**Warning**: Undefined array key "https://nctbsolution.com/joy-of-mathematics-class-8-solutions/" in

**/home/862143.cloudwaysapps.com/hpawmczmfj/public_html/wp-content/plugins/wpa-seo-auto-linker/wpa-seo-auto-linker.php**on line

**192**

**Joy of Mathematics Class 8 Solutions Chapter 23 Probability**

Welcome to NCTB Solution. Here with this post we are going to help 8th class students for the Solutions of Joy of Mathematics Class 8 Book, Chapter 23 Probability. Here students can easily find step by step solutions of all the problems for Probability. All the problem are solved with easily understandable methods so that all the students can understand easily. Here students will find solutions for Exercise 23

**Probability Exercise 23 Solution : **

**Question no – (1)**

**Solution :**

When two coins are tossed, the possible outcomes are HH, HT, TH, TT

**∴** Here total number of outcomes = 4

**(a)** Probability of getting two heads = 1/4

**(b)** Probability of getting one tail and one head

= 2/4

= 1/2

**(c)** Probability of getting two tail = 1/4

**(d)** Probability of getting at least one tail = 3/4

**Question no – (2)**

**Solution :**

When, three coins are tossed, the possible outcomes are HHH, HHT, HTH, THH, HTT, THT, TTH, TTT

**∴** Total outcome = 8

**(a)** Probability of getting three tails = 1/8

**(b)** Probability of getting no tail = 1/8

**(c)** Probability of getting at least one tail = 7/8

**(d)** Probability of getting two heads = 3/8

**Question no – (3)**

**Solution :**

When a die is rolled, all possible outcomes are 1, 2, 3, 4, 5, 6

**∴** Total number of outcomes = 6

**(a)** Favorable outcome (odd number) = 1, 3, 5

∴ P (an odd number) = 3/6 = 1/2

**(b)** Favorable outcome (number greater than 3)

= 4, 5, 6

∴ P (a number greater than) = B/6 = 1/2

**(c)** Favorable outcome (number less than 3)

= 1, 2

**∴** P (a number less than)

= 2/6

= 1/3

**(d)** Favorable outcome (a number greater than 6) = 0

**∴** P (a number greater than) = 0/6 = 0

**Question no – (4)**

**Solution :**

When two dice are rolled together the sample space is,

(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)

∴ Total number of outcomes = 36

**(a)** Probability of two number on the top of dice as less than 5

= 6/36

= 1/6

**(b)** Probability of two number on the top of dice as less than 2

= 0/36

= 0

**(c)** Probability of two number on the top of dice as greater than 12

= 0/36

= 0

**(d)** Probability of two numbers on the top of dice as 12

= 1/36

**Question no – (5)**

**Solution :**

Here, total number of outcome = 6

Number of red sector = 2

Number of black sector = 1

Number of blue sector = 2

Number of green sector= 1

**(a)** Probability of red color sector

= 2/6

= 1/3

**(b)** Probability of blue colored sector

= 2/6

= 1/3

**(c)** Probability green colored sector

= 1/6

**(d)** Probability of red or blue colored sector

= 1/3 + 1/3

= 1 + 1/3

= 2/3

**Question no – (6)**

**Solution :**

Here, the outcome = 8

Number of letter A = 2

Number of letter B = 1

Number of letter C = 2

Number of letter D = 2

Number of letter E = 1

**(a)** Probability of A = 2/8 = 1/4

**(b)** Probability of D = 2/8 = 1/4

**(c)** Probability of E = 1/8

**(d)** Probability of B = 1/8

**Question no – (7)**

**Solution :**

In a box, there are 15 cards.

The cards are = 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15

**(a)** Favorable outcomes (even number) = 2, 4, 6, 8, 10, 12

**∴** P (an even number)

= 6/15

= 2/5

**(b)** Favorable outcomes (odd number) = 1, 3, 5, 7, 9, 11, 13, 15

**∴** P (an odd number) = 8/15

**(c)** Favorable outcomes (factor of 15) = 1, 3, 5, 15

P (factor of 15) = 4/15

**(d)** Favorable outcomes (composite number) = 4, 6, 8, 9, 10, 15

**∴** P (composite number)

= 6/15

= 2/5

**Question no – (8)**

**Solution :**

Here, 15 beads in a necklace with 3 green, 4 red, 2 black and 6 white

**∴** Number of outcomes = 15

**(a)** Probability of a green bead = 3/15 = 1/5

**(b)** Probability of a red bead = 4/15

**(c)** Probability of a black bead = 2/15

**(d)** Probability of a green or red bead

= 3/15 + 4/15

= 3 +4/15

= 7/15

**Question no – (9)**

**Solution :**

A bag contains 5 red, 10 black, 15 green balls.

**∴** Here, the total number of balls

= 5 + 10 + 15

= 30

**(a)** Number of a red ball = 5

**∴** Probability of a red ball (P)

= 5/30

= 1/6

**(b)** Number of a black ball = 10

**∴** Probability of a black ball (P)

= 10/30

= 1/3

**(c)** Number of not a black ball

= 5 + 15

= 20

**∴** Probability of not a black ball (P)

= 20/30

= 2/3

**(d)** Number of not a green ball

= 5 + 10

= 15

**∴** Probability of not a green ball (P)

= 15/30

= 1/2

**Question no – (10)**

**Solution :**

Outcomes |
1 | 2 | 3 | 4 | 5 | 6 |

Frequency |
15 | 17 | 16 | 20 | 13 | 19 |

Here, total number of outcomes = 100

**(a)** Favorable outcomes (odd number) = 15, 16, 13

**∴** Probability (an odd number)

= 15 + 16 + 13/100

= 44/100

= 22/50

= 11/25

**(b)** Favorable outcomes (even number) = 17, 20, 19

**∴** Probability (an even number)

= 17 + 20 + 19/100

= 56/100

= 28/50

= 14/25

**(c)** Favorable outcomes (Prime number) = 17, 16, 13

**∴** Probability (a prime number)

= 17 + 16 + 13/100

= 46/100

= 23/50

**(d)** Favorable outcomes (number less than 7)

= 15, 17, 16, 20, 13, 19

**∴** Probability (a number less than 7))

= 15 + 17 + 16 + 20 + 13 + 19/100

= 100/100

= 1

**(e)** Favorable outcomes (more than 6) = 0

**∴** Probability (a number more than 6) = 0/100 = 0

**(f)** Favorable outcomes (composite number) = 20, 19

**∴** Probability (a composite number)

= 20 + 19/100

= 39/100

**Previous Chapter Solution : **