Probability – Exercise 5.1 – Class X

Probability – Exercise 5.1

  1. A dice is rolled. Find the probability of getting

(i) the number 4

(ii) a square number

(iii) a cube number

(iv) a number greater than 1

  1. Two coins are tossed together. What is the probability of getting

(i) no tail

(ii)utmost two balls

(ii) exactly one head

(iv) at least one tail

  1. Three coins are tossed together. Find the probability of getting

(i) at least one head

(ii)utmost two heads

(ii) no head

(iv) all heads

  1. Two dice are thrown together. Find the probability of getting

(i) a sum equal to 8

(ii) a sum less than 12

(iii)the sum divisible by 4

(iv) the product divisible by 5

(v) a product less than 20

(vi) the product divisible by 5

  1. A number is selected at random from 1 to 50. What is the probability that it is

(i) a prime number

(ii) not a perfect cube

(iii) a perfect square

(iv) a triangular number

(v) a multiple of 6

(vi) not a multiple of 2

  1. One card is drawn randomly from a well shuffled pack of 52 playing cards. Find the probability that the drawn card is

(i) a red colored card

(ii) not a black coloured card

(iii) not a diamond

(iv) not an ace

(v)a black king

(vi)a heart with number less than 10

  1. A two digit number is formed with the digits 2, 5 and 7 where repetition of digits is not allowed. Find the probability that the number so formed is

(i) a square number

(ii) divisible by 3

(iii) greater than 52

(iv) less than 57

(v) less than 25

(vi) a whole number

  1. Nine rotten number are mixed with the 30 good ones. One mango is chosen at random. What is the probability of choosing a

(i) good mango

(ii) rotten mango

  1. During holi festival, Sonali filled 7 bottles with different coloured water – red, blue, green, pink, yellow, purple and orange.

One bottle is selected at random. What is the probability of choosing the bottle with

(i) orange colour

(ii)not yellow colour

(iii) brown color

(iv) either red or green

(v) neither yellow nor pink

  1. A box contains 144 pens of which 20 are defective and the others are good. A person will buy a pen if it is good and will not buy if it is defective. The shopkeeper draws one pen from the box at random and gives it to the person. What is the probability that the person.

(i) will buy it?

(ii) will not buy it?


Probability – Exercise 5.1 – Solutions:

  1. A dice is rolled. Find the probability of getting

(i) the number 4

(ii) a square number

(iii) a cube number

(iv) a number greater than 1

Solution:

In rolling a dice, the sample space S ={1, 2, 3, 4, 5, 6}

⸫ n(S) = 6

(i) Let A be the event of getting the number 4

A = {4}

n(A) = 1

⸫ P(A) = n(A)/n(S) = 1/6

(ii) Let B is the event of getting square number

B = {1, 4}

n(B) = 2

P(B) = n(B)/n(S) = 2/6

(iii) Let C is the event getting cube number

C = {1}

n(C) = 1

P(C)= n(C)/n(S) = 1/6

(iv) Let D is the event getting a number greater than 1

D = {2, 3, 4, 5, 6}

n(D) = 5

P(D) = n(D)/n(S)  = 5/6


  1. Two coins are tossed together. What is the probability of getting

(i) no tail

(ii)utmost two balls

(ii) exactly one head

(iv) at least one tail

Solution:

When two coins are tossed, S = {HH, HT, TH, TT}

⸫n(S) = 4

(i) Let A be the event of getting no tail

A = { HH}

n(A)  = 1

P(A) = n(A)/n(S) = 1/4

(ii) Let B be the event of getting utmost tail

B= {HH, HT, TH, TT}

n(B) = 4

P(B) = n(B)/n(S) = 4/4

(ii) Let C be the event of getting exactly one head

A = {HT, TH}

n(C) = 2

P(C) = n(C)/n(S) = 2/4

(iv)Let D be the event of getting at least one tail

D = {TH, HT,TT}

n(D) = 3

P(D) = n(D)/n(S) = 3/4


  1. Three coins are tossed together. Find the probability of getting

(i) at least one head

(ii)utmost two heads

(ii) no head

(iv) all heads

Solution:

If three coins are tossed together then the sample space is given by,

S = {HHH, HHT, THH, HTH, TTH,THT, HTT, TTT}

n(S) = 8

(i) Let A be the event getting at least one head

A = {HHH,HHT,THH,HTH,TTH, THT, HTT}

n(A) = 7

P(A) = n(A)/n(S) = 7/8

(ii) Let B be the event getting utmost 2 heads

B = {HHT,THH,HTH,TTH, THT, HTT,TTT}

n(B) = 7

P(B) = n(B)/n(S) = 7/8

(iii)Let C be the event getting no head

C = {TTT}

n(C) = 1

P(C) = n(C)/n(S) = 1/8

(iv) Let D be the event getting all heads

D = {HHH}

n(D) = 1

P(D) = n(D)/n(S) = 1/8


  1. Two dice are thrown together. Find the probability of getting

(i) a sum equal to 8

(ii) a sum less than 12

(iii)the sum divisible by 4

(iv) the product divisible by 5

(v) a product less than 20

(vi) the product divisible by 5

Solution:

When a dice are thrown together, the sample space = {1, 2, 3, 4, 5, 6}

⸫ n(S) = 6

When two dice are thrown together, the sample space, S = {(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)}

⸫ n(S) = 36

(i) Let A be the event getting sum equal to 8

A = {(2, 6), (3, 5), (4, 4), (5, 3) (6, 2)}

n(A) = 5

P(A) = n(A)/n(S) = 5/36

(ii) Let B be the event getting a sum less than 12

B = {(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)}

n(B) = 35

P(B) = n(B)/n(S) = 35/36

(iii)Let A be the event getting the sum divisible by 4

C = {(1,3), (2, 2), (2, 6), (3, 1), (3, 5), (4, 4), (5, 3), (6, 2), (6, 6)}

n(C) = 9

P(C) = n(C)/n(S) = 9/36

(iv)Let D be the event getting a product of 12

D = {(2, 6), (3, 4), (4, 3), (6, 2)}

n(D) = 4

P(D) = n(D)/n(S) = 4/36

(v)Let E be the event getting the product less than 20

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

n(E) = 28

P(E) = n(E)/n(S) = 28/36

(vi) Let D be the event getting the product divisible by 5

D = {(1,5), (2, 5), (3, 5), (4, 5), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 5)}

n(D) = 11

P(D) = n(D)/n(S) = 11/36


  1. A number is selected at random from 1 to 50. What is the probability that it is

(i) a prime number

(ii) not a perfect cube

(iii) a perfect square

(iv) a triangular number

(v) a multiple of 6

(vi) not a multiple of 2

Solution:

Number from 1 to 50 , sample space , S = {1, 2, 3, 4, 5, 6, 7, 8, 9,10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50}

n(S) = 50

(i) Let A be the event getting a prime number

A = {2, 3, 5, 7, 11, 13, 17, 19, 23, 27, 31, 33, 37, 39, 41, 43, 47}

n(A) = 15

P(A) = n(A)/n(S) = 15/50

(ii) Let B be the event getting number which is not a perfect cube

B = { 2, 3, 4, 5, 6, 7, 9,10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50}

n(B) = 47

P(B) = n(B)/n(S) = 47/50

(iii)Let C be the event getting number which is perfect square

C = {1, 4, 9, 16, 25, 36, 49}

n(C) = 7

P(C) = n(C)/n(S) = 7/50

(iv)Let D be the event getting number which is a triangular number

D = {1, 3, 6, 10, 15, 21, 28, 36, 45}

n(D) = 9

P(D) = n(D)/n(S) = 9/50

(v) Let E be the event getting number which is a multiple of 6

E = {6, 12, 18, 24, 30, 36, 42, 48}

n(E) = 8

P(E) = n(E)/n(S) = 8/50

(vi)Let F be the event getting number which is not a multiple of 2

F = {1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39, 41, 43, 45, 47, 49}

n(F) = 25

P(F) = n(F)/n(S) = 25/50


  1. One card is drawn randomly from a well shuffled pack of 52 playing cards. Find the probability that the drawn card is

(i) a red colored card

(ii) not a black coloured card

(iii) not a diamond

(iv) not an ace

(v)a black king

(vi)a heart with number less than 10

Solution:

Cards of spades and clubs are back in colour. Cards of heart and diamond are red in color. Each color has 13 cards.

Given, one card is drawn randomly from a well shuffled pack of 52 playing cards. Therefore, n(S) = 52

(i) Let A be the card which is red in color.

A = {Cards of heart and diamond} = {13 + 13}

n(A) = 13+13 = 26

P(A) = n(A)/n(S) = 26/52

(ii) not a black coloured card

Let B be the card which is not a black coloured card

B = {Cards of spades and clubs}

n(B) = 13+13 = 26

P(B) = n(B)/n(S) = 26/52

(iii) Let C be the card which is not a diamond

C = {Cards of spades, hearts and clubs} = {13 + 13 + 13}

n(C) = 39

P(C) = n(C)/n(S) = 39/52

(iv) Let D be the card which is not an ace

We have four ace cards out of 52 cards

D={not an ace cards} = {52 – 4} = {48}

n(D) =  48

P(D) = n(D)/n(S) = 48/52

(v)Let E be the card which is a black king

We have two black king cards out of 52 cards

E  ={ a black king } ={Spade king and club king}

n(E) =  2

P(E) = n(E)/n(S) = 2/52

(vi)a heart with number less than 10

Let F be the card which is a heart with number less than 10

E  ={ a heart with number less than 10 }

n(E) =  8

P(E) = n(E)/n(S) = 8/52


  1. A two digit number is formed with the digits 2, 5 and 7 where repetition of digits is not allowed. Find the probability that the number so formed is

(i) a square number

(ii) divisible by 3

(iii) greater than 52

(iv) less than 57

(v) less than 25

(vi) a whole number

Solution:

A two digit number is formed with the digits 2, 5 and 7 where repetition of digits is not allowed.

A two digit number is formed with the digits 2, 5 and 7, a sample space ={25, 27, 52, 57, 72, 75}

⸫ n(S) = 6

(i) Let A be the number which is a square number

A = {25}

n(A) = 1

P(A) = n(A)/n(S) = 1/6

(ii) Let B be the number which is divisible by 3

B = {27, 25, 72, 57}

n(B) = 4

P(B) = n(B)/n(S) = 4/6

(iii)Let C be the number which is greater than 52

C = {57, 72, 75}

n(C) = 3

P(C) = n(C)/n(S) = 3/6

(iv)Let D be the number which is less than 57

D = {25, 27, 52}

n(D) = 3

P(D) = n(D)/n(S) = 3/6

 

(v) Let E be the number which is less than 25

E = {0}

n(E) = 0

P(E) = n(E)/n(S) = 0/6 = 0

(vi) Let F be the number which is a whole number

F ={25, 27, 52, 57, 72, 75}

n(F) = 6

P(F) = n(F)/n(S) = 6/6 = 1


  1. Nine rotten mangoes are mixed with 30 good ones. One mango is chosen at random. What is the probability of choosing a

(i) good mango

(ii) rotten mango

Solution:

Nine rotten mangoes are mixed with 30 good ones = 30 + 9 = 39

⸫n(S) = 39

(i)Let A be the event of choosing good mango

n(A) = 30

P(A) = n(A)/n(S) = 30/39

 

(ii) Let B be the event of choosing rotten mango

n(B) = 9

P(B) = n(B)/n(S) = 9/39


  1. During holi festival, Sonali filled 7 bottles with different coloured water – red, blue, green, pink, yellow, purple and orange.

One bottle is selected at random. What is the probability of choosing the bottle with

(i) orange colour

(ii)not yellow colour

(iii) brown color

(iv) either red or green

(v) neither yellow nor pink

Solution:

During holi festival, Sonali filled 7 bottles with different coloured water – red, blue, green, pink, yellow, purple and orange.

n(S) = 7

(i) Let A be the event of choosing orange color

n(A) = 1

P(A) = n(A)/n(S)  = 1/7

(ii) Let B the event of choosing bottles which are not yellow color

n(B) = 6

P(B) = n(B)/n(S)  = 6/7

(iii) Let C be the event of choosing brown color

n(C) = 0

P(C) = n(C)/n(S)  = 0/7 = 0

 

(iv) Let D be the event of choosing either red or green

n(D) = 2

P(D) = n(D)/n(S)  = 2/7

(v) Let E be the event of choosing neither yellow nor pink

n(E) = 5

P(E) = n(E)/n(S)  = 5/7


  1. A box contains 144 pens of which 20 are defective and the others are good. A person will buy a pen if it is good and will not buy if it is defective. The shopkeeper draws one pen from the box at random and gives it to the person. What is the probability that the person.

(i) will buy it?

(ii) will not buy it?

Solution:

A box contains 144 pens of which 20 are defective and the others are good.

n(S) = 144

(i) Let A be number of persons bought pens

n(A) = 124

P(A) = n(A)/n(S) = 124/144

(ii) Let B be number of persons bought pens

n(B) = 20

P(B) = n(B)/n(S) = 20/144


Next exercise – Probability – Exercise 5.2 – Class X


 

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