Here, n(S) = 52.
There are 13 cards of diamond (including 1 king) and there are 3 more kings. Let E = event of getting a diamond or a king.

Then, n(E) = (13 + 3)
= 16.
Therefore, P(E) = n(E) / n(S)
= 16 / 52
= 4 / 13.

• Since the product has to end in 1, 3, 7 or 9, the units place of all the 4 numbers we want to multiply can’t end in 0, 2, 4, 5, 6, 8. That means it can only end in 1, 3, 7, 9.
• This means that the probability of selecting one number that ends in 1, 3, 7 or 9 is 4/10 = 2/5.
• Therefore the probability of selecting four numbers that end in 1, 3, 7 or 9 is (2/5) x (2/5) x (2/5) x (2/5) = 16/625.
• Hence, the probability of selecting 4 numbers such that their product ends in 1, 3, 7 or 9 is 16/625.

Probability that only one of them is selected = (prob. that brother is selected) × (prob. that sister is not selected) + (Prob. that brother is not selected) × (Prob. that sister is selected)

Hence, Required Probability = 1/5 x 2/3 + 4/5 x 1/3
= 2/15 + 4/15
= 6/15
= 2/5.

• n(S) = Number of methods for drawing 2 balls out of 13 = 13C2
  = (13 x 12) / 2
  = 78.
• n(E) = No. of methods for drawing 2 balls out of 5 = 5C2
  = (5 x 4) / 2
  = 10.
• Therefore, P(E) = n(E) / n(S)
  = 10/78
  = 5/39.

n(S) = 36
n(E) = {(5,6), (6,5)} = 2
p(E) = n(E) / n(S)
= 2/36
= 1/18.

Total number of cards n(S) = 52
Number of red cards n(E) = 26.
Hence, P(E) = n(E) / n(S)
= 26/52
= 1/2.

P(A) = 7/8, P(B) = 9/10
Required Probability = P(A) x P(B)
= (7/8) x (9/10)
= 63/80.

n(S) = 52.
Let E = Event of getting a queen of club or a king of heart.
Then, n(E) = 2.
P(E) = n(E) / n(S)
= 2 / 52
= 1 / 26

Probability of getting head in one trail = 1/2.
Therefore, Required Probability of getting heads in both the trails = 1/2 x 1/2 = 1/4.

Here, S = {1, 2, 3, 4, 19, 20}.
Let E = event of getting a multiple of 3 or 5
= {3, 6 , 9, 12, 15, 18, 5, 10, 20}.
P(E) = n(E) / n(S)
= 9/20

Let S be the sample space.
Then, n(S) = ⁵²C₂ = (52 x 51) / (2 x 1) = 1326.
Let E = Event of getting 2 kings out of 4.
n(E) = ⁴C₂ = 6.
P(E) = n(E) / n(S)
= 6/1326
= 1/221 .

Here, S = {TTT, TTH, THT, HTT, THH, HTH, HHT, HHH}
Let E = event of getting at most two heads.
Then E = {TTT, TTH, THT, HTT, THH, HTH, HHT}
P(E) = n(E)/n(S)
= 7/8.

Total number of balls = (8 + 7 + 6) = 21.
Let E = event that the ball drawn is neither red nor green
= event that the ball drawn is blue.
n(E) = 7
P(E) = n(E)/n(S)
= 7/21
= 1/3

Total number of outcomes possible when a die is rolled = 6 (∵ any one face out of the 6 faces).
Hence, Total number of outcomes possible when two dice are rolled, n(S) = 6 × 6 = 36.
Let E = the event of getting two numbers whose product is even.
= {(1,2), (1,4), (1,6), (2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (3,2), (3,4), (3,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (5,2),(5,4), (5,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)}

Hence, n(E) = 27
P(E) = n(E) / n(S)
= 27/36
= 3/4

Total Number of letters in the word ASSASSINATION, n(S) = 13
Total number of consonants in the word ASSASSINATION = 7 (i.e., {S, S, S, S, N, T, N }
P(consonant) = 7/13

Total number of balls = 4 + 5 + 6 = 15
Let S be the sample space.
n(S) = Total number of ways of drawing 3 balls out of 15 = 15C3
Let E = Event of drawing 3 balls, all of them are yellow.
n(E) = Number of ways of drawing 3 balls, all of them are yellow = Number of ways of drawing 3 balls from the total 5 = 5C3 (∵ there are 5 yellow balls in the total balls)
P(E) = n(E)/n(S) = 2/91

Total number of outcomes possible when a die is rolled = 6 (∵ any one face out of the 6 faces).
Hence, Total number of outcomes possible when two dice are rolled, n(S) = 6 × 6 = 36.
Prime numbers are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29 … etc
Let E = the event that the total is a prime number
= {(1,1), (1,2), (1,4), (1,6), (2,1), (2,3), (2,5), (3,2), (3,4), (4,1), (4,3), (5,2), (5,6), (6,1), (6,5)}
Hence, n(E) = 15
P(E) = n(E)/n(S)
= 15/36 = 5/12

n(S) = Total number of ways of drawing 2 cards from 52 cards = 52C2.
Let E = event of getting 1 club and 1 diamond.
We know that there are 13 clubs and 13 diamonds in the total 52 cards.
Hence, n(E) = Number of ways of drawing one club from 13 and one diamond from 13 = ¹³C₁ × ¹³C₁ = 13/102
P(E) = n(E)/n(S)
= (¹³C₁ × ¹³C₁) / ⁵²C₂
= 13/102.