OP Malhotra Class 10 Maths Solutions Chapter 20 Probability Ex 20(b)

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S Chand Class 10 ICSE Maths Solutions Chapter 20 Probability Ex 20(b)

Question 1.
Find the probability of getting a number less than 5 in a single throw of a die.
Solution:
∵ A die has numbers from 1 to 6
∴ Total possible outcomes = 6
Now favourable outcomes = 1, 2, 3, 4 = 4
∴ Probability p (E) = \(\frac{\text { Number of favourable outcomes }}{\text { Number of possible outcomes }}\) = \(\frac { 4 }{ 6 }\) = \(\frac { 2 }{ 3 }\)

Question 2.
A jar contains 3 white, 4 blue, 5 red and 2 green marbles. If a marble is drawn at random from the jar, what is the probability that the marble is (i) white ?, (ii) blue ?, (iii) red ?, (iv) green ?, (v) not white ?, (vi) not red?
Solution:
∴ Total number of marbles = 3 white + 4 blue + 5 red + 2 green = 14 marble
∴ Total possible outcomes = 14
We know that P(E) = \(\frac{\text { Total number of favourable outcomes }}{\text { Total possible outcomes }}\)
∴ Probability of

(i) The marble being white = P (E) = \(\frac { 3 }{ 14 }\)
(ii) The marble being blue = P (E) = \(\frac { 4 }{ 14 }\) = \(\frac { 2 }{ 7 }\)
(iii) The marble being red = P (E) = \(\frac { 5 }{ 14 }\)
(iv) The marble being green = P (E) = \(\frac { 5 }{ 14 }\) = \(\frac { 1 }{ 7 }\)
(v) The marble is not white = P (E) = \(\frac { 14 – 3 }{ 14 }\) = \(\frac { 11 }{ 14 }\) (∵ white marble are 3)
(vi) The marble which is not red = P (E) = \(\frac { 14 – 5 }{ 14 }\) = \(\frac { 9 }{ 14 }\) (∵ red marbles are 5)

Question 3.
If a card is drawn at random from a standard deck of playing cards, what is the probability that it is
(i) a nine?
(ii) an ace?
(iii) black?
(iv) red?
(v) a heart?
(vi) a diamond?
(vii) a face card?
(viii) not a face card?
(ix) the ace of diamonds?
(x) a red face card?
(xi) a black ace?
(xii) a red king?
Solution:
We know that a standard pack of playing cards contains 52 cards, of 4 suits, Hearts and Diamonds are red and spades and clubs are black. Each suit has 13 cards, four of which are face cards (ace, king, queen and jack). Therefore,
Total possible outcomes = 52
Probability of being card

(i) a nine, there are four 9’s in the pack of cards, P (E) = \(\frac{\text { Number of favourable outcomes }}{\text { Number 0f possible outcomes }}\) = \(\frac { 4 }{ 2 }\) = \(\frac { 1 }{ 3 }\)

(ii) an ace, there are 4 aces in the pack of cards P (E) = \(\frac{\text { Number of favourable outcomes }}{\text { Number 0f possible outcomes }}\) = \(\frac { 4 }{ 2 }\) = \(\frac { 1 }{ 3 }\)

(iii) black, there are 26 black cards in the pack of cards
∴ P (E) = \(\frac{\text { Number of favourable outcomes }}{\text { Number of possible outcomes }}\) = \(\frac { 26 }{ 52 }\) = \(\frac { 1 }{ 2 }\)

(iv) Red, there are 26 red cards in the pack of cards
∴ P (E) = \(\frac{\text { Number of favourable outcomes }}{\text { Number of possible outcomes }}\) = \(\frac { 26 }{ 52 }\) = \(\frac { 1 }{ 2 }\)

(v) a heart, there are 13 cards of ‘heart suit’
∴ P (E) = \(\frac{\text { Number of favourable outcomes }}{\text { Number of possible outcomes }}\) = \(\frac { 13 }{ 52 }\) = \(\frac { 1 }{ 4 }\)

(vi) a diamond, there are 13 cards of diamond suit
∴ P (E) = \(\frac{\text { Number of favourable outcomes }}{\text { Number of possible outcomes }}\) = \(\frac { 13 }{ 52 }\) = \(\frac { 1 }{ 4 }\)

(vii) a face card, there are 4 × 4= 16 face cards in the pack of cards
∴ P (E) = \(\frac{\text { Number of favourable outcomes }}{\text { Number of possible outcomes }}\) = \(\frac { 16 }{ 52 }\) = \(\frac { 4 }{ 13}\)

(viii) not a face cards
There are 9 × 4= 36 cards which are not face cards
∴ P (E) = \(\frac{\text { Number of favourable outcomes }}{\text { Number of possible outcomes }}\) = \(\frac { 36 }{ 52 }\) = \(\frac { 9 }{ 13 }\)

(ix) The ace of diamond, there is only one ace of diamond
∴ P (E) = \(\frac{\text { Number of favourable outcomes }}{\text { Number of possible outcomes }}\) = \(\frac { 1 }{ 52 }\)

(x) A red face card
Number of red face cards in a deck of 52 cards = 4 × 2 = 8
∴ P (E) = \(\frac{\text { Number of favourable outcomes }}{\text { Number of possible outcomes }}\) = \(\frac { 8 }{ 52 }\) = \(\frac { 2 }{ 13 }\)

(xi) A black ace
Number of black aces = 2
∴ P (E) = \(\frac{\text { Number of favourable outcomes }}{\text { Number of possible outcomes }}\) = \(\frac { 2 }{ 52 }\) = \(\frac { 1 }{ 26}\)

(xii) A red king
Number of red kings = 2
∴ P (E) = \(\frac{\text { Number of favourable outcomes }}{\text { Number of possible outcomes }}\) = \(\frac { 2 }{ 52 }\) = \(\frac { 1 }{ 26}\)

Question 4.
A bag contains 3 red bails, 5 black balls and 4 white balls. A ball is drawn at random from the bag. What is the probability that the ball drawn is :
(i) White
(ii) red
(iii) black?
Solution:
Total balls in the bag = 3 red + 5 black + 4 white = 12 balls
∴ Number of possible outcomes = 12

(i) Probability of a white ball
P (E) = \(\frac{\text { Number of favourable outcomes }}{\text { Number of possible outcomes }}\) = \(\frac { 4 }{ 12 }\) = \(\frac { 1 }{ 3}\)

(ii) Probability of a red ball
P (E) = \(\frac{\text { Number of favourable outcomes }}{\text { Number of possible outcomes }}\) = \(\frac { 3 }{ 12 }\) = \(\frac { 1 }{ 4}\)

(iii) Probability of a black ball
P (E) = \(\frac{\text { Number of favourable outcomes }}{\text { Number 0f possible outcomes }}\) = \(\frac { 5 }{ 12 }\)

Question 5.
1000 tickets of a lottery were sold and there are 5 prizes on these tickets. If Salman has purchased on lottery ticket, what is the probability of
(i) winning a prize?
(ii) not winning a prize?
Solution:
Total number of tickets = 1000
Number of prizes = 5

(i) ∴ Probability of a ticket which bring a prize
P (E) = \(\frac{\text { Number of favourable outcomes }}{\text { Number of possible outcomes }}\) = \(\frac { 5 }{ 1000 }\) = \(\frac { 1 }{ 200 }\) = 0.005

(ii) Number of tickets which do not give prize = 1000 – 5 = 995
Probability of a ticket getting not a prize
P (E) = \(\frac{\text { Number of favourable outcomes }}{\text { Number of possible outcomes }}\) = \(\frac { 995 }{ 1000 }\) = \(\frac { 199 }{ 200 }\) = 0.995

Question 6.
The numbers from 1-15 are each printed on a counter and the 15 counters are placed in a bag. If one counter is picked at random, what is the probability that the number on it is : (i) an odd number (ii) an even number (iii) a prime number (iv) a multiple of 4
Solution:
Counters having numbers from 1 to 15 are 15
If one counter is drawn at random, then

(i) Odd number are 1, 3, 5, 7, 9, 11, 13, 15 = 8
∴ Probability of being an odd number
P (E) = \(\frac{\text { Number of favourable outcomes }}{\text { Number of possible outcomes }}\) = \(\frac { 8 }{ 15 }\)

(ii) Even numbers are 2, 4, 6, 8, 10, 12, 14 = 7
∴ Probability of even number
P (E) = \(\frac{\text { Number of favourable outcomes }}{\text { Number of possible outcomes }}\) = \(\frac { 7 }{ 15 }\)

(iii) Prime numbers are 2, 3, 5, 7, 11, 13 = 6
∴ Probability of being a prime number
P (E) = \(\frac{\text { Number of favourable outcomes }}{\text { Number of possible outcomes }}\) = \(\frac { 6 }{ 15 }\) = \(\frac { 2 }{ 5 }\)

(iv) A multiple of 4 are 4,8, 12 = 3
Probability of being a multiple of 4 = \(\frac{\text { Number of favourable outcomes }}{\text { Number of possible outcomes }}\) = \(\frac { 3 }{ 15 }\) = \(\frac { 1 }{ 5 }\)

Question 7.
Instead of numbers, the letters in the word CHANCE were stuck on a die. Find the probability of rolling:
(i) letter C
(ii) a vowel
(iii) a consonant
(iv) any letter except C.

Solution:
There are 6 letters on the die making the word CHANCE which are C, H, A, N, C, E

(i) Probability of being a letter being C = P (E) = \(\frac{\text { Number of favourable outcomes }}{\text { Number of possible outcomes }}\)
(There are 2 C’s) = \(\frac { 2 }{ 6 }\) = \(\frac { 1 }{ 3 }\)

(ii) Probability of being the letter a vowel which are A and E (2)
P (E) = \(\frac{\text { Number of favourable outcomes }}{\text { Number of possible outcomes }}\) = \(\frac { 2 }{ 6 }\) = \(\frac { 1 }{ 3 }\)

(iii) Probability of being a consonant letter = C, H, N, C = 4
P (E) = \(\frac{\text { Number of favourable outcomes }}{\text { Number of possible outcomes }}\) = \(\frac { 4 }{ 6 }\) = \(\frac { 2 }{ 3 }\)

(iv) Probability of being any letter except which are H, A, N, E = 4
P (E) = \(\frac{\text { Number of favourable outcomes }}{\text { Number of possible outcomes }}\) = \(\frac { 4 }{ 6 }\) = \(\frac { 2 }{ 3 }\)

Question 8.
A bag contains 5 white balls, 7 red balls, 4 black balls and 2 blue balls. One ball is drawn at random from the bag. What is the probability that the ball drawn is :
(i) white
(ii) red or black
(iii) not blue
(iv) neither white nor blue.
Solution:
Total number of balls in bag 5 white + 7 red + 4 black + 2 blue = 18 balls

(i) Probability of being a white ball
P (E) = \(\frac{\text { Number of favourable outcomes }}{\text { Number of possible outcomes }}\) = \(\frac { 5 }{ 18 }\)

(ii) Probability of being a red or black ball
P (E) = \(\frac{\text { Number of favourable outcomes }}{\text { Number of possible outcomes }}\) = \(\frac { 7+4 }{ 18 }\) = \(\frac { 11 }{ 18 }\)

(iii) Probability of being not a blue ball
P (E) = \(\frac{\text { Number of favourable outcomes }}{\text { Number of possible outcomes }}\) = \(\frac { 18-2 }{ 18 }\) = \(\frac { 16 }{ 18 }\) = \(\frac { 8 }{ 9 }\)

(iv) Probability of a ball which is neither white nor a blue
P (E) = \(\frac{\text { Number of favourable outcomes }}{\text { Number of possible outcomes }}\) = \(\frac { 18-(5+2) }{ 18 }\) = \(\frac { 18-7 }{ 18 }\) = \(\frac { 11 }{ 18 }\)

Question 9.
Find the probability of drawing an ace or a jack from a pack of 52 cards.
Solution:
Number of possible outcomes = 52 cards
Number of favourable outcomes = 4 aces + 4 jacks = 8 cards
∴ Probability of being an ace or a jack
P (E) = \(\frac{\text { Number of favourable outcomes }}{\text { Number of possible outcomes }}\) = \(\frac { 8 }{ 52 }\) = \(\frac { 2 }{ 13 }\)

Question 10.
Tickets numbered 1 to 25 are mixed up and a ticket is drawn at random. What is the probability that the ticket drawn has a number which is a multiple of both 2 and 3?
Solution:
Total number of tickets = 25
Multiples of 2 and 3 both are 6, 12, 18, 24 =4
Probability of ticket which is a multiple of both 2 and 3
P (E) = \(\frac{\text { Number of favourable outcomes }}{\text { Number of possible outcomes }}\) = \(\frac { 4 }{ 25 }\)

Question 11.
A card is drawn at random from a pack of 52 cards. Find the probability that the card drawn is
(i) black
(ii) red and a queen
(iii) neither heart nor a king
(iv) an ace, jack or a king
(v)‘10’ of diamonds
(vi) club or an ace
Solution:
Number of possible outcomes = 52
(i) Number of black cards = 26
∴ Probability of being a black card
P (E) = \(\frac{\text { Number of favourable outcomes }}{\text { Number of possible outcomes }}\) = \(\frac { 26 }{ 52 }\) = \(\frac { 1 }{ 2 }\)

(ii) Number of cards being a red and a queen = 2
∴ Probability P (E) = \(\frac{\text { Number of favourable outcomes }}{\text { Number of possible outcomes }}\) = \(\frac { 2 }{ 52 }\) = \(\frac { 1 }{ 26 }\)

(iii) Number of cards which are neither heart nor a king = 52 – (13 +3 other kings) = 52 – 16 = 36
∴ Probability P (E) = \(\frac { 36 }{ 52 }\) = \(\frac { 9 }{ 13 }\)

(iv) Number of cards which is an ace, jack or a king = 4 + 4 + 4 = 12
∴ Probability P (E) = \(\frac{\text { Number of favourable outcomes }}{\text { Number of possible outcomes }}\) = \(\frac { 12 }{ 52 }\) = \(\frac { 3 }{ 13 }\)

(v) Number of cards being 10’s of diamond = 1
∴ Probability P (E) = \(\frac{\text { Number of favourable outcomes }}{\text { Number of possible outcomes }}\) = \(\frac { 1 }{ 52 }\)

(vi) Number of cards which are club or an ace =13 + 3 = 16
∴ Probability P (E) = \(\frac{\text { Number of favourable outcomes }}{\text { Number of possible outcomes }}\) = \(\frac { 16 }{ 52 }\) = \(\frac { 4 }{ 13 }\)

Question 12.
A box contains 20 balls bearing numbers 1, 2, 3, 4,……………20. A ball is drawn at random from the box. What is the probability that the number on the ball is
(i) an odd number
(ii) divisible by 2 or 3
(iii) prime number
(iv) not divisible by 10.
Solution:
Total number of balls in the box bearing numbers 1 to 20 = 20
(i) Odd number between 1 to 20 = 1, 3, 5, 7, 9, 11, 13, 15, 17, 19 = 10
∴ Probability P (E) = \(\frac{\text { Number of favourable outcomes }}{\text { Number of possible outcomes }}\) = \(\frac { 10 }{ 20 }\) = \(\frac { 1 }{ 2 }\)

(ii) Number which are divisible by 2 or 3 are 2, 3, 4, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20 = 13
∴ Probability P (E) = \(\frac{\text { Number of favourable outcomes }}{\text { Number of possible outcomes }}\) = \(\frac { 13 }{ 20 }\)

(iii) Numbers which are prime = 2, 3, 5, 7, 11, 13, 17, 19 = 8
∴ Probability P (E) = \(\frac{\text { Number of favourable outcomes }}{\text { Number 0f possible outcomes }}\) = \(\frac { 8 }{ 20 }\) = \(\frac { 2 }{ 5 }\)

(iv) Numbers which are not divisible by 10.will be 1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 12, 13, 14, 15, 16, 17,18,19 = 18
∴ Probability P (E) = \(\frac{\text { Number of favourable outcomes }}{\text { Number of possible outcomes }}\) = \(\frac { 18 }{ 20 }\) = \(\frac { 9 }{ 10 }\)

Question 13.
A die is rolled. If the outcome is an even number, what is the probability that it is a prime number?
Solution:
A die has 6 numbers from 1 to 6 and a prime number which is even is 2 = 1
∴ Probability P (E) = \(\frac{\text { Number of favourable outcomes }}{\text { Number of possible outcomes }}\) = \(\frac { 1 }{ 6 }\)

Question 14.
A single letter is selected at random from the word “PROBABILITY”. What is.the probability of it being a vowel?
Solution:
In the word “PROBABILITY” there are vowels: O, A, I, I = 4 and total number of letter = 11
Number of favourable outcomes 4
∴ Probability P (E) = \(\frac{\text { Number of favourable outcomes }}{\text { Number of possible outcomes }}\) = \(\frac { 4 }{ 11 }\)

Question 15.
A survey of 500 families shows the following results :

No. of children in the family	1	2	3	0
No. of families	250	125	75	50

Out of these, one family is chosen at random. Find the probability that the chosen family has 2 children.
Solution:
Total number of families = 500
Number of families which has 2 children each = 125
∴ Probability P (E) = \(\frac{\text { Number of favourable outcomes }}{\text { Number of possible outcomes }}\) = \(\frac { 125 }{ 500 }\) = \(\frac { 1 }{ 4 }\)

Question 16.
If a spinner illustrated is spun, what is the probability of getting
(i) an even number?
(ii) a 3 or a 5?
(iii) a number greater than 4?
(iv) a multiple of 2?

Solution:
In a spinner given here has numbers 1 to 8 = 8
(i) Even numbers are 2, 4, 6, 8 = 4
∴ Probability P (E) = \(\frac{\text { Number of favourable outcomes }}{\text { Number of possible outcomes }}\) = \(\frac { 4 }{ 8 }\) = \(\frac { 1 }{ 2 }\)

(ii) Probability of a number which is either 3 or 5 = 2 = \(\frac{\text { Number of favourable outcomes }}{\text { Number of possible outcomes }}\) = \(\frac { 2 }{ 8 }\) = \(\frac { 1 }{ 4 }\)

(iii) Numbers greater than 4 are 5, 6, 7, 8 = 4
∴ Probability P (E) = \(\frac{\text { Number of favourable outcomes }}{\text { Number of possible outcomes }}\) = \(\frac { 4 }{ 8 }\) = \(\frac { 1 }{ 2 }\)

(iv) Numbers which are multiple of 2 are 2, 4, 6, 8 = 4
∴ Probability P (E) = \(\frac{\text { Number of favourable outcomes }}{\text { Number of possible outcomes }}\) = \(\frac { 4 }{ 8 }\) = \(\frac { 1 }{ 2 }\)

Question 17.
A fete is being held in a large hall which has 9 doors. By mistake, the caretaker has left 3 of them locked. If someone tries a door at random, what is the probability that it will be :
(i) locked
(ii) not locked
Solution:
Total number of doors (possible outcomes) = 9
Number of locked door = 3
Number of doors which are not locked = 9-3 = 6

(i) Probability of door which is locked
∴ Probability P (E) = \(\frac{\text { Number of favourable outcomes }}{\text { Number of possible outcomes }}\) = \(\frac { 3 }{ 9 }\) = \(\frac { 1 }{ 3 }\)

(ii) Probability of door which is not locked
∴ Probability P (E) = \(\frac{\text { Number of favourable outcomes }}{\text { Number of possible outcomes }}\) = \(\frac { 6 }{ 9 }\) = \(\frac { 2 }{ 3 }\)

Question 18.
A child has a block in the shape of a cube with one letter written on each face as shown.

The cube is thrown once. What is the probability of getting (i) A ? (ii) D?
Solution:
Numbers letters on the block = 6
Which are A, B, C, D, E, A

(i) Probability of coming letter ‘A’ = \(\frac{\text { Number of favourable outcomes }}{\text { Number of possible outcomes }}\) = \(\frac { 2 }{ 6 }\) = \(\frac { 1 }{ 3 }\)

(ii) Probability of coming letter ‘A’ = \(\frac{\text { Number of favourable outcomes }}{\text { Number of possible outcomes }}\) = \(\frac { 1 }{ 6 }\)

Question 19.
A card is drawn from a well-shuffled deck of playing cards. Find the probability of drawing
(i) a face card
(ii) a red face card
Solution:
Number of total cards in a deck of playing cards = 52

(i) Number of face cards in the deck = 4 × 4 = 16
∴ Probability of getting face card
Probability P (E) = \(\frac{\text { Number of favourable outcomes }}{\text { Number of possible outcomes }}\) = \(\frac { 16 }{ 52 }\) = \(\frac { 4 }{ 13 }\)

(ii) Number of red face cards = 2 × 4 = 8
∴ Probability of coming red face card
P (E) = \(\frac{\text { Number of favourable outcomes }}{\text { Number of possible outcomes }}\) = \(\frac { 8 }{ 52 }\) = \(\frac { 2 }{ 13 }\)

Question 20.
A bag contains 5 red balls and some bine balls. If the probability of drawing a blue ball is double that of a red ball, find the number of blue balls in the bag.
Solution:
Number of red balls = 5
Let number of blue balls = x
∴ Total balls in the bag = (5 + x)
Probability of drawing a blue ball is double that of a red ball
Now probability of a blue ball = \(\frac { x }{ 5 + x }\)
Probability of a red ball = \(\frac { 5 }{ 5 + x }\)
According to the condition,
\(\frac{x}{5+x}\) = \(\frac{2 \times 5}{5+x}\) ⇒ \(\frac{x}{5+x}\) = \(\frac{10}{5+x}\) ⇒ 5x + x² = 50 + 10x ⇒ x² + 5x – 10x = 50 = 0
⇒ x² – 5x – 50 = 0 ⇒ x² – 10x + 5x – 50 = 0

⇒ x (x – 10) + 5 (x – 10) = 0 ⇒ (x – 10) (x + 5) = 0
Either x – 10 = 0, then x = 10
or x + 5 = 0, then x = -5 which is not possible being negative
∴ Number of blue balls = 10

Question 21.
The king, queen and jack of clubs are removed from a deck of 52 playing cards and then well shuffled. One card is selected from the remaining cards. Find the probability of getting (i) a heart, (ii) a king, (iii) a club, (iv) the ‘10’ of hearts.
Solution:
Total number of cards in a deck = 52
Number of cards which have been removed = 3 (king, queen, jack of clubs)
∴ Remaining cards in the deck = 52 – 3 = 49
(i) Number of heart cards = 13
∴ Probability of a heart card
P (E) = \(\frac{\text { Number of favourable outcomes }}{\text { Number of possible outcomes }}\) = \(\frac { 13 }{ 49 }\)

(ii) Number of Kings = 3
∴ Probability of a King = \(\frac{\text { Number of favourable outcomes }}{\text { Number of possible outcomes }}\) = \(\frac { 3 }{ 49 }\)

(iii) Number of club cards in the deck = 13 – 3 = 10
∴ Probability of being a club card
P (E) = \(\frac{\text { Number of favourable outcomes }}{\text { Number of possible outcomes }}\) = \(\frac { 10 }{ 49 }\)

(iv) ‘10’ of hearts = 1
∴ Probability of ‘ 10’ of hearts
P (E) = \(\frac{\text { Number of favourable outcomes }}{\text { Number of possible outcomes }}\) = \(\frac { 1 }{ 49 }\)

Question 22.
If a coin is tossed twice, what is the probability of getting ‘head’ at least once?
Solution:
A coin has two sides head and tail i.e.
∴ By tossing the coin twice, the following outcomes are possible
HH, HT, TH, TT = 4
Number of getting head (H) at least once = 3
∴ Probability P (E) = \(\frac{\text { Number of favourable outcomes }}{\text { Number of possible outcomes }}\) = \(\frac { 3 }{ 4 }\)

Question 23.
Fill in the blanks with the appropriate correct answer.
(i) Chance of throwing 6 with a single die is _______
(ii) A pair of fair dice is thrown and one die shows a four. The probability that the other die shows 5 is _______
(iii) Probability of a sure event is _______
(iv) Probability of an impossible event is _______
(v) The probability of an event (other than sure and an impossible event) lies between.
(vi) A die is rolled once. The probability of getting a prime number is _______
Solution:
(i) \(\frac { 1 }{ 6 }\)
(ii) \(\frac { 1 }{ 36 }\)
(iii) 1
(iv) 0
(v) 0 and 1
(vi) \(\frac { 3 }{ 6 }\) = \(\frac { 1 }{ 2 }\)
Note for (ii): By throwing a pair of dice, the outcomes can be 6 × 6 = 36.

Self-Evaluation and Revision (LATEST ICSE QUESTIONS)

Question 1.
A die is thrown once. What is the probability that the
(i) number is even?
(ii) number is greater than 2?
Solution:
Dice is thrown once
Sample space = {1, 2, 3, 4, 5, 6}

(i) No. of ways in favour = 3
Total ways = 6
∴ Probability = \(\frac { 3 }{ 6 }\) = \(\frac { 1 }{ 2 }\)

(ii) No. of ways in favour = 4
Total ways = 6
∴ Probability = \(\frac { 4 }{ 6 }\) = \(\frac { 2 }{ 3 }\)

Question 2.
Cards marked with numbers 1, 2, 3, 4, ……. 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?
Solution:
(i) P (a prime number) = \(\frac { 8 }{ 20 }\) = \(\frac { 2 }{ 5 }\)
[∵ From 1 to 20, there are total ‘8’ prime numbers, 2, 3, 5, 7, 11, 13, 17, 19]

(ii) P (divisible by 3) = \(\frac { 6 }{ 20 }\) = \(\frac { 3 }{ 10 }\)
[∵ There are (3, 6, 9, 12, 15, 18) 6 numbers divisible by 3 from 1 to 20]

(iii) P (a perfect square) = \(\frac { 4 }{ 20 }\) = \(\frac { 1 }{ 5 }\)
[∵ There are (1, 4, 9, 16) 4 numbers which are perfect square from 1 to 20]

Question 3.
From a pack of 52 playing cards all cards whose numbers are multiples of 3 are removed. A card is now drawn at random. What is the probability that the card drawn is
(i) a face card (King, Jack or Queen)?
(ii) an even numbered red card?
Solution:
(i) n (S) = 52 – 12 = 40
n(E₁) = 12
P(E₁) = \(\frac{n\left(\mathrm{E}_1\right)}{n(\mathrm{~S})}\) = \(\frac { 12 }{ 40 }\) = \(\frac { 3 }{ 10 }\)

(ii) n (E₂) = 10
P(E₂) = \(\frac{n\left(\mathrm{E}_2\right)}{n(\mathrm{~S})}\) = \(\frac { 10 }{ 40 }\) = \(\frac { 1 }{ 4 }\)

Question 4.
Two coins are tossed once. Find the probability of getting :
(i) 2 heads
(ii) at least 1 tail.
Solution:
Total possible outcomes are : HH, HT, TT, TH, i.e., 4
(i) Favourable outcomes are HH, i.e., 1 So, P(2 heads)
= \(\frac{\text { no. of favourable outcomes }}{\text { total no. of possible outcomes }}\) = \(\frac { 1 }{ 4 }\)

(ii) Favourable outcomes are HT, TT, TH, i.e., 3
So, P (at least one tail) = \(\frac { 3 }{ 4 }\)

Question 5.
A box contains some black balls and 30 white balls. If the probability of drawing a black ball is two-fifths of a white ball, find the number of black balls in the box.
Solution:
No. of white balls = 30
No. of black balls = x
Total ball = x + 30
∴ P (white ball) = \(\frac { 30 }{ x + 30 }\)
P (black ball) = \(\frac { x }{ x + 30 }\)
According to question,
P (black ball) = \(\frac { 2 }{ 5 }\) × p (white ball)
⇒ \(\frac { x }{ x + 30 }\) = \(\frac { 2 }{ 5 }\) × \(\frac { 30 }{ x + 30 }\)
⇒ x = \(\frac { 2 }{ 5 }\) × 30
x = 12
∴ No. of black ball = 12

Question 6.
A die has 6 faces marked by the given numbers as shown below:

The die is thrown once. What is the probability of getting
(i) a positive integer.
(ii) an integer greater than -3.
(iii) the smallest integer.
Solution:
Total outcomes n(S)= 6
(i) a positive integer = (1, 2, 3)
No. of favourable n(E) = 3
Probability = \(\frac{n(\mathrm{E})}{n(\mathrm{~S})}\) = \(\frac { 3 }{ 6 }\) = \(\frac { 1 }{ 2 }\)

(ii) Integer greater than -3
= (1,2, 3, -1,-2)
No. of favourable n(E) = 5
Probability = \(\frac{n(\mathrm{E})}{n(\mathrm{~S})}\) = \(\frac { 5 }{ 6 }\)

(iii) Smallest integer = -3
No. of favourable n(E) = 1
Probability = \(\frac{n(\mathrm{E})}{n(\mathrm{~S})}\) = \(\frac { 1 }{ 6 }\)

Question 7.
A bag contains 5 white balls, 6 red balls and 9 green balls. A ball is drawn at random from the bag. Find the probability that the ball drawn is:
(i) a green ball
(ii) a white or a red ball
(iii) is neither a green ball nor a white ball.
Solution:
(i) Total number of balls = 5 + 6 + 9 = 20
P(Green ball) = \(\frac{\text { Number of green balls }}{\text { Total number of balls }}\) = \(\frac { 9 }{ 20 }\)

(ii) P(White ball or Red ball) = P(White ball) + P(Red ball)
= \(\frac{\text { Number of white balls }}{\text { Total number of balls }}\) + \(\frac{\text { Number of red balls }}{\text { Total number of balls}}\)
= \(\frac { 5 }{ 20 }\) + \(\frac { 6 }{ 20 }\) = \(\frac { 11 }{ 20 }\)

(iii) P(Neither green ball nor white ball) = P(Red ball)
= \(\frac{\text { Number of red balls }}{\text { Total number of balls }}\) = \(\frac { 6 }{ 20 }\) = \(\frac { 3 }{ 10 }\)

Question 8.
A game of numbers has cards marked with 11, 12, 13, ………, 40. A card is drawn at random. Find the probability that the number on the card drawn is :
(i) A perfect square
(ii) Divisible by 7
Solution:
The perfect squares lying between 11 and 40 are 16, 25 and 36.
So the number of possible outcomes is = 3
Total number of cards from 11 to 40 is 40 – 11 + 1 = 30

(i) Probability that the number on the card drawn is a perfect square = \(\frac{\text { Number of possible outcomes }}{\text { Total number of outcomes }}\) = \(\frac { 3 }{ 30 }\) = \(\frac { 1 }{ 10 }\)
So, the probability that the number on the card drawn is a perfect square is \(\frac { 1 }{ 10 }\)

(ii) The numbers from 11 to 40 that are divisible by 7 are 14, 21, 28 and 35. So the number of possible outcomes is 4. The total number of cards from 11 to 40 is 30. The probability that the number on the card drawn is divisible by 7
= \(\frac{\text { Number of possible outcomes }}{\text { Total number of outcomes }}\) = \(\frac { 4 }{ 30 }\) = \(\frac { 2 }{ 15 }\)
So, the probability that the number on the card drawn is divisible by 7 is \(\frac { 2 }{ 15 }\).