Q1. Cards marked with numbers 13,14,15,…,60 are placed in a box and mixed thoroughly. One card is drawn at random from the box. Find the probability that number on the card drawn is

(i) divisible by 5

(ii) a number is a perfect square

answer:

Total number of possible outcomes n(S)=60-12=48,

(i)

*E*→ event of getting no divisible by 5

Number of favorable outcomes n(

*E*)=60/5-2=10, excluding {5,10}.

(ii)

*E*→ event of getting a perfect square.

Number of favorable outcomes n(

*E*)=4, they are

Q2. Hundred identical cards are numbered from 1 to 100. The cards The cards are well shuffled and then a card is drawn. Find the probability that the number on card drawn is:

(i) a multiple of 5

(ii) a multiple of 6

(iii) between 40 and 60

(iv) greater than 85

(v) less than 48

answer:

There are 100 cards from which one card is drawn.

Total number of elementary events =n(S)=100

(i) From numbers 1 to 100. there are 20 numbers which are multiple of 5 i.e. {5,10,15,…,95,100}.

Number of favorable events =n(*E*)=20.

Probability of selecting a card with a multiple of 5

(ii) From numbers 1 to 100. there are 16 numbers which are multiple of 6 i.e. {6,12,18,…,90,96}.

Number of favorable events =n(

*E*)=16.

Probability of selecting a card with a multiple of 6

(iii) From numbers 1 to 100. there are 19 natural numbers which are between 40 and 60, n(

*E*)=19.

Probability of selecting a card between 40 and 60

(iv) From numbers 1 to 100. there are 15 natural numbers which are greater than 85, n(

*E*)=15.

Probability of selecting a card with a number greater than 85

(v) From numbers 1 to 100. there are 47 natural numbers which are less than 48, n(

*E*)=47.

Probability of selecting a card with a number less than 48

Q3. From 25 identical cards, numbered 1, 2, 3,…, 24, 25: one card is drawn at random. Find the probability that the number on the card drawn is a multiple of:

(i) 3 (ii) 5 (iii) 3 and 5 (iv) 3 or 5

answer:

There are 25 cards from which one card is drawn.

Total number of elementary events =n(S)=25

(i) From numbers 1 to 25, there are 8 numbers which are multiple of 3 i.e. {3,6,9,12,15,18,21,24}→n(*E*)=8.

Probability of selecting a card with a multiple of 3

(ii) From numbers 1 to 25. there are 5 numbers which are multiple of 5 i.e. {5,10,15,20,25}→n(

*E*)=8.

Probability of selecting a card with a multiple of 5

(iii) From numbers 1 to 25. there is only one number which is multiple of 3 and 5 i.e. {15}→n(

*E*)=1.

Probability of selecting a card with a multiple of 3 and 5

(iv) From numbers 1 to 25. there are 12 numbers which are multiple of 3 or 5 i.e. {3,5,6,9,10,12,15,18,20,21,24,25}

*E*)=12

Probability of selecting a card with a multiple of 3 or 5 =

Q4. Thirty identical cards are marked with numbers 1 to 30. If one card is drawn at random, find the probability that it is:

(i) a multiple of 4 or 6

(ii) a multiple of 3 and 5

(iii) a multiple of 3 or 5

answer:

There are 30 cards from which one card is drawn.

Total number of elementary events =n(S)=30

(i) From numbers 1 to 30, there are 10 numbers which are multiple of 4 or 6 i.e.

*E*={4,6,8,12,16,18,20,24,28,30}

Probability of selecting a card with a multiple of 4 or 6

(ii) From numbers 1 to 30. there are 2 numbers which are multiple of 3 and 5 i.e.

*E*={15,30}.

Probability of selecting a card with a multiple of 3 and 5

(iii) From numbers 1 to 30, there are 14 numbers which is multiple of 3 or 5 i.e.

*E*={3,5,6,9,10,12,15,18,20,21,24,25,27,30}

Probability of selecting a card with a multiple of 3 or 5

Q5. A bag contains 100 identical marble stones which are numbered 1 to 100. If one stone is drawn at random from the bag, find the probability that it bears:

(i) a perfect square number

(ii) a number divisible by 4

(iii) a number divisible by 5

(iv) a number divisible by 4 and 5

(v) a number divisible by 4 or 5

answer:

Total number of possible outcomes =100

(i) Numbers which are perfect squares

Number of favorable outcomes =10

(ii) Numbers which are divisible by 4 =100/4=25

*A*={4,8,12,…,96,100}

P(number divisible by 4) =25/100=¼

(iii) Numbers which are divisible by 5 =100/5=20

P(number divisible by 5) =20/100=⅕

(iv) Numbers which are divisible by 4 and 5

P(number divisible by 4 and 5) =5/100=0.05

(v) Numbers which are divisible by 4 or 5 =[Numbers which are divisible by 4+Numbers which are divisible by 5-Numbers which are divisible by 4 and 5]

They are denoted by

*)=n(*

*A∪B**A*)+n(

*B*)-n(

*)*

*A∩B*n(

*)=25+20-5=40*

*A∪B*P(number divisible by 4 or 5) =40/100=0.40

Q6. Cards marked with numbers 1, 2, 3, …, 20 are well shuffled

and a card is drawn at random. What is the probability that

the number on the card is:

(i) a prime number

(ii) divisible by 3

(iii) a perfect square

answer:

Total possible outcomes =20

(i) Favorable outcomes for a prime number =2, 3, 5, 7, 11, 13, 17, 19.

Number of favorable outcomes =8

(ii) Favorable outcomes for a number divisible by 3 are 3, 6, 9, 12, 15, 18. Number of favorable outcomes =6

(iii) Favorable outcomes for a perfect square are 1, 4, 9, 16.

Number of favorable outcomes =4

Q7. A bag contains cards which are numbered from 2 to 90. A card is drawn at random from the bag. Find the probability that it bears.

(i) a two digit number

(ii) a number which is a perfect square

answer:

Total number of possible outcomes n(S)=89, they are {2, 3, 4, …., 90}

(i) Let *E* event of getting a 2 digit number

Number of favorable outcomes n(*E*)=81, they are {10, 11, 12, 13, …. 80}

(ii)

*E*→ event of getting a number which is perfect square

Number of favorable outcomes n(

*E*)=8, they are {4, 9, 16, 25, 36, 49, 64, 81}

Q8. What is the probability that a number selected at random from the number 1,2,2,3,3,3,4,4,4,4 will be their average?

answer:

Given numbers are 1,2,2,3,3,3,4,4,4,4. n(*E*)=10

The average (1+2+2+3+3+3+4+4+4+4):10=3

*E*→ event of getting 3

Number of favorable outcomes n(*E*)=3, (3,3,3)

Q9. Find the probability that a number selected at random from the numbers 1, 2, 3, …, 35 is a

(i) Prime number (ii) Multiple of 7 (iii) Multiple of 3 or 5

answer:

Total number of possible outcomes n(S)=35.

(i) *E*→ event of getting a prime number.

Number of favorable outcomes n(*E*)=11, they are

(ii)

*E*→ event of getting number which is multiple of 7

Number of favorable outcomes n(

*E*)=5, they are

(iii)

*E*→ event of getting number which is multiple of 3 or 5

Number of favorable outcomes n(

*E*)=16, they are

*A*→event of getting number which is multiple of 3

*B*→event of getting number which is multiple of 5

*A∩B*→event of getting number which is multiple of 3 and 5

*A∪B*→event of getting number which is multiple of 3 or 5

*)=n(*

*A∪B**A*)+n(

*B*)-n(

*)*

*A∩B*n(

*)=11+7-2=16*

*A∪B*P(

*)=16/35*

*A∪B*Q10. A box contains 90 discs which are numbered from 1 to 90. If one discs is drawn at random from the box, find the probability that it bears

(i) a two digit number

(ii) a perfect square number

(iii) (iii) a number divisible by 5.

answer:

Total number of possible outcomes n(S)=90, they are

(i)

*E*→ event of getting 2 digit number

Number of favorable outcomes n(

*E*)=81, they are

(ii)

*E*→ event of getting a perfect square.

Number of favorable outcomes n(

*E*)=9, they are

(iii)

*E*→ event of getting a number divisible by 5.

Number of favorable outcomes n(

*E*)=90/5=18, they are