LOADING . . .
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
The king, queen and jack of clubs are removed from a deck of 52 playing cards and then well shuffled. One card is selected from the remaining cards, find the probability of getting a queen.
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
The king, queen and jack of clubs are removed from a deck of 52 playing cards and then well shuffled. One card is selected from the remaining cards, find the probability of getting a queen.
જવાબ : 3/49
જવાબ : 1/2
જવાબ : 1/5
જવાબ : 1/2
જવાબ :
Two customers Shyam and Ram are visiting a particular shop in the same week (Tuesday to Saturday). Each is equally likely to visit the shop on any day as on another day. What is the probability that both will visit the shop on
(i) the same day?
(ii) consecutive days?
(iii) different days?
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
Number of favourable outcomes = 5
Hence required probability = 5/25 =1/5
Number of favourable outcomes = 8
Hence required probability =8/25
Number of favourable outcomes = 20
Hence required probability =20/25=4/5
જવાબ :
જવાબ : Total no. of possible outcomes = 25
(i)favourable cases are 2,3,5,7,11,13,17,19,23 which are 9 in number Required probability = 9/25 (ii)Multiple of 3 or 5જવાબ : 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
1^{2}, 2^{2}, …, 8^{2} = 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
1 |
Total outcomes for a dice |
A |
2 |
2 |
Total outcomes for 2 dice |
B |
6 |
3 |
Total outcomes for a coin |
C |
4 |
4 |
Total outcomes for 2 coin |
D |
36 |
જવાબ :
1-B, 2-D, 3-A, 4-C
Probability for 2 coins tossed at a time
1 |
P(2 heads) |
A |
4 |
2 |
P(Total outcomes) |
B |
1/4 |
3 |
P(one head & one tail) |
C |
3/4 |
4 |
P(atleast one head) |
D |
1/2 |
જવાબ :
1-B, 2-A, 3-D, 4-C
Probability for 3 coins tossed at a time
1 |
P(2 heads) |
A |
1/2 |
2 |
P(atleast 2 tails) |
B |
7/8 |
3 |
P(3 heads) |
C |
3/8 |
4 |
P(atleast 1 tail) |
D |
1/8 |
જવાબ :
1-C, 2-A, 3-D, 4-B
Probability for 1 dice rolled
1 |
P(6) |
A |
2/6 |
2 |
P(even) |
B |
1/6 |
3 |
P(divisible of 3) |
C |
4/6 |
4 |
P(>2) |
D |
3/6 |
જવાબ :
1-B, 2-D, 3-A, 4-C
Probability for 2 dice rolled at a time
1 |
P(2 three’s) |
A |
16/36 |
2 |
P(atleast 2 two’s) |
B |
4/36 |
3 |
P(total atleast six) |
C |
11/36 |
4 |
P(total of 5) |
D |
1/36 |
જવાબ :
1-D, 2-C, 3-A, 4-B
1 |
Total outcomes for 3 dice |
A |
8 |
2 |
Total outcomes for 4 dice |
B |
196 |
3 |
Total outcomes for 3 coin |
C |
16 |
4 |
Total outcomes for 4 coin |
D |
1296 |
Hide | Show
જવાબ :
1-B, 2-D, 3-A, 4-C
Probability for 2 dice rolled at a time
1 |
P(2 TWO’s) |
A |
16/36 |
2 |
P(atleast 2 SIX’s) |
B |
3/36 |
3 |
P(total atleast six) |
C |
11/36 |
4 |
P(total of 4) |
D |
1/36 |
જવાબ :
1-D, 2-C, 3-A, 4-B
Probability for 1 dice rolled
1 |
P(<6) |
A |
1/6 |
2 |
P(odd prime no.) |
B |
5/6 |
3 |
P(divisible of 2) |
C |
2/6 |
4 |
P(2) |
D |
3/6 |
જવાબ :
1-B, 2-C, 3-D, 4-A
Probability for 2 coins tossed at a time
1 |
P(2 Tails) |
A |
1 |
2 |
P(either head or tail) |
B |
1/4 |
3 |
P(atleast one tails) |
C |
0 |
4 |
P(neither head nor tail) |
D |
3/2 |
જવાબ :
1-B, 2-A, 3-D, 4-C
Probability for 3 coins tossed at a time
1 |
P(3 tails) |
A |
4/8 |
2 |
P(less than 2 tails) |
B |
7/8 |
3 |
P(1 tails) |
C |
1/8 |
4 |
P(atleast 1 tails) |
D |
3/8 |
જવાબ :
1-C, 2-A, 3-D, 4-B
Chapter 15 : Probability
.આ પ્રકરણને લગતા વિવિધ એનિમેશન વિડીયો, હેતુલક્ષી પ્રશ્નો, ટૂંકા પ્રશ્નો, લાંબા પ્રશ્નો, પરિક્ષામાં પુછાઈ ગયેલા પ્રશ્નો તેમજ પરિક્ષામાં પુછાઈ શકે તેવા અનેક મુદ્દાસર પ્રશ્નો જોવા અમારી વેબસાઈટ પર રજીસ્ટર થાઓ અથવા અમારી App ફ્રી માં ડાઉનલોડ કરો.
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