A number is selected at random from the set {1, 2, 3, ........, 50}. The probability that it is a prime, is

n(S) = 49
Favourable numbers are 11, 21, 31, 41.
? Required probability = 4/49

n(S) = 2³ = 8
Let E = Event of getting exactly two heads
= {(H, H, T), (H, T ,H), (T, H, H)}
? n(E) = 3
? required probability = 3/8

Total ways = 52
There is one queen of club and one king of heart.
? Favorable ways = 1 + 1 = 2
? Required probability = 2/52 = 1/26

Required probability = 3/52 + 13/52 = 16/52 =4/13
[Hint Why 13/52 because there are 13 spades and why 3/52 instead of 4/52 (there are four kings) because one king is already counted in spades.]

If the product of the four numbers ends in one of the digits 1, 3, 7 or 9, each number should have the last digit as one of these 4 digits.
? The number of favourable cases = 4⁴
Total number of all possible cases = 10⁴
Hence, the required probability =4⁴/10⁴
= 2⁴/5⁴ = 16/625

Required probability = (²C₁ x ³C₂ + ²C₂ x ³C₁) / (⁵C₃) = 9/10

Number of ways to select 3 marbles out of 7 marbles = n(s) = ⁷C₃ = 35
Probability that 2 are green and 1 is red = n(E) = ⁴C₂ x ³C₁ = 18
? Required probability = 18/35

Total number of marbles = 6 + 4 + 2 + 3 = 15
Ways of selection of two red marbles = n(E) = ⁶C₂
Ways of selection of two marbles = n(S) = ¹⁵C₂
So, required probability = (⁶C₂) / (¹⁵C₂)
= (6 x 5) / (15 x 14) = 1/7

Ways of selection of two blue marbles = ⁴C₂
Ways of selection of one yellow marble = ³C₁
Ways of selection of three marbles = ¹⁵C₃
So, required probability = (⁴C₂ x ³C₁) / (¹⁵C₃)
= [4!/{2! (4 - 2)!} x 3!/{1! (3 - 1)!}] / [15! / {3! (15 - 3)!}]
= [(4 x 3 x 2 x 1) / (2 x 1 x 2 x 1)] x [(3 x 2 x 1) / (1 x 2 x 1)] / [ (15 x 14 x 13 x 12!) / (3 x 2 x 12!)]
= (6 x 3) / (5 x 7 x 13)
= 18 / 455

Probability that none is blue from four marbles
= ^{15 - 4}C₄ / ¹⁵C₄
= ¹¹C₄ / ¹⁵C₄ = 22/91

So, probability that at least one is blue from four marbles = 1 - 22/91 = 69/91