Answer: (d) P (A ∩ B) = P (A) × P (B)
Events are said to be independent if the occurrence or non-occurrence of one event does not affect the probability of the occurrence or non-occurrence of the other.
Now, by the multiplication theorem,
P(A ∩ B) = P(A) × P(B/A) ………… 1
Since A and B are independent events,
So, P(B/A) = P(B)
From equation 1, we get
P(A ∩ B) = P(A) × P(B)

Answer: (b) 3/11
There are 11 letters in the word PROBABILITY out of which 1 can be selected in ¹¹C₁ ways.
So, exhaustive number of cases = 11
There are 3 vowels i.e. A, I, O
So, the favorable number of cases = 3
Hence, the required probability = 3/11

Answer: (d) 1/6
When a die is rolled, total number of outcomes = 6 (1, 2, 3, 4, 5, 6)
Total even number = 3 (2, 4, 6)
Number of even prime number = 1 (2)
So, the probability that an even prime number is obtained = 1/6

Answer: (c) 247/256
Let x be number a discrete random variable which denotes the number of heads obtained in n (in this question n = 8)
The general form for probability of random variable x is
P(X = x) = nCx × px × qn-x
Now, in the question, we want at least two heads
Now, p =q = 1/2
So, P(X ≥ 2) = ⁸C₂ × (1/2)² × (1/2)^8-2
⇒ P(X ≥ 2) = ⁸C₂ × (1/2)² × (1/2)⁶
⇒ 1 – P(X < 2) = ⁸C₀ × (1/2)⁰ × (1/2)⁸ + ⁸C₁ × (1/2)¹ × (1/2)^8-1
⇒ 1 – P(X < 2) = (1/2)⁸ + 8 × (1/2)¹ × (1/2)⁷
⇒ 1 – P(X < 2) = 1/256 + 8 × (1/2)⁸
⇒ 1 – P(X < 2) = 1/256 + 8/256
⇒ 1 – P(X < 2) = 9/256
⇒ P(X < 2) = 1 – 9/256
⇒ P(X < 2) = (256 – 9)/256
⇒ P(X < 2) = 247/256

Answer: (d) 1/7
In a leap year, total number of days = 366 days.
In 366 days, there are 52 weeks and 2 days.
Now two days may 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
Now in total 7 possibilities, Sunday and Monday both come together is 1 time.
So probabilities of 53 Sunday and Monday in a leap year = 1/7

Answer: (d) 0.9
Given, A and B are two mutually exclusive events.
So, P(A ∩ B) = 0
Again given P(A) = 0.5 and P(B ̅) = 0.6
P(B) = 1 – P(B ̅) = 1 – 0.6 = 0.4
Now, P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
⇒ P(A ∪ B) = P(A) + P(B)
⇒ P(A ∪ B) = 0.5 + 0.4 = 0.9

Answer: (b) 7/15
While placing 7 while balls in a row, total gaps = 8
3 black balls can be placed in 8 gaps = C = (8 × 7 × 6)/(3 × 2 × 1) = 8 × 7 = 56
So, the total number of ways of arranging white and black balls such that no two black balls are adjacent = 56 × 3! × 7!
Actual number of arrangement possible with 7 white and 3 black balls = (7 + 3)! = 10!
So, the required Probability = (56 × 3! × 7!)/10!
= (56 × 3! × 7!)/(10 × 9 × 8 × 7!)
= (56 × 3!)/(10 × 9 × 8)
= (56 × 3 × 2 × 1)/(10 × 9 × 8)
= (7 × 3 × 2 × 1)/(10 × 9)
= (7 × 2)/(10 × 3)
= 7/(5 × 3)
= 7/15

Answer: (a) [1/3, 1/2]
P(A) = (3x + 1)/3
P(B) = (x – 1)/4
P(C) = (1 – 2x)/4
These are mutually exclusive events.
⇒ -1 ≤ 3x ≤ 2, -3 ≤ x ≤ 1, -1 ≤ 2x ≤ 1
⇒ -1/3 ≤ x ≤ 2/3, -2 ≤ x ≤ 1, -1/2 ≤ x ≤ 1/2
Also, 0 ≤ (3x + 1)/3 + (x – 1)/4 + (1 – 2x)/4 ≤ 1
⇒ 1/3 ≤ x ≤ 13/3
⇒ max {-1/3, -3, -1/2, 1/3} ≤ x ≤ min {2/3, 1/2, 1, 13/3}
⇒ 1/3 ≤ x ≤ 1/2
⇒ x ∈ [1/3, 1/2]

Answer: (a) 10/21
Total number of balls = 2 + 3 + 2 = 7
Two balls are drawn.
Now, P(none of them is blue) = ⁵C₂ / ⁷C₂
= {(5 × 4)/(2 × 1)}/{(7 × 6)/(2 × 1)}
= (5 × 4)/(7 × 6)
= (5 × 2)/(7 × 3)
= 10/21

Answer: (b) 2/5
Given digits are 0, 1, 3, 5, 7
Now we have to form 4 digit numbers greater than 5000.
So leftmost digit is either 5 or 7.
When digits are repeated
Number of ways for filling left most digit = 2
Now remaining 3 digits can be filled = 5 × 5 × 5
So total number of ways of 4 digits greater than 5000 = 2 × 5 × 5 × 5 = 250
Again a number is divisible by 5 if the unit digit is either 0 or 5. So there are 2 ways to fill the unit place.
So total number of ways of 4 digits greater than 5000 and divisible by 5 = 2 × 5 × 5 × 2 = 100
Now probability of 4 digit numbers greater than 5000 and divisible by 5
= 100/250
= 2/5

Answer: (a) P (A U B) = P (A) + P (B)
If A and B are mutually exclusive events,
Then P(A ∩ B) = 0
Now, by the addition theorem,
P(A U B) = P(A) + P(B) – P(A ∩ B)
⇒ P(A U B) = P(A) + P(B)

Answer: (a) 4/5
Total number of ways of choosing two numbers out of six = ⁶C₂ = (6 × 5)/2 = 3 × 5 = 15
If smaller number is chosen as 3 then greater has choice are 4, 5, 6
So, total choices = 3
If smaller number is chosen as 2 then greater has choice are 3, 4, 5, 6
So, total choices = 4
If smaller number is chosen as 1 then greater has choice are 2, 3, 4, 5, 6
So, total choices = 5
Total favourable case = 3 + 4 + 5 = 12
Now, required probability = 12/15 = 4/5

Answer: (a) 1/4
Since m and n are selected between 1 and 100,
Hence total sample space = 100 × 100
Again, 71 = 7, 72 = 49, 73 = 343, 74 = 2401, 75 = 16807, etc
Hence 1, 3, 7 and 9 will be the last digit in the power of 7.
Now, favourable number of case are
→ 1,1 1,2 1,3 …………. 1,100
2,1 2,2 2,3 …………. 2,100
3,1 3,2 3,3 …………. 3,100
……………….
………………
100,1 100,2 100,3 …………. 100,100
Now, for m = 1, n = 3, 7, 11, ………, 97
So, favourable cases = 25
Again for m = 2, n = 4, 8, 12, ………, 100
So, favourable cases = 25
Hence for every m, favourable cases = 25
So, total favourable cases = 100 × 25
Required Probability = (100 × 25)/(100 × 100)
= 25/100
= 1/4

Answer: (d) 48/425
Total number of cards = 52 and one card is lost.
Case 1: if lost card is a diamond card
Total number of cards = 51
Number of diamond cards = 12
Now two cards are drawn.
P(both cards are diamonds) = ¹²C₂ / ⁵¹C₂
Total number of cards = 52 and one card is lost.
Case 2: If lost card is not a diamond card
Total number of cards = 51
Number of diamond cards = 13
Now two cards are drawn.
P(both cards are diamonds) = ¹³C₂ / ⁵¹C₂
Now probability that both cards are diamond = ¹²C₂ / ⁵¹C₂ + ¹³C₂ / ⁵¹C₂
= (¹²C₂ + ¹³C₂) / ⁵¹C₂
= {(12 × 11)/(2 × 1) + (13 × 12)/(2 × 1)}/{(51 × 50)/(2 × 1)}
= (12 × 11 + 13 × 12)/(51 × 50)
= (132 + 156)/2550
= 288/2550
= 96/850 (288 and 2550 divided by 3)
= 48/425 (96 and 850 divided by 2)
So probability that both cards are diamond is 48/425

Answer: (c) 9/1547
Total number of cards = 52
Number of king card = 4
Now, 7 cards are drawn from 52 cards.
P (3 cards are king) = {⁴C₃ × ⁴⁸C₄}/ ⁵²C₇
= {4 × (48 × 47 × 46 × 45)/(4 × 3 × 2 × 1)}/{(52 × 51 × 50 × 49 × 48 × 47 × 46)/(7 × 6 × 5 × 4 × 3 × 2 × 1)}
= {4 × (48 × 47 × 46 × 45) × (7 × 6 × 5 × 4 × 3 × 2 × 1)}/{(4 × 3 × 2 × 1) × {(52 × 51 × 50 × 49 × 48 × 47 × 46)}
= (7 × 6 × 5 × 4 × 45)/(52 × 51 × 50 × 49)
= (6 × 5 × 4 × 45)/(52 × 51 × 50 × 7)
= (6 × 4 × 45)/(7 × 52 × 51 × 10)
= (6 × 45)/(7 × 13 × 51 × 10)
= (6 × 3)/(7 × 13 × 17 × 2)
= (3 × 3)/(7 × 13 × 17)
= 9/1547

Answer: (a) 1/2
When a die is rolled, total number of outcomes = 6 (1, 2, 3, 4, 5, 6)
Total even number = 3 (2, 4, 6)
So, the probability that an even number is obtained = 3/6 = 1/2

Answer: (a) 1/462
Given, 6 boys and 6 girls sit in a row at random.
Then, the total number of arrangement of 6 boys and 6 girls = arrangement of 12 persons = 12!
Now, boys and girls sit alternatively.
So, the total number of arrangement = 2 × 6! × 6!
Now, P(boys and girls sit alternatively) = (2 × 6! × 6!)/12!
= (2× 6 × 5! × 6!)/(12 × 11!)
= (5! × 6!)/11!
= (5 × 4 × 3 × 2 × 1 × 6!)/(11 × 10 × 9 × 8 × 7 × 6!)
= (5 × 4 × 3 × 2)/(11 × 10 × 9 × 8 × 7)
= (4 × 3)/(11 × 9 × 8 × 7)
= 3/(11 × 9 × 2 × 7)
= 1/(11 × 3 × 2 × 7)
= 1/462

Answer: (d) 21
When two dice are thrown, then total outcome = 6 × 6 = 36
A: Getting an odd number on the first die.
A = {(1, 1), (1, 2), (1, 3), (1, 4),(1, 5), (1, 6), (3, 1), (3, 2), (3, 3), (3, 4),(3, 5), (3, 6), (5, 1), (5, 2), (5, 3), (5, 4),(5, 5), (5, 6)}
Total outcome = 18
B: Getting a total of 7 on the two dice.
B = {(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)}
Total outcome = 6
C: Getting a total of greater than or equal to 8 on the two dice.
C = {(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)}
Total outcome = 15
Now n(A U B) = n(A) + n(B) – n(A ∩ B)
⇒ n(A U B) = 18 + 6 – 3
⇒ n(A U B) = 21

Answer: (d) 0.9
Given, A and B are two mutually exclusive events.
So, P(A ∩ B) = 0
Again given P(A) = 0.5 and P(B ̅) = 0.6
P(B) = 1 – P(B ̅) = 1 – 0.6 = 0.4
Now, P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
⇒ P(A ∪ B) = P(A) + P(B)
⇒ P(A ∪ B) = 0.5 + 0.4 = 0.9

Answer: (c) e^-1/2 /8
This question is based on Poisson distribution.
Now, λ = np = 500×(1/1000) = 500/1000 = 1/2
Now, P(x = 2) = {e^-1/2 × (1/2)²}/2! = e^-1/2 /(4 × 2) = e^-1/2 /8