A card is drawn at random from a well-shuffled deck of playing cards.

1.	A card is drawn at random from a well-shuffled deck of playing cards. Find the probability that the card was drawn is(i) A card of a spade or an Ace(ii) A red king(iii) Either a king or queen(iv) Neither a king nor the queen
Answer» Total number of all possible outcomes = 52 (i) Number of space card = 13 Number of aces = 4 (including 1 of space) Therefore, Number of space cards and aces = (13 + 4 - 1) = 16 Therefore, P(getting a space or an ace card) = \(\frac{16}{52}\) = \(\frac{4}{13}\) (ii) Number of red kings = 2 Therefore, P(getting a red king) = \(\frac{2}{52}\) = \(\frac{1}{26}\) (iii) Total Number of kings = 4 Total Number of queens = 4 let E be the event of getting either a king or a queen Then, the favorable outcomes = 4 + 4 = 8 Therefore, P(getting a king or a queen) = P(E) = \(\frac{8}{52}\) = \(\frac{2}{13}\) (iv) Let E be the event of getting either a king or a queen. Then,(not E) is the event that drawn card is neither a king nor a queen. Then, P(getting a king or a queen) = \(\frac{2}{13}\) Now, P(E) + P(not E) = 1 Therefore, P(getting a king nor a queen) = 1 - \(\frac{2}{13}\) = \(\frac{11}{13}\)

A card is drawn at random from a well-shuffled deck of playing cards. Find the probability that the card was drawn is(i) A card of a spade or an Ace(ii) A red king(iii) Either a king or queen(iv) Neither a king nor the queen

Answer»

Total number of all possible outcomes = 52

(i) Number of space card = 13

Number of aces = 4 (including 1 of space)

Therefore, Number of space cards and aces = (13 + 4 - 1) = 16

Therefore, P(getting a space or an ace card) = \(\frac{16}{52}\) = \(\frac{4}{13}\)

(ii) Number of red kings = 2

Therefore, P(getting a red king) = \(\frac{2}{52}\) = \(\frac{1}{26}\)

(iii) Total Number of kings = 4

Total Number of queens = 4

let E be the event of getting either a king or a queen

Then, the favorable outcomes = 4 + 4 = 8

Therefore, P(getting a king or a queen) = P(E) = \(\frac{8}{52}\) = \(\frac{2}{13}\)

(iv) Let E be the event of getting either a king or a queen. Then,(not E) is the event that drawn card is neither a king nor a queen.

Then, P(getting a king or a queen) = \(\frac{2}{13}\)

Now, P(E) + P(not E) = 1

Therefore, P(getting a king nor a queen) = 1 - \(\frac{2}{13}\) = \(\frac{11}{13}\)