**Class 8 ICSE Maths Solutions Chapter 23 Probability (Selina Concise)**

Welcome to NCTB Solutions. Here with this post we are going to help 8th class students for the Solutions of Selina Class 8 ICSE Math Book, Chapter 23, Probability. Here students can easily find step by step solutions of all the problems for Probability, Exercise 23 Also here our mathematics teacher’s are solved all the problems with easily understandable methods with proper guidance so that all the students can understand easily. Here in this post students will get chapter 23 solutions. Here in this post all the solutions are based on ICSE latest Syllabus.

**Probability Exercise 23 Solution :**

**Question no – (1) **

**Solution : **

**(i)** a prime number,

A die has 1, 2, 3, 4, 5, 6

No of possible out comes = 6

Prime no = 1, 3, 5.

p (E) = 3/6

= 1/2

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

No of favorable outcome = 5 and 6.

So, P(E) = 2/6

= 1/3

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

No of favorable outcome not greater than 4 are, 1, 2, 3, 4

So, p (E) = 4/6

= 2/3

**Question no – (2) **

**Solution : **

**(i)** Getting tail favorable outcomes = 1

no of favorable outcome = 2

**∴** P(E) = 1/2

**(ii)** A head,

**∴** P (E) = 1/2

Therefore, probability of getting a tail will be 1/2 and a head will be 1/2.

**Question no – (3) **

**Solution : **

**(i)** Exactly one head,

Exactly one head, possible no of favorable outcome = 2

Total no of possible outcome = 4

**∴** P(E) = 2/4

= 1/2

**(ii)** Exactly one tail,

Exactly one tail

possible no of favorable outcomes = 2

Total no of possible outcome = 4

**∴** P(E) = 2/4

= 1/2

**(iii)** Two tails,

Two tails,

Favorable outcomes = 1

Total no of possible outcomes = 4

**∴** P(E) = 1/4

**(iv)** Two heads,

Possible no of favorable outcomes = 1

Total no of possible outcomes = 4

**∴** P(E) = 1/4

**Question no – (4) **

**Solution : **

In ‘PENCIL’ total word = 6

Total consonant = 4 (P, E, N, C, I, L)

**∴** P (E) = 4/6

= 2/3

Therefore, the probability that the letter chosen is a consonant will be 2/3.

**Question no – (5) **

**Solution : **

**(i)** A red ball

Total no of possible outcome = 3

a red ball possible outcome = 1

**∴** p(E) = 1/3

**(ii)** Not a red ball

Number of favorable outcome = 1 + 1

= 2

**∴** P(E) = 2/3

**(iii)** A white ball

Number of favorable outcome = 0

**∴** P(E) = 0/3

= 0

**Question no – (6) **

**Solution : **

**(i)** Greater than 2,

Die has 6 numbers = 1, 2, 3, 4, 5, 6

Possible outcome = 6

**∴** P(E) = 4/6

= 2/3

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

Favorable = 1, 2

**∴** P(E) = 2/6

= 1/3

**(iii)** Not greater than 2,

Favorable outcome = 1, 2

**∴** P(E) = 2/6

= 1/3

Therefore, the probability of getting a number not greater than 2 will be 1/3

**Question no – (7) **

**Solution : **

In a bag 3 balls white, 2 are red, 5 are black.

**(i)** a black ball

Total ball = (2 + 5 + 3)

= 10

No of favorable outcome of 1 black ball = 5

**∴** P (E) = 5/10

= 1/2

**(ii)** a red ball

One red ball favorable outcome = 2

Total = 10

**∴** P(E) = 2/10

= 1/5

**(iii)** a white ball

No of outcome white ball = 10

Favorable outcome = 3

**∴** P (E) = 3/10

**(iv)** not a red ball

Number of favorable outcome

= (3 + 5)

= 8

Not red ball.

**∴** P (E) = 8/10

= 4/5

**(v)** not a black ball

No of favorable outcome not a black ball = (3 + 2)

= 5

**∴** P (E) = 5/10

= 1/2

**Question no – (8) **

**Solution : **

**(i)** Will be an even number

A dies, number of favorable outcome of an even number

= 2, 4, 6 = 3

**∴** P (E) = 3/6

= 1/2

**(ii)** Will be an odd number

**∴** P (E) = 3/6

= 1/2

**(iii)** Will not an even number.

**∴** P (E) = 3/6

= 1/2

**Question no – (9) **

**Solution : **

**(i)** 8

Favorable outcome = 0

(8 is not possible)

**∴** P (E) = 0/6

= 0

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

Number of greater than 8 will be zero

**∴** P (E) = 0/6

= 0

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

Number less than 8 will, 1, 2, 3, 4, 5, 6

**∴** P (E) = 6/6

= 1

**Question no – (10) **

**Solution : **

(ii) 3.8

(iv) -0.8

(vi) -2/5

**(ii), (iv)** and **(vi)** number cannot be the probability an event.

**Question no – (11) **

**Solution : **

Total black balls = 6.

**(i)** White ball probability = 0

**∴** P (E) = 0

**(ii)** Black ball,

Favorable outcome = 1

**∴** P (E) = 1/6

**Question no – (12) **

**Solution : **

**(i)** Here, outcomes = 8

(H, H, H), (T, T, T), (H, H, T), (H, T, H), (T, H, H), (T, T, H) (H, T, T), (T, H, T)

**∴** Getting all heads, P (H) = 1/8

**(ii)** exactly two heads ?

Favorable outcomes = 3

**∴** P (E) = 3/8

**(iii)** exactly one head ?

Favorable outcomes = 3

**∴** P (E) = 3/8

**(iv)** no head ?

Here, favorable outcomes = 1

**∴** P (E) = 1/8

**Question no – (13) **

**Solution : **

Total page of book = 92

Possible outcomes = 92

**∴** Number of pages whose sum of its page 9,

= 10 that are, 9, 18, 27, 36, 45, 54, 63, 72, 81, 90

**∴** P (E) = 10/92

= 5/46

Therefore, the probability that sum of digits in the page number is 9 is 5/46.

**Question no – (14) **

**Solution : **

Coin has = H and T.

Two coin tossed so, no of coins = 2 × 2 = 4

HH, TT, HT, TH

**(i)** at least one head ?

At least one head, then no of outcomes = 3.

**∴** P (E) = 3/4

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

When both head or both tails, then no of outcomes = 2

**∴** P (E) = 2/4

= 1/2

**Question no – (15) **

**Solution : **

**(i)** Total outcomes = 10

1, 2, 3, 4, 5, 6, 7, 8, 9, 10

**∴ **P (E) = 5/10

= 1/2

**(ii)** Favorable outcomes = 3

**∴** P (E) = 3/10

**(iii)** Favorable outcome = 1

**∴** P (E) = 1/10

**(iv)** Favorable outcomes = 7

**∴** P (E) = 7/10

[that are 2, 3, 4, 6, 8, 9, 10]

**Question no – (17) **

**Solution : **

Total outcome = 6

**(i)** A prime number,

Favorable outcome = 3

**∴** P (E) = 3/6

= 1/2

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

Favorable outcome = 3

**∴** P (E) = 3/6

= 1/2

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

Favorable outcome = 4

**∴** P (E) = 4/6

= 2/3

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

Favorable outcome = 5

**∴** P (E) = 5/6

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

Favorable outcome = 0

**∴** P (E) = 0/6

**Question no – (18) **

**Solution : **

Here, total outcome = 4

(HH, TT, HT, TH)

**(i) **Favorable outcome = 2

**∴** P(E) = 2/4

= 1/2

**(ii)** at least one head

Favorable outcome = 3

(HH, HT, TH)

P (E) = 3/4

**(iii)** no head

Favorable outcome = 1

**∴** P (E) = 1/4

**(iv)** at most one head

Favorable outcome = 3

**∴** P (E) = 3/4

**Previous Chapter Solution :**

👉 Chapter 1 👈