**1. A die is thrown, find the probability of getting:**

**(i) a prime number**

**(ii) a number greater than 4**

**(iii) a number not greater than 4.**

**Solution**

A die has six numbers : 1, 2, 3, 4, 5, 6

∴ Number of possible outcomes = 6

**(i)**A prime number

Number of favourable outcomes = a prime number = 1, 3, 5 which are in numbers

P(E) = (Numbers of favourable outcome)/(Numbers of all outcome)

= 3/6 = ½

**(ii)**Numbers of favourable outcome = greater than four i.e two number 5 and 6

∴ P(E) = (Numbers of favourable outcome)/(Number of all possible outcome)

= 2/6 = 1/3

**(iii)**Number of favourable outcome = not graeter than 4 or numbers will be 1, 2, 3, 4 which are 4 in numbers

∴ P(E) = (Numbers of favourable outcome)/(Number of all outcome)

= 4/6

= 2/3

**2. A coin is tossed. What is the probability of getting:**

**(i) a tail?**

**(ii) ahead?**

**Solution**

On tossing a coin once,

Number of possible outcome = 2

**(i)**Favourable outcome getting a tail = 1

⇒ number of favorable outcome = 2

∴ P(E) = (Number of favourable outcome)/(Number of all possible outcome)

= ½

**(ii)**a head

Similarly, favourable outcome getting a head = 1

But number of possible outcome = 2

∴ P(E) = (Number of favourable outcome)/(Number of all possible outcome)

= ½

**3. A coin is tossed twice. Find the probability of getting:**

**(i) exactly one head (ii) exactly one tail**

**(iii) two tails (iv) two heads**

**Solution**

**(i)**Exactly one head

Possible number of favourable outcomes = 2

(i.e. TH and HT)

Total number of possible outcomes = 4

∴ P(E) = (Number of favourable outcomes)/(Total number of possible outcomes)

= 2/4

= ½

**(ii)**Exactly one tail

Possible number of favourable outcomes = 2

(i.e. TH and HT)

Total number of possible outcomes = 4

P(E) = (Number of favourable outcomes)/(Total number of possible outcomes)

= 2/4

= ½

**(iii)**Two tails

Possible number of favourable outcomes = 1

(i.e., TT)

Total number of possible outcome s = 4

∴ P(E) = (Number of favourable outcomes)/(Total number of possible outcomes)

= ¼

(iv) Two heads

Possible number of favourable outcomes = 1

(i.e. HH)

Total number of possible outcomes = 4

∴ P(E) = ¼

**4. A letter is chosen from the word ‘PENCIL’ what is the probability that the letter chosen is a consonant?**

**Solution**

Total no. of letters in the word ‘PENCIL = 6

Total Number of Consonant = ‘PNCL’ i.e. 4

P(E) = (Total No. of consonants)/(Total No. of letters in the word PENCIL)

= 4/6

= 2/3

**5. A bag contains a black ball, a red ball and a green ball, all the balls are identical in shape and size. A ball is drawn from the bag without looking into it. What is the probability that the ball drawn is:**

**(i) a red ball**

**(ii) not a red ball**

**(iii) a white ball.**

**Solution**

**(i)**Total number of possible outcomes = 3

∴ P(E) = 1/3

**(ii)**Not a red ball

Number of favourable outcomes = Green ball + Black Ball

= 1 + 1 + 2

∴ P(E) = 2/3

**(iii)**A white ball

Number of favourable outcome = 0

∴ P(E) = 0/3

= 0

**6. In a single throw of a die, find the probability of getting a number**

**(i) greater than 2**

**(ii) less than or equal to 2**

**(iii) not greater than 2.**

**Solution**

**(i)**A die has six numbers = 1, 2, 3, 4, 5, 6

∴ Number of possible outcomes = 6

∴ P(E) = 4/6 = 2/3

**(ii)**Less than or equal to 2

Number of favourable outcome = 1, 2

∴ P(E) = 2/6

= 1/3

**(iii)**Not greater than 2

Number of favourable outcomes = 1, 2

∴ P(E) = 2/6

= 1/3

**7. A bag contains 3 white, 5 black and 2 red balls, all of the same shape and size.**

**A ball is drawn from the bag without looking into it, find the probability that the ball drawn is:**

**(i) a black ball.**

**(ii) a red ball.**

**(iii) a white ball.**

**(iv) not a red ball.**

**(v) not a black ball.**

**Solution**

In a bag, 3 balls are white

2 balls are red

5 balls are black

Total number of balls = 3 + 2 + 5 = 10

**(i)**Numbers of possible outcome of one black ball = 10

And number of favourable outcome of one black ball = 5

∴ P(E) = (Number of favourable outcome)/(Number of all possible outcome)

= 5/10

= 1/2

**(ii)**Number of possible outcome of one red ball = 10

And number of favourable outcome = 2

∴ P(E) = (Number of favourable outcome)/(Number of all possible outcome)

= 2/10

= 1/5

**(iii)**Number of possible outcome of white ball = 10

And number of favourable outcome = 3

∴ P(E) = (Number of favourable outcome)/(Number of all possible outcome)

= 3/10

**(iv)**Number of possible outcome = 10

Number of favourable outcome = 3 + 5 = 8

Not a red ball

∴ P(E) = (Number of favourable outcome)/(Number of all possible outcome)

= 8/10

= 4/5

**(v)**Number of possible outcomes = 10

Number of favourable outcome not a black ball = 3 + 2 = 5

∴ P(E) = (Number of favourable outcome)/(Number of all possible outcome) = 5/10

= 1/2

**8. In a single throw of a die, find the probability that the number:**

**(i) will be an even number.**

**(ii) will be an odd number.**

**(iii) will not be an even number.**

**Solution**

A die has six numbers : 1, 2, 3, 4, 5, 6

∴ Number of possible outcome = 6

**(i)**Number of favourable outcome = an even number i.e. 2, 4, 6 which are 3 in numbers

∴ P(E) = (Numbers of favourable outcome)/(Number of all possible outcome)

= 3/6

= 1/2

**(ii) & (iii)**Number of favourable outcome = not an even number i.e. odd numbers : 1, 3, 5 which are 3 in numbers

∴ P(E) = (Number of favourable outcome)/(Number of all possible outcome)

= 3/6

= 1/2

**9. In a single throw of a die, find the probability of getting :**

**(i) 8**

**(ii) a number greater than 8**

**(iii) a number less than 8**

**Solution**

On a die the numbers are 1, 2, 3, 4, 5, 6 i.e. six.

∴ Number of possible outcome = 6

**(i)**Number of favourable outcome = 0

**(∵ 8 is not possible)**

∴ P(E) = (Number of favourable outcome)/(Number of all possible outcome)

= 0/6 = 0

**(ii)**Number greater than 8 will be 0

∴ P(E) = (Number of favourable outcome)/(Number of all possible outcome)

= 0/6

= 0

**(iii)**Number less than 8 wil be 1, 2, 3, 4, 5, 6

∴ P(E) = (Number of favourable outcome)/(Number of all possible outcome)

= 6/6

= 1

**10. Which of the following can not be the probability of an event?**

**(i) 2/7**

**(ii) 3.8**

**(iii) 37%**

**(iv) - 0.8**

**(v) 0.8**

**(vi) -2/5**

**(vii) 7/8**

**Solution**

(i) The probability of an event cannot be

(ii) 3.8 i.e. the probability of an even cannot exceed 1.

(iv) i.e. -0.8 and

(vi) -2/5, This is because probability of an even can never be less than 1.

**11. A bag contains six identical black balls. A child withdraws one ball from the bag without looking into it. What is the probability that he takes out:**

**(i) a white ball,**

**(ii) a black ball**

**Solution**

∵ There are 6 black balls in a bag

∴ number of possible outcome = 6

**(i)**A white ball

As there is no white ball in the bag

∴ Its probability is zero (0) = or P(E) = 0

**(ii)**a black ball

∴ Number of favourable outcome = 1

∴ P(E) = (Number of favourable outcome)/(Number of all possible outcome)

= 1/6

**12. Three identical coins are tossed together. What is the probability of obtaining:**

**all heads?**

**exactly two heads?**

**exactly one head?**

**no head?**

**Solution**

Total outcomes = 8

i.e., (H, H, H), (H, H, T), (H, T, H), (T, T, T), (T, H, H), (T, T, H), (H, T, T), (T, H, T)

**(i)**Favourable outcome = i.e., (H, H, H)

∴ P (of getting all heads) = 1/8

**(ii)**Favourable outcome = 3 (H, H, T), (H, T, H), (T, H, H)

∴ P(E) = 3/8

**(iii)**Favourable outcome = 3 (H, T, H), (T, T, H), (H, T, T)

∴ P(E) = 3/8

**(iv)**Favourable outcomes = 1 i.e., (T, T, T)

∴ P(E) = 1/8

**13. A book contains 92 pages. A page is chosen at random. What is the probability that the sum of the digits in the page number is 9?**

**Solution**

Number of pages of the book = 92

Which are from 1 to 92

Number of possible outcomes = 92

∴ Number of pages whose sum of its page is 9 = 10

i.e. 9, 18, 27, 36, 45, 54, 63, 72, 81, 90

∴ P(E) = 10/92

= 5/46

**14. Two coins are tossed together. What is the probability of getting:**

**(i) at least one head**

**(ii) both heads or both tails.**

**Solution**

∵ A coins has two faces Head and Tailor H, T

∴ Two coins are tossed

∴ Number of coins = 2 x 2 = 4

which are HH, HT, TH, TT

**(i)**At least one head, then

Number of outcomes = 3

∴ P(E) = (Number of favourable outcome)/(Number of all possible outcome)

= 3/4

**(ii)**When both head or both tails, then

Number of outcomes = 2

∴ P(E) = (Numberof favourable outcome)/(Number of all possible outcome)

= 2/4

= 1/2

**15. From 10 identical cards, numbered 1, 2, 3, …… , 10, one card is drawn at random. Find the probability that the number on the card drawn is a multiple of:**

**(i) 2 (ii) 3**

**(iii) 2 and 3 (iv) 2 or 3**

**Solution**

Total outcomes = 10

i.e. 1, 2, 3, 4, 5, 6, 7, 8, 9, 10

**(i)**Favourable outcomes = 5 i.e., 2, 4, 6, 8, 10

P(E) = 5/10 = 1/2

**(ii)**Favourable outcomes = 3 i.e., 3, 6, 9

P(E) = 3/10

**(iii)**Favourable outcomes = 1 i.e., 6

P(E) = 1/10

**(iv)**Favourable outcome = 7

i.e., 2, 3, 4, 6, 8, 9, 10

P(E) = 7/10

**16. Two dice are thrown at the same time. Find the probability that the sum of the two numbers appearing on the top of the dice is:**

**(i) 0**

**(ii) 12**

**(iii) less than 12**

**(iv) less than or equal to 12**

**Solution**

Total outcomes = 36 i.e.

(1, 1) (1, 2) (1, 3) (1, 4) (1,5) (1, 6)

(2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6)

(3, 1) (3, 2) (3, 3) (3 ,4) (3, 5) (3, 6)

(4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6)

(5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6)

(6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6)

**(i)**Favourable outcomes = 0

P(E) = 0/36 = 0

**(ii)**Favourable outcomes = i.e., (6,6)

P(E) = 1/36

**(iii)**Favourable outcomes = 35

**[Except (6, 6)]**

P(E) = 35/36

**(iv)**Favourable outcome = 36

P(E) = 36/36

= 1

**17. A die is thrown once. Find the probability of getting:**

**(i) a prime number**

**(ii) a number greater than 3**

**(iii) a number other than 3 and 5**

**(iv) a number less than 6**

**(v) a number greater than 6.**

**Solution**

Total outcomes = 6

i.e., 1, 2, 3, 4, 5 and 6

**(i)**Favourable outcomes = 3 i.e., 2, 3, 5

P(E) = 3/6

= 1/2

**(ii)**Favourable outcomes = 3 i.e., 4, 5, 6

P(E) = 3/6

= 1/2

**(iii)**Favourable outcomes = 4 i.e., 1, 2, 4, 6

P(E) = 4/6

= 2/3

**(iv)**Favourable outcomes = 5

i.e., 1, 2, 3, 4 and 5

P(E) = 5/6

**(v)**Favourable outcomes = 0

P(E) = 0/6 = 0

**18. Two coins are tossed together. Find the probability of getting:**

**(i) exactly one tail**

**(ii) at least one head**

**(iii) no head**

**(iv) at most one head**

**Solution**

Total outcomes = 4

i.e., HH, HT, TT, TH

**(i)**Favourable outcomes = 2 i.e., HT and TH

P(E) = 2/4

= 1/2

**(ii)**Favourable outcomes = 3

i.e., HH, HT and TH

P(E) = 3/4

**(iii)**Favourable outcomes = 1

i.e., TT

P(E) = 1/4

**(iv)**Favourable outcomes = 3

i.e., HH, HT and TH

P(E) = 3/4