Probability: Play with numbers (part 2)

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,

{13,14,15,…,60}
(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}
→n(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
P(a perfect square) =10/100=1/10
(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
B={5,10,15,…,95,100}
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)=n(A)+n(B)-n(A∩B)
n(A∪B)=25+20-5=40
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

P(a prime number)=8/20=⅖
(ii) Favorable outcomes for a number divisible by 3 are 3, 6, 9, 12, 15, 18. Number of favorable outcomes =6
P(divisible by 3) =6/20=3/10=0.3
(iii) Favorable outcomes for a perfect square are 1, 4, 9, 16.
Number of favorable outcomes =4
P(a perfect square)=4/20=⅕

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)=n(A)+n(B)-n(A∩B)
n(A∪B)=11+7-2=16
P(A∪B)=16/35

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

{1,2,3,…,90}
(i) E→ event of getting 2 digit number
Number of favorable outcomes n(E)=81, they are