1. In a through of a coin find the probability of getting a tail.
Solution:
In this case sample space, S = { H, T } , Event E = { T }
p(e) = n(E) / n(S)=1/2
2. An unbiased die is tossed. Find the probability of getting of getting a multiple of 2.
Solution:
Here Sample space S = { 1, 2, 3, 4, 5, 6 }, Event E = { 2, 4, 6 } multiple of 2
p(e) = n(E) / n(S)=3/6
=1/2
3. An unbiased die is tossed. Find the probability of getting a number less than or equal to 4.
Solution :
Here Sample space S = { 1, 2, 3, 4, 5, 6 }, Event E = { 1, 2, 3, 4 } number less than or equal to 4.
p(e) = n(E) / n(S)=4/6 =2/3
4. Two coins are tossed. What is the probability of getting
(a) At most one head ?
Solution :
n(S) = { (HH), (HT), (TH), (TT) } = 4
n(E) = { HT, TH, TT } = 3 at most one head
p(e) = n(E) / n(S)=3/4
(b) At most two heads ?
Solution : n(S) = { (HH), (HT), (TH), (TT) } = 4
n(E) = { (HH), (HT), (TH), (TT) }= 4
p(e) = n(E) / n(S)=4/4 =1
5. What is the chance that a leap year selected randomly will will have 53 sundays ?
Solution :
A leap year has 366 days, out of which there are 52 weeks and 2 more days.
2 more days can be (Sunday, Monday) (Monday, Tuesday) (Tuesday, Wednesday)
(Wednesday, Thrusday) (Thrusday, Friday) (Friday, Saturday) (Saturday, Sunday)
= n(S) = 7
p(e) = n(E) / n(S) =2/7
6. What is the chance that a normal year selected randomly will will have 53
sundays ?
Solution :
A normal year has 365 days, out of which there are 52 weeks and 1 more day
So, extra day can be Sunday, Monday, Tuesday, Wednesday, Thrusday, Friday, Saturday
So, n(S) = 7 , n (E) = 1
p(e) = n(E) / n(S) =1/7
7. When two dice are thrown, what is the probability that
(a) Sum of numbers appeared is less than equal to 4
Solution :
E = { (1,1) (1,2) (1,3) (2,1) (2,2) (3,1) }
n(E) = 6 and n(S) = 36
= n(E) / n(S) =6/36 =1/6
(b) Sum of numbers is a multiple of 4
Solution :
E= { (1,3) (2,2) (2,6) (3,1) (3,5) (4,4) (5,3) (6,2) (6, 6) }
n(E) = 9, n (S) = 36
= n(E) / n(S) =9/36 =1/4
(c) Numbers appeared are equal
Solution :
E = { (1,1) (2,2) (3,3) (4,4) (5,5) (6,6) }
n(E) = 6, n(S) = 36
= n(E) / n(S) =6/36 =1/6
8. A card is drawn at random from a pack of 52 cards, What is the probability that it is
(a) A card of Red Suit ?
Solution :
There are 26 cards of Red Suit
n(s)=52C1 ,n(e)=26C1
= 26C1/52C1 =26/25 =1/2
(b) An honour card of Black suit ?
Solution :
There are 16 honour cards out of which 8 are of Black suit and 8 are of Red Suit.
So n(E) = 8 , n(S)=52
= 8C1/52C1
=8/52 =2/13
(c) A card is drawn and its number is multiple of 2
Solution :
E = 4 (2)'s + 4 (4)'s + 4 (6)'s + 4 (8)'s + 4 (10)'s
So, n(E) = 20, n (S) = 52
= 20C1/52C1
=20/52 =5/13
(d) A king or a queen ?
Solution :
There are 4 kings and 4 Queens in 52 cards
p(a king or queen) =4/52 +4/52
=1/13
(e) A king of black suit ?
Solution :
There are 2 kings in black suit ( King of Spade and King of Club )
=2/52 =1/26
9. A bag contains 4 red, 3 yellow and 5 green balls. 3 balls are drawn randomly. What is the probability that balls drawn contain
(a) Balls of different colors ?
Solution :
Toatal numbers of balls = 12
n(s)=12C3 =220
(b) Exactly two Red Balls ?
Solution :
Here only three balls are to be drawn out of which condition is of Exactly two Red balls,
2 balls can be selected in 4C2 ways
remaing 12 - 4=8 balls other than red selected in 8C1 ways
=4C2 * 8C1 / 12C3
=48/220 =12/55
(c) No Red balls ?
Solution :
Now three balls can be selected from 3 Y + 5 G balls
=8C3 /12C3
=56/220 =14/15
10. A bag contains 4 Red balls and 5 Green balls. Two balls are drawn at random. Find the probability that they are of the same colour ?
Solution :
Let S be the sample space and E be the event, so
n(E) = ( Number of ways of drawing 2 balls of Red ) OR
( Number of ways of drawing 2 balls of Green )
n(E)= 4C2 +5C2
=6 +10 =16
= n(E)/n(S) = 16/60
=4/15
11. A three-digit number is formed with the digits 1, 2, 3, 4, 5 at random. What is probability that number formed is
(a) Divisible by 2
Solution:
From the given digits 1, 2, 3, 4, 5 numbers formed is :
n(S) =5P3 =60
For divisibility with 2, even number or 0 should appear at unit place, here 2, 4 are even numbers and can occupy unit place in 2 ! ways, Rest 2 place can be filled in
n(E)= 4P2 *2 =12* 2 =24
= n(E)/n(S) = 24/60 =2/5
(b)Not divisible by 2 ?
Solution :
P (Not divisible by 2 ) = 1 - P (Divisible by 2 )
=1 - 2/5 =3/5
(c)Divisible by 5 ?
Solution :
A number ends with 5, 0 then the number will be divisible by 5 Here only 5 is present, end place will be fixed by 5 so,
n(E)= 4P2
n(S) =5P3 = n(E)/n(S)
=4P2 / 5P3 =1/5
12. The letters of the word CASTIGATION is arranged in different ways randomly. What is the chance that vowels occupy the even places ?
Solution:
Vowels are A I A I O,
A = { 3, 6, 9, 12, 15, 18, 21, 24, 27, 30 }, so n(A)=10
B = { 7, 14, 21, 28 }, n(B)= 4
B = { 13, 26 }, n (B) = 2
P(A ∪ B)=P(A) + P(B)
= 6/30 + 2/30 =4/15
So, probability that a number is divisible by 5 or 13 is 4 / 15
15. The odds favouring the event of a person hitting a target are 3 to 5. The odds against the event of another person hitting the target are 3 to 2. If each of them fire once at the target, find the probability that both of them hit the target.
Solution :
Let A be the event of first person hitting the target,
P(A)= 3/3+5 =3/8 (odds in favour)
Let B be the event of Second person hitting a target.
P(B) = 2 /3+2 =2/5(odds against)
Since both events are independent and both will hit the target so
P(A ∪ B)=P(A)P(B)= 3/8 * 2/5
=3/20
16. In the above example find the probability that at least one one of them hit the target.
Solution :
For At least one one of them hit the target.
P(A ∪ B)=P(A) + P(B) - P(A ∩ B)
=3/8 + 2/8 - 3/20
=(15+16-6) / 40 =5/8
17. The probabilities that drivers A, B and C will drive home safely after consuming liquor are 2 / 5 , 3 / 7 and 3 / 4, respectively. What is the probability that they will drive home safely after consuming liquor ?
Solution :
Let A be the event of driver A drive safely after consuming liquor.
Let B be the event of driver B drive safely after consuming liquor.
Let C be the event of driver C drive safely after consuming liquor.
P(A)=2/5, P(B)=3/7, P(C)=3/4
The events A, B and C are independent . Therefore,
P(A ∩ B ∩ C)=P(A)P(B)P(C)
=(2/5) (3/7) (3/4) =9/70
Therefore, The probability that all the drivers will drive home safely after consuming liquor is 9 / 10
18. The probabilities that A and B will tell the truth are 2 / 3 and 4 / 5 respectively . What is the probability that they agree with each other ?
Solution :
Let A be the event of A will tell truth. B be the event of B tell truth
When both agree then they say true or they say false together, that is
19. In the above problem find out the probability that both contradict each other ?
Solution :
They will contradict if A tells truth and B tells lies or B tells truth and A tells lies, So
20. A bag contains 4 white, 5 red and 6 blue balls. Three balls are drawn at random from the bag. The probability that all of them are red, is:
Option 1 : 1/22
Option 2 : 3/22
Option 3 : 2/91
Option 4 : 2/77
21. A box contains 20 electric bulbs, out of which 4 are defective. Two bulbs are chosen at random from this box. The probability that at least one of these is defective, is:
Option 1 : 4/19
Option 2 : 7/19
Option 3 : 12/19
Option 4 : 21/95
22.In a class, 30% of the students offered English, 20% offered Hindi and 10% offered both. If a student is selected at random, what is the probability that he has offered English or Hindi ?
Option 1 : 2/5
Option 2 : 3/4
Option 3 : 3/5
Option 4 : 3/10
24. A box contains 6 red balls, 7 green balls and 5 blue balls. Each ball is of a different size. The probability that the red ball being selected is the smallest red ball, is
Option 1 : 1/18
Option 2 : 1/3
Option 3 : 1/6
Option 4 : 2/3
25. If A and B are 2 independent events and P(A)=0.5 and P(B) = 0.4, find P(A/B):
Option 1 : 0.5
Option 2 : 0.4
Option 3 : 0.88
Option 4 : None of these
26. A 5-digit number is formed by the digits 1,2,3,4 and 5 without repetition. What is the probability that the number formed is a multiple of 4?
Option 1 : 1/4
Option 2 : 1/5
Option 3 : 2/5
Option 4 : 1/120
Option 5 : 4
27. In a single throw of dice, what is the probability to get a number greater or equal to 4?
Option 1 : 1/3
Option 2 : 2/3
Option 3 : 1/2
Option 4 : None of these
28. A bag contains 5 oranges, 4 bananas and 3 apples. Rohit wants to eat a banana or an apple. He draws a fruit from the bag randomly. What is the probability that he will get a fruit of his choice?
Option 1 : 3.5/12
Option 2 : 7/12
Option 3 : 5/12
Option 4 : None of these
29.There are two boxes A and B. Box A has three red and four blue balls. Box B has five red and two blue balls. Anya draws a ball from each bag randomly. What is the probability that both balls are red?
Option 1 : 4/7
Option 2 : 8/49
Option 3 : 7/8
Option 4 : 15/49
30.Ravi has a bag full of 10 Nestle and 5 Cadbury chocolates. He draws two chocolates. What is the probability that he got at least one Nestle chocolate?
Option 1 : 2/3
Option 2 : 3/7
Option 3 : 2/21
Option 4 : None of these
31.The probability of having at least one tail in 5 throws of a coin is
Option 1 : 1/32
Option 2 : 31/32
Option 3 : 1/5
Option 4 : None of these
32.A bag contains 5 yellow and 4 brown pencils. If two pencils are drawn, what is the probability that the pencils are of the same color?
Option 1 : 5/108
Option 2 : 1/6
Option 3 : 5/18
Option 4 : 4/9
33.A single letter is drawn at random from the word, "ASPIRATION", the probability that it is a vowel is?
Option 1 : 1/2
Option 2 : 1/3
Option 3 : 3/5
Option 4 : 2/5
34.The probability that a man can hit a target is 3/4. He tries 5 times. The probability that he will hit the target at least three times is:
Option 1 : 291/364
Option 2 : 371/464
Option 3 : 471/502
Option 4 : 459/512
35.An unbiased dice is rolled 3 times. The probability that the value on the dice is not more than 4 in any of the 3 rolls is:
Option 1 : 8/27
Option 2 : 1/27
Option 3 : 26/27
Option 4 : 2/3
36.Probability of occurrence of event A is 0.5 and that of event B is 0.2. The probability of occurrence of both A and B is 0.1. What is the probability that none of A and B occur?
Option 1 : 0.3
Option 2 : 0.4
Option 3 : 0.7
Option 4 : None of these
37.An unbiased coin is tossed 5 times. If tail appears on first four tosses, then probability of tail appearing on the fifth toss is:
Option 1 : 1/2
Option 2 : 1
Option 3 : 0
Option 4 : 4/5
38. X and Y are two independent events. The probability that X and Y occur is 1/12, and the probability that neither occur is 1/2, the probability of occurrence of X can be:
Option 1 : 1/3
Option 2 : 1/5
Option 3 : 1/2
Option 4 : 1/10
39. An unbiased coin is tossed n times. If the probability of getting 4 tails equals the probability of getting 7 tails, then the probability of getting two tails is:
Option 1 : 55/2048
Option 2 : 3/4096
Option 3 : 1/1024
Option 4 : None of these
40. Sudhanshu and Pankaj stand in a circle with 10 other persons. If the arrangement of the person is at random, then the probability that there are exactly 3 persons between Sudhanshu and Pankaj is?
Option 1 : 9/11
Option 2 : 2/11
Option 3 : 1/11
Option 4 : None of these
Solution:
In this case sample space, S = { H, T } , Event E = { T }
p(e) = n(E) / n(S)=1/2
2. An unbiased die is tossed. Find the probability of getting of getting a multiple of 2.
Solution:
Here Sample space S = { 1, 2, 3, 4, 5, 6 }, Event E = { 2, 4, 6 } multiple of 2
p(e) = n(E) / n(S)=3/6
=1/2
3. An unbiased die is tossed. Find the probability of getting a number less than or equal to 4.
Solution :
Here Sample space S = { 1, 2, 3, 4, 5, 6 }, Event E = { 1, 2, 3, 4 } number less than or equal to 4.
p(e) = n(E) / n(S)=4/6 =2/3
4. Two coins are tossed. What is the probability of getting
(a) At most one head ?
Solution :
n(S) = { (HH), (HT), (TH), (TT) } = 4
n(E) = { HT, TH, TT } = 3 at most one head
p(e) = n(E) / n(S)=3/4
(b) At most two heads ?
Solution : n(S) = { (HH), (HT), (TH), (TT) } = 4
n(E) = { (HH), (HT), (TH), (TT) }= 4
p(e) = n(E) / n(S)=4/4 =1
5. What is the chance that a leap year selected randomly will will have 53 sundays ?
Solution :
A leap year has 366 days, out of which there are 52 weeks and 2 more days.
2 more days can be (Sunday, Monday) (Monday, Tuesday) (Tuesday, Wednesday)
(Wednesday, Thrusday) (Thrusday, Friday) (Friday, Saturday) (Saturday, Sunday)
= n(S) = 7
p(e) = n(E) / n(S) =2/7
6. What is the chance that a normal year selected randomly will will have 53
sundays ?
Solution :
A normal year has 365 days, out of which there are 52 weeks and 1 more day
So, extra day can be Sunday, Monday, Tuesday, Wednesday, Thrusday, Friday, Saturday
So, n(S) = 7 , n (E) = 1
p(e) = n(E) / n(S) =1/7
7. When two dice are thrown, what is the probability that
(a) Sum of numbers appeared is less than equal to 4
Solution :
E = { (1,1) (1,2) (1,3) (2,1) (2,2) (3,1) }
n(E) = 6 and n(S) = 36
= n(E) / n(S) =6/36 =1/6
(b) Sum of numbers is a multiple of 4
Solution :
E= { (1,3) (2,2) (2,6) (3,1) (3,5) (4,4) (5,3) (6,2) (6, 6) }
n(E) = 9, n (S) = 36
= n(E) / n(S) =9/36 =1/4
(c) Numbers appeared are equal
Solution :
E = { (1,1) (2,2) (3,3) (4,4) (5,5) (6,6) }
n(E) = 6, n(S) = 36
= n(E) / n(S) =6/36 =1/6
8. A card is drawn at random from a pack of 52 cards, What is the probability that it is
(a) A card of Red Suit ?
Solution :
There are 26 cards of Red Suit
n(s)=52C1 ,n(e)=26C1
= 26C1/52C1 =26/25 =1/2
(b) An honour card of Black suit ?
Solution :
There are 16 honour cards out of which 8 are of Black suit and 8 are of Red Suit.
So n(E) = 8 , n(S)=52
= 8C1/52C1
=8/52 =2/13
(c) A card is drawn and its number is multiple of 2
Solution :
E = 4 (2)'s + 4 (4)'s + 4 (6)'s + 4 (8)'s + 4 (10)'s
So, n(E) = 20, n (S) = 52
= 20C1/52C1
=20/52 =5/13
(d) A king or a queen ?
Solution :
There are 4 kings and 4 Queens in 52 cards
p(a king or queen) =4/52 +4/52
=1/13
(e) A king of black suit ?
Solution :
There are 2 kings in black suit ( King of Spade and King of Club )
=2/52 =1/26
9. A bag contains 4 red, 3 yellow and 5 green balls. 3 balls are drawn randomly. What is the probability that balls drawn contain
(a) Balls of different colors ?
Solution :
Toatal numbers of balls = 12
n(s)=12C3 =220
(b) Exactly two Red Balls ?
Solution :
Here only three balls are to be drawn out of which condition is of Exactly two Red balls,
2 balls can be selected in 4C2 ways
remaing 12 - 4=8 balls other than red selected in 8C1 ways
=4C2 * 8C1 / 12C3
=48/220 =12/55
(c) No Red balls ?
Solution :
Now three balls can be selected from 3 Y + 5 G balls
=8C3 /12C3
=56/220 =14/15
10. A bag contains 4 Red balls and 5 Green balls. Two balls are drawn at random. Find the probability that they are of the same colour ?
Solution :
Let S be the sample space and E be the event, so
n(E) = ( Number of ways of drawing 2 balls of Red ) OR
( Number of ways of drawing 2 balls of Green )
n(E)= 4C2 +5C2
=6 +10 =16
= n(E)/n(S) = 16/60
=4/15
11. A three-digit number is formed with the digits 1, 2, 3, 4, 5 at random. What is probability that number formed is
(a) Divisible by 2
Solution:
From the given digits 1, 2, 3, 4, 5 numbers formed is :
n(S) =5P3 =60
For divisibility with 2, even number or 0 should appear at unit place, here 2, 4 are even numbers and can occupy unit place in 2 ! ways, Rest 2 place can be filled in
n(E)= 4P2 *2 =12* 2 =24
= n(E)/n(S) = 24/60 =2/5
(b)Not divisible by 2 ?
Solution :
P (Not divisible by 2 ) = 1 - P (Divisible by 2 )
=1 - 2/5 =3/5
(c)Divisible by 5 ?
Solution :
A number ends with 5, 0 then the number will be divisible by 5 Here only 5 is present, end place will be fixed by 5 so,
n(E)= 4P2
n(S) =5P3 = n(E)/n(S)
=4P2 / 5P3 =1/5
12. The letters of the word CASTIGATION is arranged in different ways randomly. What is the chance that vowels occupy the even places ?
Solution:
Vowels are A I A I O,
C A S T I G A T I O N
(O) (E) (O) (E) (O) (E) (O) (E) (O) (E) (O)
So there are 5 even places in which five vowels can be arranged and in rest of 6 places 6 constants can be arranged as follows :
13. A number is selected at random from the numbers 1 to 30. What is the probability that it is divisible by either 3 or 7 ?
Solution:
Let A be event of selecting a number divisible by 3. B be the event of selecting a number divisible by 7.
A = { 3, 6, 9, 12, 15, 18, 21, 24, 27, 30 }, so n(A)=10
B = { 7, 14, 21, 28 }, n(B)= 4
Since A and B are not mutually exclusive So :
14. In the above problem what is the probability that the number selected is divisible by 5 or 13 ?
Solution :
Let A be event of selecting a number divisible by 5. B be the event of selecting a number
divisible by 13
n(S)=30
A = { 5, 10, 15, 20, 25, 30 } , n (A) = 6 B = { 13, 26 }, n (B) = 2
P(A ∪ B)=P(A) + P(B)
= 6/30 + 2/30 =4/15
So, probability that a number is divisible by 5 or 13 is 4 / 15
15. The odds favouring the event of a person hitting a target are 3 to 5. The odds against the event of another person hitting the target are 3 to 2. If each of them fire once at the target, find the probability that both of them hit the target.
Solution :
Let A be the event of first person hitting the target,
P(A)= 3/3+5 =3/8 (odds in favour)
Let B be the event of Second person hitting a target.
P(B) = 2 /3+2 =2/5(odds against)
Since both events are independent and both will hit the target so
P(A ∪ B)=P(A)P(B)= 3/8 * 2/5
=3/20
16. In the above example find the probability that at least one one of them hit the target.
Solution :
For At least one one of them hit the target.
P(A ∪ B)=P(A) + P(B) - P(A ∩ B)
=3/8 + 2/8 - 3/20
=(15+16-6) / 40 =5/8
17. The probabilities that drivers A, B and C will drive home safely after consuming liquor are 2 / 5 , 3 / 7 and 3 / 4, respectively. What is the probability that they will drive home safely after consuming liquor ?
Solution :
Let A be the event of driver A drive safely after consuming liquor.
Let B be the event of driver B drive safely after consuming liquor.
Let C be the event of driver C drive safely after consuming liquor.
P(A)=2/5, P(B)=3/7, P(C)=3/4
The events A, B and C are independent . Therefore,
P(A ∩ B ∩ C)=P(A)P(B)P(C)
=(2/5) (3/7) (3/4) =9/70
Therefore, The probability that all the drivers will drive home safely after consuming liquor is 9 / 10
18. The probabilities that A and B will tell the truth are 2 / 3 and 4 / 5 respectively . What is the probability that they agree with each other ?
Solution :
Let A be the event of A will tell truth. B be the event of B tell truth
When both agree then they say true or they say false together, that is
19. In the above problem find out the probability that both contradict each other ?
Solution :
They will contradict if A tells truth and B tells lies or B tells truth and A tells lies, So
20. A bag contains 4 white, 5 red and 6 blue balls. Three balls are drawn at random from the bag. The probability that all of them are red, is:
Option 1 : 1/22
Option 2 : 3/22
Option 3 : 2/91
Option 4 : 2/77
21. A box contains 20 electric bulbs, out of which 4 are defective. Two bulbs are chosen at random from this box. The probability that at least one of these is defective, is:
Option 1 : 4/19
Option 2 : 7/19
Option 3 : 12/19
Option 4 : 21/95
22.In a class, 30% of the students offered English, 20% offered Hindi and 10% offered both. If a student is selected at random, what is the probability that he has offered English or Hindi ?
Option 1 : 2/5
Option 2 : 3/4
Option 3 : 3/5
Option 4 : 3/10
24. A box contains 6 red balls, 7 green balls and 5 blue balls. Each ball is of a different size. The probability that the red ball being selected is the smallest red ball, is
Option 1 : 1/18
Option 2 : 1/3
Option 3 : 1/6
Option 4 : 2/3
25. If A and B are 2 independent events and P(A)=0.5 and P(B) = 0.4, find P(A/B):
Option 1 : 0.5
Option 2 : 0.4
Option 3 : 0.88
Option 4 : None of these
26. A 5-digit number is formed by the digits 1,2,3,4 and 5 without repetition. What is the probability that the number formed is a multiple of 4?
Option 1 : 1/4
Option 2 : 1/5
Option 3 : 2/5
Option 4 : 1/120
Option 5 : 4
27. In a single throw of dice, what is the probability to get a number greater or equal to 4?
Option 1 : 1/3
Option 2 : 2/3
Option 3 : 1/2
Option 4 : None of these
28. A bag contains 5 oranges, 4 bananas and 3 apples. Rohit wants to eat a banana or an apple. He draws a fruit from the bag randomly. What is the probability that he will get a fruit of his choice?
Option 1 : 3.5/12
Option 2 : 7/12
Option 3 : 5/12
Option 4 : None of these
29.There are two boxes A and B. Box A has three red and four blue balls. Box B has five red and two blue balls. Anya draws a ball from each bag randomly. What is the probability that both balls are red?
Option 1 : 4/7
Option 2 : 8/49
Option 3 : 7/8
Option 4 : 15/49
30.Ravi has a bag full of 10 Nestle and 5 Cadbury chocolates. He draws two chocolates. What is the probability that he got at least one Nestle chocolate?
Option 1 : 2/3
Option 2 : 3/7
Option 3 : 2/21
Option 4 : None of these
31.The probability of having at least one tail in 5 throws of a coin is
Option 1 : 1/32
Option 2 : 31/32
Option 3 : 1/5
Option 4 : None of these
32.A bag contains 5 yellow and 4 brown pencils. If two pencils are drawn, what is the probability that the pencils are of the same color?
Option 1 : 5/108
Option 2 : 1/6
Option 3 : 5/18
Option 4 : 4/9
33.A single letter is drawn at random from the word, "ASPIRATION", the probability that it is a vowel is?
Option 1 : 1/2
Option 2 : 1/3
Option 3 : 3/5
Option 4 : 2/5
34.The probability that a man can hit a target is 3/4. He tries 5 times. The probability that he will hit the target at least three times is:
Option 1 : 291/364
Option 2 : 371/464
Option 3 : 471/502
Option 4 : 459/512
35.An unbiased dice is rolled 3 times. The probability that the value on the dice is not more than 4 in any of the 3 rolls is:
Option 1 : 8/27
Option 2 : 1/27
Option 3 : 26/27
Option 4 : 2/3
36.Probability of occurrence of event A is 0.5 and that of event B is 0.2. The probability of occurrence of both A and B is 0.1. What is the probability that none of A and B occur?
Option 1 : 0.3
Option 2 : 0.4
Option 3 : 0.7
Option 4 : None of these
37.An unbiased coin is tossed 5 times. If tail appears on first four tosses, then probability of tail appearing on the fifth toss is:
Option 1 : 1/2
Option 2 : 1
Option 3 : 0
Option 4 : 4/5
38. X and Y are two independent events. The probability that X and Y occur is 1/12, and the probability that neither occur is 1/2, the probability of occurrence of X can be:
Option 1 : 1/3
Option 2 : 1/5
Option 3 : 1/2
Option 4 : 1/10
39. An unbiased coin is tossed n times. If the probability of getting 4 tails equals the probability of getting 7 tails, then the probability of getting two tails is:
Option 1 : 55/2048
Option 2 : 3/4096
Option 3 : 1/1024
Option 4 : None of these
40. Sudhanshu and Pankaj stand in a circle with 10 other persons. If the arrangement of the person is at random, then the probability that there are exactly 3 persons between Sudhanshu and Pankaj is?
Option 1 : 9/11
Option 2 : 2/11
Option 3 : 1/11
Option 4 : None of these
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