Let A be the event that X is selected and B is the event that Y is selected.

P(A) = 1/7,  P(B) = 2/9.

Let C be the event that both are selected.

P(C) = P(A) × P(B) as A and B are independent events: 

       = (1/7) × (2/9)  = 2/63

Let X be the event that cards are in a club which is not king and other is the king of club.
Let Y be the event that one is any club card and other is a non-club king.
Hence, required probability:
=P(A)+P(B)

= 12 C 1 * 1 C 1 52 C 2 + 13 C 1 * 3 C 1 52 C 2

= 12 * 2 52 * 51 + 13 * 3 * 2 52 * 51=  24 + 78 52 * 51 =  1 26

Given number of balls = 3 + 5 + 7 = 15

One ball is drawn randomly = 15C1

probability that it is either pink or red =  7 C 1 + 3 C 1 15 C 1 = 7 + 3 15 = 10 15 = 2 3

The possible outcomes are as follows :

5H, 5T, (H, 4T), (T, 4H), (2H, 3T) (3H, 2T), i.e. 6 outcomes in all.

Therefore the probability that head appears an odd number of times = 3/6 =1/2 (In only three outcomes out of the six outcomes, head appears an odd number of times).

The probability of an impossible event is 0.

The event is known ahead of time to be not possible, therefore by definition in mathematics, the probability is defined to be 0 which means it can never happen.

The probability of a certain event is 1.

Find the number of cases in which none of the digits show a '6'.

i.e. all three dice show a number other than '6', 5×5×5=125 cases.

Total possible outcomes when three dice are thrown = 216.

The number of outcomes in which at least one die shows a '6' = Total possible outcomes when three dice are thrown - Number of outcomes in which none of them show '6'.

=216?125=91

The required probability = 91/256

P( only one of them will be selected) = p[(E and not F) or (F and not E)] 

 =  P E ? F ? F ? E 

=  P E P F + P F P E

 = 1 7 × 4 5 + 1 5 × 6 7 = 2 7

Total number of elementary events =  10 C 5

Number of ways of selecting no defective bulbs i.e., 5 non-defective bulbs out of 7 is 7 C 5.

So,required probability = 7 C 5/  10 C 5 = 1/12.