start with 5 =

= 1 × 3 × 2 × 1 = 6

Number of ways to form a number greater than 5000, divisible by 5 that start with 7 =

= 1 × 2 × 3 × 2 = 12

Number of ways to form a number greater than 5000, divisible by 5 = 6 + 12 = 18

P (number divisible by 5 is formed without repetition of digits) = 18/48

**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**

Total number of digits = 10

Number of ways to select 4 different digits = ^{10}C_{4}

Number of ways to arrange the 4 selected digits = 4!

Number of possible sequences = 4! × ^{10}C_{4}

= 5040

P (Opening the suitcase) = 1/5040

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