Class 8 RD Sharma Solutions – Chapter 26 Data Handling IV (Probability) – Exercise 26.1 | Set 1

Question 1. The probability that it will rain tomorrow is 0.85. What is the probability that it will not rain tomorrow?

Solution:

Let event of raining tomorrow be P (A)

The probability of raining tomorrow is P (A) = 0.85

Probability of not raining is given by P (A) = 1 â€“ P (A)

Therefore, probability of not raining tomorrow = P (A) = 1 â€“ 0.85

= 0.15

Question 2. A die is thrown. Find the probability of getting:(i) a prime number

Solution:

Outcomes of a die are 1, 2, 3, 4, 5, 5 and 6

Total outcomes = 6

Prime numbers are 1, 3 and 5

Favorable outcomes = 3

Probability of getting a prime number = Number of favorable outcomes/Total outcomes

= 3/6

= 1/2

Therefore, probability of getting a prime number = 1/2

(ii) 2 or 4

Solution:

Outcomes of a die are 1, 2, 3, 4, 5, 5 and 6

Total outcomes = 6

Favorable outcomes=2

Probability of getting 2 and 4 = Favorable outcomes/Total outcomes

= 2/6

= 1/3

Therefore, probability of getting 2 and 4 is 1/3

(iii) a multiple of 2 or 3

Solution:

Outcomes of a die are 1, 2, 3, 4, 5, 5 and 6

Total outcomes = 6

Multiples of 2 and 3 are 2, 3, 4 and 6

Favorable outcomes = 4

Total number of multiples are 4

Probability of getting a multiple of 2 or 3 = Favorable outcomes/Total number of outcomes

= 4/6

= 2/3

Therefore, probability of getting a multiple of 2 or 3 is 2/3

(i)8 as the sum

Solution:

Possible outcomes when a pair of dice is rolled are

(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)

(where first number shows the number on first dice and second number shows the number on second dice.)

Total number of outcomes = 36

Number of outcomes having 8 as sum are (6, 2), (5, 3), (4, 4), (3, 5) and (2, 6)

Numbers of favorable outcomes = 5

Probability of getting numbers of outcomes having 8 as sum = Favorable outcomes/Total outcomes

= 5/36

Therefore, probability of getting numbers of outcomes having 8 as sum is 5/36

(ii)a doublet

Solution:

Total number of outcomes = 36

Number of outcomes as doublet are (1, 1), (2, 2), (3, 3), (4, 4), (5, 5) and (6, 6)

Number of favorable outcomes = 6

Probability of getting numbers of outcomes as doublet = Favorable outcomes/Total outcomes

= 6/36

= 1/6

Therefore, probability of getting numbers of outcomes as doublet is 1/6

(iii)a doublet of prime numbers

Solution:

Total number of outcomes = 36

Number of outcomes as doublet of prime numbers are (1, 1), (3, 3), (5, 5)

Number of favorable outcomes = 3

Probability of getting numbers of outcomes as doublet of prime numbers = Favorable outcomes/Total outcomes

= 3/36

= 1/12

Therefore, probability of getting numbers of outcomes as doublet of prime numbers is 1/12

(iv)a doublet of odd numbers

Solution:

Total number of outcomes = 36

Number of outcomes as doublet of odd numbers are (1, 1), (3, 3), (5, 5)

Number of favorable outcomes = 3

Probability of getting numbers of outcomes as doublet of odd numbers = Favorable outcomes/Total outcomes

= 3/36

= 1/12

Therefore, probability of getting numbers of outcomes as doublet of odd numbers is 1/12

(v)a sum greater than 9

Solution:

Total number of outcomes = 36

Number of outcomes having sum greater than 9 are (4, 6), (5, 5), (5, 6), (6, 6), (6, 4), (6, 5)

Number of favorable outcomes = 6

Probability of getting numbers of outcomes having sum greater than 9 = Favorable outcomes/Total outcomes

= 6/36

= 1/6

Therefore, probability of getting numbers of outcomes having sum greater than 9 is 1/6

(vi)An even number on first

Solution:

Total number of outcomes = 36

Number of outcomes having an even number on first are:

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

Number of favorable outcomes = 18

Probability of getting numbers of outcomes having an even number on first = Favorable outcomes/Total outcomes

= 18/36

= 1/2

Therefore, probability of getting numbers of outcomes having an even number on first is 1/2

(vii)An even number on one and a multiple of 3 on the other

Solution:

Total number of outcomes = 36

Number of outcomes having an even number on one and a multiple of 3 on the other are (2, 3), (2, 6), (4, 3), (4, 6), (6, 3) and (6, 6)

Number of favorable outcomes = 6

Probability of getting an even number on one and a multiple of 3 on the other is = Favorable outcomes/Total outcomes

= 6/36

= 1/6

Therefore, probability of getting an even number on one and a multiple of 3 on the other is 1/6

(viii)Neither 9 nor 11 as the sum of the numbers on the faces

Solution:

Total number of outcomes = 36

Number of outcomes having 9 nor 11 as the sum of the numbers on the faces are (3, 6), (4, 5), (5, 4), (5, 6), (6, 3) and (6, 5)

Number of favorable outcomes for 9 nor 11 as the sum of the numbers on the faces are 6

Probability of getting 9 nor 11 as the sum of the numbers on the faces is = Favorable outcomes/Total outcomes

= 6/36

= 1/6

Probability of outcomes having 9 nor 11 as the sum of the numbers on the faces P (E) = 1/6

Probability of outcomes not having 9 nor 11 as the sum of the numbers on the faces is given by P (E) = 1 â€“ 1/6 = (6 – 1)/5 = 5/6

Therefore, probability of outcomes not having 9 nor 11 as the sum of the numbers on the faces is 5/6

(ix)A sum less than 6

Solution:

Total number of outcomes = 36

Number of outcomes having a sum less than 6 are (1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (4, 1)

Number of favorable outcomes = 10

Probability of getting a sum less than 6 is = Favorable outcomes/Total outcomes

= 10/36

= 5/18

Therefore, probability of getting sum less than 6 is 5/18

(x)A sum less than 7

Solution:

Total number of outcomes = 36

Number of outcomes having a sum less than 7 are (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (2, 1), (2, 2), (2, 3), (2, 4), (3, 1), (3, 2), (3, 3), (4, 1), (4, 2), (5, 1)

Number of favorable outcomes = 15

Probability of getting a sum less than 7 is = Favorable outcomes/Total outcomes

= 15/36

= 5/12

Therefore, probability of getting sum less than 7 is 5/12

(xi)A sum more than 7

Solution:

Total number of outcomes = 36

Number of outcomes having a sum more than 7 are

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

Number of favorable outcomes = 15

Probability of getting a sum more than 7 is = Favorable outcomes/Total outcomes

= 15/36

= 5/12

Therefore, probability of getting sum more than 7 is 5/12

(xii)At least once

Solution:

Total number of outcomes = 36

Number of favorable outcomes =11

Probability of getting outcomes for at least once is = Favorable outcomes/Total outcomes

= 11/36

Therefore, probability of getting outcomes for at least once is 11/36

(xiii) A number other than 5 on any dice.

Solution:

Total number of outcomes = 36

Number of outcomes having 5 on any die are

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

Number of favorable outcomes having 5 on any die = 15

Probability of getting 5 on any die is = Favorable outcomes/Total outcomes

= 11/36

Therefore, probability of getting 5 on any die is 11/36

Probability of not getting 5 on any die P (E) = 1 â€“ P (E)

= 1 â€“ 11/36

= (36 – 11)/36

= 25/36

Therefore, probability of not getting 5 on any die is 25/36

4. Three coins are tossed together. Find the probability of getting:(i) exactly two heads

Solution:

Possible outcome of tossing three coins are HTT, HHT, HHH, HTH, TTT, TTH, THT, THH

Total outcomes = 8

Number of outcomes of exactly two heads are HHT,HTH,THH

Favorable outcomes = 3

Probability of getting exactly two heads is = Favorable outcomes/Total outcomes

= 3/8

Therefore, probability of getting exactly two heads is 3/8

Solution:

Possible outcome of tossing three coins are: HTT, HHT, HHH, HTH, TTT, TTH, THT, THH

Total outcomes = 8

Number of outcomes of at least two heads are HHT,HHH,HTH,THH

Favorable outcomes = 4

Probability of getting at least two heads = Favorable outcomes/Total outcomes

= 4/8

= 1/2

Therefore, probability of getting at least two heads is 1/2

(iii) at least one head and one tail

Solution:

Possible outcome of tossing three coins are: HTT, HHT, HHH, HTH, TTT, TTH, THT, THH

Total outcomes = 8

Number of outcomes of at least one head and one tail are HTT, HHT, HTH, TTH, THT, THH

Favorable outcomes = 6

Probability of getting at least one head and one tail = Favorable outcomes/Total outcomes

= 6/8

= 3/4

Therefore, probability of getting at least one head and one tail is 3/4

(iv) no tails

Solution:

Possible outcome of tossing three coins are: HTT, HHT, HHH, HTH, TTT, TTH, THT, THH

Total outcomes = 8

Number of outcomes of no tails are HHH

Favorable outcomes = 1

Probability of getting no tails = Favorable outcomes/Total outcomes

= 1/8

Therefore, probability of getting no tails is 1/8

(i)A black king

Solution:

Total number of cards are 52

Number of black king cards are 2

Probability of getting black king cards is = Favorable outcomes/Total outcomes

= 2/52

= 1/26

Therefore, probability of getting black king cards is 1/26

(ii)Either a black card or a king

Solution:

Total number of cards are 52

Number of either a black card or a king = 28

Probability of getting either a black card or a king is = Favorable outcomes/Total outcomes

= 28/52

= 7/13

Therefore, probability of getting either a black card or a king is 7/13

(iii)Black and a king

Solution:

Total number of cards are 52

Number of black and a king are 2

Probability of getting black and a king is = Favorable outcomes/Total outcomes

= 2/52

= 1/26

Therefore, probability of getting black and a king is 1/26

(iv) a jack, queen or a king

Solution:

Total number of cards are 52

Number of a jack, queen or a king = 12

Probability of getting a jack, queen or a king is = Favorable outcomes/Total outcomes

= 12/52

= 3/13

Therefore, probability of getting a jack, queen or a king is 3/13

(v) Neither a heart nor a king

Solution:

Total numbers of cards = 52

Total number of heart cards = 13

Probability of getting a heart is = Favorable outcomes/Total outcomes

= 13/52

= 1/4

Total number of king cards = 4

Probability of getting a king is = Favorable outcomes/Total outcomes

= 4/52

= 1/13

One card is common in heart and king (king of heart)

Total probability of getting a heart and a king = 1/4+ 1/13 â€“ 1/52

= (13 + 4 – 1)/52

= (17 – 1)/52

= 16/52

= 4/13

Therefore, probability of getting neither a heart nor a king = 1 â€“ 4/13 = (13 – 4)/13 = 9/13

Solution:

Total numbers of cards = 52

Number of spade cards = 13

Probability of getting spade cards is = Favorable outcomes/Total outcomes

= 13/52

= 1/4

Number of ace cards = 4

Probability of getting ace cards is = Favorable outcomes/Total outcomes

= 4/52

= 1/13

One card is common in both ace and spade (ace of spade) = 1/52

Probability of getting an ace or spade cards is = 1/4 + 1/13 â€“ 1/52

= (13 + 4 – 1)/52

= (17 – 1)/52

= 16/52

= 4/13

Therefore, probability of getting an ace or spade cards is = 4/13

(vii)Neither an ace nor a king

Solution:

Total numbers of cards = 52

Number of king cards = 4

Number of ace cards = 4

Total number of cards = 4 + 4 = 8

Total number of neither an ace nor a king are = 52 â€“ 8 = 44

Probability of getting neither an ace nor a king is = Favorable outcomes/Total outcomes

= 44/52

= 11/13

Therefore, probability of getting neither an ace nor a king is 11/13

(viii)Neither a red card nor a queen

Solution:

Total numbers of cards = 52

Red cards include hearts and diamonds

Number of hearts in a deck of 52 cards = 13

Number of diamonds in a deck of 52 cards = 13

Number of queen in a deck of 52 cards = 4

Total number of red card and queen = 13 + 13 + 2 = 28 (since queen of heart and queen of diamond are already considered)

Number of card which is neither a red card nor a queen = 52 â€“ 28 = 24

Probability of getting neither a king nor a queen is = Favorable outcomes/Total outcomes

= 24/52

= 6/13

Therefore, probability of getting neither a king nor a queen is 6/13

(ix)Other than an ace

Solution:

Total numbers of cards = 52

Total number of ace cards = 4

Total number of non-ace cards = 52 – 4 = 48

Probability of getting non-ace is = Favorable outcomes/Total outcomes

= 48/52

= 12/13

Therefore, probability of getting non-ace card is 12/13

(x)A ten

Solution:

Total numbers of cards are 52

Total number of ten cards = 4

Probability of getting ten cards is = Favorable outcomes/Total outcomes

= 4/52

= 1/13

Therefore, probability of getting ten- card is 1/13

Solution:

Total numbers of cards = 52

Total number of spade cards = 13

Probability of getting spade is = Favorable outcomes/Total outcomes

= 13/52

= 1/4

Therefore, probability of getting a spade is 1/4

(xii)A black card

Solution:

Total numbers of cards = 52

Cards of spades and clubs are black cards.

Number of clubs = 13

Total number of black card out of 52 cards = 13 + 13 = 26

Probability of getting black cards is = Favorable outcomes/Total outcomes

= 26/52

= 1/2

Therefore, probability of getting a black card is 1/2

(xiii)The seven of clubs

Solution:

Total numbers of cards = 52

Total number of the seven of club cards = 1

Probability of getting the seven of clubs cards is = Favorable outcomes/Total outcomes

= 1/52

Therefore, probability of the seven of club card is 1/52

(xiv)Jack

Solution:

Total numbers of cards = 52

Total number of jack cards = 4

Probability of getting jack cards is = Favorable outcomes/Total outcomes

= 4/52

= 1/13

Therefore, probability of the jack card is 1/13

Solution:

Total numbers of cards = 52

Total number of the ace of spades cards = 1

Probability of getting ace of spade cards is = Favorable outcomes/Total outcomes

= 1/52

Therefore, probability of the ace of spade card is 1/52

(xvi)A queen

Solution:

Total numbers of cards = 52

Total number of queen cards = 4

Probability of getting queen cards is = Favorable outcomes/Total outcomes

= 4/52

= 1/13

Therefore, probability of a queen card is 1/13

(xvii)A heart

Solution:

Total numbers of cards = 52

Total number of heart cards = 13

Probability of getting queen cards is = Favorable outcomes/Total outcomes

= 13/52

= 1/4

Therefore, probability of a heart card is 1/4

(xviii)A red card

Solution:

Total numbers of cards = 52

Total number of red cards = 13+13 = 26

Probability of getting queen cards is = Favorable outcomes/Total outcomes

= 26/52

= 1/2

Therefore, probability of a red card is 1/2

Question 6. An urn contains 10 red and 8 white balls. One ball is drawn at random. Find the probability that the ball drawn is white.

Solution:

Total number of red balls = 10

Total number of red white balls = 8

Total number of balls = 10 + 8 = 18

Probability of getting a white ball is = Total number of white balls/Total numbers of balls

= 8/18

= 4/9

Therefore, probability of a white ball is 4/9

Question 7. A bag contains 3 red balls, 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?

Solution:

Total numbers of red balls = 3

Number of black balls = 5

Number of white balls = 4

Total number of balls = 3 + 5 + 4 = 12

Probability of getting a white ball is = Total number of white balls/Total numbers of balls

= 4/12

= 1/3

Therefore, probability of a white ball is 1/3

(ii)red?

Solution:

Total numbers of red balls = 3

Number of black balls = 5

Number of white balls = 4

Total number of balls = 3 + 5 + 4 = 12

Probability of getting a red ball is = Total number of red balls/Total numbers of balls

= 3/12

= 1/4

Therefore, probability of a red ball is 1/4

(iii)black?

Solution:

Total numbers of red balls = 3

Number of black balls = 5

Number of white balls = 4

Total number of balls = 3 + 5 + 4 = 12

Probability of getting a black ball is = Total number of black balls/Total numbers of balls

= 5/12

Therefore, probability of a black ball is 5/12

(iv) not red?

Solution:

Total numbers of red balls = 3

Number of black balls = 5

Number of white balls = 4

Total number of Non -red balls = 5 + 4 = 9

Probability of getting a not red ball is = Total number of not red balls/Total numbers of balls

= 9/12

= 3/4

Therefore, probability of not a red ball is 3/4

Question 8. What is the probability that a number selected from the numbers 1, 2, 3, â€¦, 15 is a multiple of 4?

Solution:

Total numbers are 15

Multiples of 4 are 4, 8, 12

Favorable outcomes=3

Probability of getting a multiple of 4 is = Favorable outcomes/Total outcomes

= 3/15

= 1/5

Therefore, probability of getting multiples of 4 is 1/5

Question 9. A bag contains 6 red, 8 black and 4 white balls. A ball is drawn at random. What is the probability that ball drawn is not black?

Solution:

Total numbers of red balls = 6

Number of black balls = 8

Number of white balls = 4

Total number of balls = 6 + 8 + 4 = 18

Number of non-black balls are = 6 + 4 = 10

Probability of getting a non-black ball is = Total number of non-black balls/Total number of balls

= 10/18

= 5/9

Therefore, probability of getting a non-black ball is 5/9

Question 10. A bag contains 5 white and 7 red balls. One ball is drawn at random. What is the probability that ball drawn is white?

Solution:

Total numbers of red balls = 7

Number of white balls = 5

Total number of balls = 7 + 5 = 12

Probability of getting a non-white ball is = Total number of white balls/Total number of balls

= 5/12

Therefore, probability of getting a white ball is 5/12

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