CBSE Solutions for Class 10 English

જવાબ : ​​​​​​The sample space is given by
S = {HH, HT, TH, TT}
Total events = 4
exactly one head = {HT, TH} = 2
P(one head) = 2/4 = 1/2

જવાબ : 2/3

જવાબ : ​​​​​(1,0)

જવાબ : 0≤P(E)≤1

જવાબ :

જવાબ : 1/26

જવાબ : 1

જવાબ :

જવાબ :

જવાબ :

જવાબ :

જવાબ : 4/9

જવાબ : 0

જવાબ :

Total number of favourable outcomes = 132 + 12 = 144 Number of favourable outcomes = 132

Hence, P (getting a good pencil) =

જવાબ :

Let E be the event of having the same birthday  

P(E) = 0.99  

But P(E) + P = 1

P  = 1 – P(E) = 1 – 0.99 = 0.01

જવાબ :

Since P(E) + P (not E) = 1  

P (not E) = 1 – P(E) = 1 – 0.04 = 0.96

જવાબ : 0,1

જવાબ : 1

જવાબ : 1, Certain Event

જવાબ : 0, impossible event

જવાબ : 1

જવાબ : The sample space is given by
S = {HH, HT, TH, TT}
Total events = 4

atmost one head = {HT, TH, TT} = 3
P(atmost one head) = ¾

જવાબ : S = {HH, HT, TH, TT}, i.e., 4
∴ P (atleast one head) = 3/4 

જવાબ : S = {HHH, HHT, HTH, THH, TTH, THT, HTT, TTT} = 8
P(exactly two heads) = 3/8

જવાબ : S = {HH, HT, TH, TT}, i.e., 3
P(at least one tail) = ¾

જવાબ : P (neither an ace nor a king)
= 1 – P (either an ace or a king)
= 1 – [P (an ace) + P (a king)]  

જવાબ : P (neither a king nor a queen)
= 1 – P (king or queen)

જવાબ : Total number of cards = 52
Numbers of jacks = 4
Numbers of aces = 4
Card is neither a jack nor an ace
= 52 – 4 – 4 = 44
∴ Required probability = 44/52 = 11/13

જવાબ : Two dice can be thrown as 6 x 6 = 36 ways
“a doublet” can be obtained by (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6), i.e., 6 ways
P(a doublet) = 6/3}= 1/6

જવાબ : S = {1, 2, 3, 4, 5, 6} = 6
‘an even number or multiple of 3’ are 2, 3, 4, 6, i.e., 4 numbers
∴ Required probability = 4/6 = 2/3

જવાબ : Two dice can be thrown as 6 × 6 = 36 ways
The probability of number on each die is even are

જવાબ : Two dice can be thrown as 6 × 6 = 36 ways
 

જવાબ : S = {HH, HT, TH, TT}
Total number of ways = 4
Atleast one head = {HH, HT, TH}, i.e., 3 ways
∴ P (atleast one head) =3/4

જવાબ : S = {HH, HT, TH, TT} = 4
Favourable cases are HT, TH, TT i.e., 3 cases
∴ P (not more than 1 head) = 3/4

જવાબ : Total number of possible outcomes = 21 = 23 = 8
(HHH, TIT, HHT, THH, THT, HTH, TTH, HTT)
Possible outcomes of at least two heads = 4
(HHT, THH, HHH, HTH)
∴ P(at least two heads) = 4/8 =1/2

જવાબ : Total number of possible outcomes = 21 = 23 = 8
(HHH, TIT, HHT, THH, THT, HTH, TTH, HTT)
Possible outcomes of at most two heads = 7
(HHT, TTT, THH, THT, HTH, TTH, HTT)
∴ P(at most two heads) = 7/8 

જવાબ : S {HHH, HHT, HTH, THH, TTH, THT, HTT, TTT}
Total number of ways = 8
 (i) Exactly two heads
= HHT, HTH, THH, i.e., 3 ways
∴ P (exactly two heads) = 3/8

જવાબ : S {HHH, HHT, HTH, THH, TTH, THT, HTT, TTT}
Total number of ways = 8
Atleast two heads= HHT, HTH, THH, HHH i.e., 4 ways
∴ P (atleast two heads) = 4/8=1/2

જવાબ : S {HHH, HHT, HTH, THH, TTH, THT, HTT, TTT}
Total number of ways = 8
Atleast two tails= TTH, THT, HTT,TTT i.e., 4 ways
∴ P (atleast two tails) = 4/8=1/2

જવાબ : S = {HH, HT, TH, TT) = 4
P (both head or both tail)
= P (both heads) + P (both tails)
= 1/4+ ¼ = 1/2

જવાબ : 2/3

જવાબ : ​​​​​(1,0)

જવાબ : ​​​​​0≤P(E)≤1

જવાબ : 1/2

જવાબ : 1/26

જવાબ : 1

જવાબ : 3/49

જવાબ : 1/2

જવાબ : 1/5

જવાબ : 1/2

જવાબ :

જવાબ :

Total favourable outcomes associated to the random experiment of visiting a particular shop in the same week (Tuesday to Saturday) by two customers Shyam and Ram are:
 
(T, T) (T, W) (T, TH) (T, F) (T, S)

Total number of favourable outcomes = 25

1. The favourable outcomes of visiting on the same day are (T, T), (W, W), (TH, TH), (F, F) and (S, S).

Number of favourable outcomes = 5

Hence required probability = 5/25 =1/5

2. The favourable outcomes of visiting on consecutive days are (T, W), (W, T), (W, TH), (TH, W), (TH, F), (F, TH), (S, F) and (F, S).

Number of favourable outcomes = 8

Hence required probability =8/25

3. Number of favourable outcomes of visiting on different days are 25 – 5 = 20

Number of favourable outcomes = 20

Hence required probability =20/25=4/5

જવાબ :

જવાબ : Total no. of possible outcomes = 25

જવાબ : number of cards = 52
Black face cards = 6
Remaining cards = 52 – 6 = 46

જવાબ : (i) cards in a deck = 52
no. of kings = 4
no. of red kings = 2

જવાબ : x can be any one of 1, 4, 9, or 16, i.e., 4 ways
y can be any one of 1, 2, 3, or 4, i.e., 4 ways
Total number of cases of xy = 4 × 4 = 16 ways
Number of cases, where product is more than 16
(9, 2), (9, 3), (9, 4), (16, 2), (16, 3), (16, 4), i.e., 6 ways
∴ Required probability = 6/16 = 3/8

જવાબ : X can be any one of 1, 2, 3, and 4 i.e., 4 ways
Y can be any one of 1, 4, 9, and 16 i.e., 4 ways
Total no. of cases of XY = 4 × 4 = 16 ways
No. of cases, where product is less than 16 (1, 1), (1, 4), (1, 9), (2, 1), (2, 4), (3, 1), (3, 4), (4,1) i.e., 8 ways
∴ P (product x & y less then 16) = 8/16 = 1/2

જવાબ : Total numbers = 8
(i) “Odd numbers” are 1, 3, 5, 7, i.e., 4
∴ P(an odd number) = 4/8 =1/2

જવાબ : Total number of outcomes = 25
(i) Possible outcomes of numbers divisible by 3 or 5 in numbers 1 to 25 are (3, 6, 9, 12, 15, 18, 21, 24, 5, 10, 20, 25) = 12
∴P(No. divisible by 3 or 5) = 12/25

(ii) Possible outcomes of numbers which are a perfect square = 5, i.e., (1, 4, 9, 16, 25)
∴ P(a perfect square no.) = 5/25 = 1/5

જવાબ : (i) Numbers divisible by 2 or 3 from 1 to 20 are 2, 4, 6, 8, 10, 12, 14, 16, 18, 3, 9, 15, 20 = 13
Total outcomes = 20
Possible outcomes = 13
∴ P(divisible by 2 or 3) = 13/20

જવાબ : Total number of cards = 49
(i) Odd numbers are 1, 3, 5, …., 49, i.e., 25
∴ P(an odd number) = 25/49

(ii) ‘A multiple of 5’ numbers are 5, 10, 15, ……., 45, i.e., 9
∴ P(a multiple of 5) = 9/49

 (iii) “A perfect square” numbers are 1, 4, 9, …….., 49, i.e., 7
∴ P(a perfect square number) = 7/49 = 1/7

(iv) “An even prime number” is 2, i.e., only one number
∴ P(an even prime number) = 1/49

જવાબ : In a leap year, total number of days = 366
∴ 366 days
= 52 complete weeks + 2 extra days
Thus, a leap year always has 52 Mondays and extra 2 days.
Extra 2 days can be,
(i) Sunday and Monday
(ii) Monday and Tuesday
(iii) Tuesday and Wednesday
(iv) Wednesday and Thursday
(v) Thursday and Friday
(vi) Friday and Saturday
(vii) Saturday and Sunday
Let E be the event that a leap year has 53 Mondays.
∴ E = {Sun and Mon, Mon and Tues}
∴ P(E) = 2/7

જવાબ : Total number of cards = 18
Prime numbers are: 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, i.e., 11
∴ P(Prime number) = 11/180

જવાબ : Extremely patient = 3     …..[given in ques.
Extremely honest = 6
Extremely kind = 12 – 3 – 6 = 3

જવાબ : (i) Perfect squares till 70 are
1², 2², …, 8² = 8
∴ P(a perfect square) = 8/70= 4/35
(ii) Number divisible by 2 and 3 till 70 are:
6, 12, 18, …, 66, i.e., 11 nos.
∴ P(a no. divisible by 2 and 3) = 11/70

જવાબ : Number of blue marbles = 35
Number of white marbles = 25
Number of red marbles = 40
Total Number of marbles = 35 + 25 + 40 = 100(iii)P(red) = 40/100 = 2/5

જવાબ :

P(bicycles) = 20/100 = 0.2

P(motorbikes) = 50/100 = 0.5

P(cars) = 30/100 = 0.3

જવાબ :

1-B, 2-D, 3-A, 4-C

જવાબ :

1-B, 2-A, 3-D, 4-C

જવાબ :

1-C, 2-A, 3-D, 4-B

જવાબ :

1-B, 2-D, 3-A, 4-C

જવાબ :

1-D, 2-C, 3-A, 4-B

જવાબ :

1-B, 2-D, 3-A, 4-C

જવાબ :

1-D, 2-C, 3-A, 4-B

જવાબ :

1-B, 2-C, 3-D, 4-A

જવાબ :

1-B, 2-A, 3-D, 4-C

જવાબ :

1-C, 2-A, 3-D, 4-B